Redox Reactions

Option (iv) is not a redox reaction as the oxidation number of Ba and S in the reaction equation is +2 and +4 respectively in both reactant and product side is same. As redox reaction is only related to the change in oxidation states of reactants the selected option does not show reduction and oxidation.

Options (i)& (ii) are example of a displacement redox reaction

Option (iii) is a form of combination redox reaction.

For an element to be strong oxidizing agent it should itself readily undergo reduction. So among given elements we find which one will undergo reduction more spontaneously.

The given are the reduction potential of elements for which the half-cell reactions are:

We know that, higher the reduction potential of any element the more readily the given element will undergo reduction as it gives out more energy in the form of electric potential reaching a more stable state. (Here the stable state is element at lower energy)

Since among all the elements Ag has the highest potential it will undergo reduction itself more readily and thus will act as a strong oxidizing agent.

Reducing agents should have the ability to undergo oxidation that implies the property to easily loose electrons. Consequently the coupling element should have a greater tendency to attract the donated electron. As I^- and Br^- already are reduced ions so they cannot further undergo reduction. Between Ag and Br we can find from electrochemical series that the reduction potential of Br₂ is more as compared to Ag.

Thus Cu will reduce Br₂ as follows:

Corresponding [potential for the developed cell is

Feasibility of a redox reaction is decided by the electric potential of the cell made out of two elements. So the net E°_cell for the cell should be positive. The relative position of Ag and Fe³⁺ in electrochemical series and the reduction potential of cell should be determined for all cases.

(i)

The

Since the value is positive the reaction proceeds spontaneously and is feasible.

∴ We can calculate for remaining three reduction potential similarly,

(ii)

(iii)

(iv)

Since the value of potential of cell for Ag and Fe³⁺ is negative we can say that the reaction is non-spontaneous and is not feasible as the redox reaction resists undergoing changes under mentioned configuration of Ag which undergoes oxidation, as its emf is less as compared to Fe³⁺, it readily undergoes reduction. Thus reaction in Option (iv) is not feasible.

One of the reasons for the difference in the oxidation states of sulphur in both reactions is due to difference in reactivity of Bromine and Iodine, as Iodine is a bigger molecule thus it has more tendency to share electrons than to donate as compared to bromine. Whereas bromine with higher ionic character tends to donate electron rather than sharing. So the more readily an atom donates electron the more easily it gives out energy in the form of electric potential due to ions formed.

Another reason is, as Br (Br₂/Br^- = 1.09) has higher reduction potential as compared to I₂ (I₂/I^- = 0.54) it more readily undergoes reduction itself which in turn increases the driving and thus acting as a stronger oxidizing agent than I₂.

(i) The oxidation number of hydrogen is +1 incase if it’s bonded with non-metals and have higher electronegativity than hydrogen. When hydrogen with metals or in some cases with atoms of similar electronegativity as compared hydrogen the oxidation number is -1 (eg. Aluminium hydride, Lithium hydride)

(ii) The algebraic sum of all the oxidation numbers in a neutral compound is zero, for a charged ion or molecular compound the sum is equal to the charge on the ion.

(iii) This option holds true for all cases.

(iv) Fluorine has the highest reduction potential of +2.87 in the entire electrochemical series, thus its tendency is always to undergo reduction and acts as the oxidizing agent. Thus it cannot give away any electrons, so it always accepts one electron to complete its outermost shell. Thus it always has a oxidation number of -1.

We find out the oxidation number for Nitrogen for all options by using the unknown oxidation state method.

(i) NH₂OH

→ x= -1

(ii) NH₄NO₃

In this compound there are two nitrogen atoms bonded to two different types of groups. So we consider one molecular group at a time. We first consider NH₄⁺ as an ion and find out oxidation number for the other N atom. The oxidation number of NH₄ is +1.

∴ → x= +5

Now we know that for NO₃ the oxidation number is -1

∴ → x= -3

Thus N has two oxidation states -3 and +5. This was possible because there were two kids of molecules available for nitrogen to bond.

As in case of option (iii) & (iv) nitrogen is bonded to only hydrogen so by same calculations as above the oxidation number for N₂H₄ is -2 and that for N₃H is -1/3. The fraction -1/3 shows that one valence electron is being shared between three N atoms in N₃H.

The oxidation numbers for all above ionic molecules can be calculated using assigning variable for unknown oxidation state and following the rule that algebraic sum of all oxidation states of a compound is equal to the charge on the compound.

CrO^-₂ → → x= +3

ClO^-₃→ → x= +5

CrO^2-₄→ → x= +6

MnO^-₄→ → x= +7

The ascending order of increasing oxidation state of the given molecules is: CrO^-₂ <ClO^-₃<CrO^2-₄<MnO^-₄

The correct option suggesting that is option (i).

*Caution must be taken in noticing the charge on the molecule and not taking it as zero, along with that the rules of finding oxidation number should not be violated.

Oxidation number is basically the ability of an atom to loose or gain electrons in order to gain stability or in the influence of any other element. This strongly suggests that the electronic configuration influences whether the concerned atom can accept or donate electrons under appropriate conditions.

The valence electrons are directly related to oxidation number, thus more the valence electrons more is the possibility to donate electrons, and if it has a tendency to accept electrons then the number of empty orbitals increases the oxidation number.

For example, N has Z=7, with valence electronic configuration as 2s² 2p³, so it has tendency to either fulfill its external orbit by 3 electrons to complete its octate, or it can donate 3 electrons to gain positive charge an attain duplet thus gaining stability.

Thus from above option the maximum possible electrons that can be donated is 7 which is shown by option (iv).

*there is one more abstract possibility that option (ii) 3d³ 4s² shows tendency to accept 7 electrons to completely fill the d-orbital and one can even expect an oxidation state of -7, but due to instability of ion developed and low feasibility that kind of oxidation state is not exhibited.

When a single element undergoes simultaneous oxidation and reduction in a reaction to give two or more products showing different change in oxidation state is called a Disproportionation reaction. For a molecule to undergo disproportionation reaction the number of free or valence electrons should have suitable electronic configuration to give more the one oxidation states.

In option (i), Carbon only undergoes combustion to give carbon dioxide and water, this does not give more than one carbon products and for that of oxygen the oxidation number is same.

For option (ii) there is only free radical substitution reaction of hydrogen getting replaced by chlorine so there is no change in oxidation state of carbon or chlorine.

For option (iii) the compound under consideration contains fluorine, but according to rule for oxidation numbers, fluorine cannot attain any other oxidation number than -1, so disproportionation reaction are not possible with fluorine compounds.

Thus option (iv) is correct, the disproportionation reaction can be shown as follows:

The compound under consideration contains fluorine, but according to rule for oxidation numbers, fluorine cannot attain any other oxidation number than -1, so disproportionation reaction are not possible with fluorine compounds.

Another reason is the electronic configuration of fluorine,

EC = 2s² 2p⁵

The possibility of electrons being removed from fluorine can be considered by supplying external energy, but as removing 2 or 5 electrons would require large amount of external energy which is not generally present during a chemical reaction which only proceeds using the enthalpy of the reaction. Thus instead of giving any of its electrons, F can just accept 1 e^- and complete its octate and attain stability. This process is spontaneous as shown by the high reduction potential of 2.87 V.

E° = 2.87 V

Thus the possibility of F giving electron is diminished hence it cannot undergo disproportionation reaction.

(i) Potassium is undergoing oxidation

Potassium has a general oxidation number of +1 as it as a group 1 alkali member. In KClO₃ and KCl the charge on potassium does not change so it is not going under oxidation but has same ON in both the compounds.

(ii) Chlorine is undergoing oxidation

The oxidation number of Cl in KClO₃ is

→

And that in KCl is -1, so the change is from +5 to -1 which shoes reduction in oxidation number thus the statement is wrong.

(iii) Oxygen is reduced

As only atom having high electronegativity in periodic table is F, only with fluorine compounds oxygen shows positive charge or oxidation number, thus oxygen here is not reduced as also justified that both reactant and product oxygen molecule shows -2 oxidation number.

(i) Zinc is undergoing oxidation in the above reaction so statement one is wrong

(ii) Chlorine does not have any change in oxidation number since in HCl and ZnCl₂ the ON is -1

(iii) Hydrogen has reduced itself from +1 to 0, thus acting as an oxidising agent, i.e. oxidant

(iv) Zinc underwent oxidation from ON of 0 to +2, implying it being a reducing agent i.e. reductant.

The electronic configuration shows the number of electrons available for donation or the number of empty orbitals to accept electrons.

Option (i) has a 3s¹ configuration which allows the donation of 1 electron is possible to gain +1 oxidation number. Also it can accept on electron and gain ON = -1, but being an s block element, alkali earth metals, are highly resistive to reduction by accepting electron. Also the have very high negative electrode potential suggesting that it can only donate one electron having only one oxidation state.

In Option (ii) the given configuration is for element Scandium. Now we consider the outer s orbital which can give 2 electrons and exhibit an oxidation state of +2, the following lone d electron can be further donated in order to make the d-orbital empty thus attaining a lower energy state. This removal of this electron gives an oxidation state of +3. Thus Sc exhibits +2, +3 oxidation state.

Option (iii) similar to the said example the subsequent valence orbital of 4s can lose two electrons giving a +2 ON, while the d-orbital electron is given away in the presence of a strong oxidizing agent. Hence Ti can exhibit three Oxidation states based on configuration i.e. +2, +3, +4.

Option (iv) this configuration suggests that, there are 3 empty p orbitals that can accommodate 3 foreign electrons thus implying a -3 ON, whereas it can also donate those 3 orbital to acquire lower energy giving 3+ ON. The inner 3s orbital can loose electrons to further give an oxidation state of +5.

The element with 3s² 3p³configuration is phosphorous, which is seen to exhibit many compounds of hydrides and oxides, which have large application in fertilizer industry. So study of oxidation number plays an important role.

The above reaction shows that P₄ undergoes reduction and oxidation simultaneously showing an example of disproportionation reaction. The oxidation number in P₄ is zero as there are all elemental toms of phosphorous bonded in a pyramidal geometry. The oxidation reaction in results in the formation of Hypophosphite ion H₂PO^-₂, which has an oxidation state of

Thus showing oxidation of P atom from 0 to +1. In case of phosphine the hydrogen involves has +1 ON number thus P having ON of -3, showing P is undergoing reduction.

In option (iv) it is stated that hydrogen is neither oxidant nor reductant but for H₂O, H₂ being in +1 state throughout the entire reaction it neither undergoes oxidation nor reduction, thus Option (iv) is correct.

According to thermodynamics, an element undergoes changes spontaneously only if it attains a stable state as a product, or after that change it attains lower energy state, or reaches equilibrium with its surrounding. This process of reaching lower energy in the case of ion formation occurs along with its release in the form of electrode potential energy. This is considered as an primary thermodynamic stability criteria.

As far by the convention the release in energy from the system is explained by positive sign. Thus the option with positive EMF will undergo spontaneous changes and the reduction reaction is favoured. The Standard Hydrogen Electrode (SHE) has a potential of 0.0 V as considered as a reference also called a Reference Electrode. There are many reference electrodes whose initial half-cell electrode potential is known. When another element is connected as the second electrode it may undergo reduction or oxidation, depending on whether it can do that in accordance with thermodynamics stability criteria mentioned earlier.

So for the electrodes containing negative potential, the reduction reaction are not favoured, as it shows the half-cell reactions are non-spontaneous and requires input of energy from external source.

∴ Elements having negative potential will readily undergo oxidation and those with positive electrode potential wil undergo reduction.

With hydrogen as one electrode, from the given options Al and Fe will act as anode and undergo oxidation. This configuration will give a positive E°_cell.

Al and H₂ cell representation

Similar can be built for Fe and H₂.

The positive EMF value shows the feasibility of the configuration chosen.

Bleaching is a process in which the colors of the treated substance disappear and are whitened by using reactants called as bleaching agents. Cl₂ was discovered by Swedish chemist Carl Wilhelm Scheele in 1774, which is used as a bleaching agent in pulp industry. Claude Louis Berthelot was the first to use Cl as a bleaching agent.

Cl₂ is produced in-situ along with sodium hydroxide using electrolysis of aqueous NaCl

The oxidation state of Cl₂ is zero, while that in the hypochlorite ion is +1, thus undergoing oxidation. This method is used to remove color or whiten the wood pulp in paper industry. This chlorine generated oxidizes the chromopohores present in wood pulp, which are the main color substances that can reflect, transmit or absorb wavelengths in the visible spectrum present in living things.

Note: In the above mentioned reaction, Cl₂ is an oxidizing agent which in presence of hydroxide forms a hypochlorite which is harmful substance. Along with it, 2,3,7,8 – tetrachlorodibenzo-p-dioxin (TCDD), which is the most harmful substance produced in the process, so now the practice of Totally Chlorine Free (TCF) bleaching methods are used i.e. by using ClO₂.

There are two kinds of explanation for which this deviation is observed. One is based on thermodynamic study and other based on electrochemistry. For now we will consider the electrochemistry, the important criteria which decides whether a substance will undergo disproportionation reaction or not is its ability to undergo multiple oxidation state.

As Mn can undergo multiple oxidation state, it can undergodisproportionation reaction, but along with oxidation state it should have definite configuration or availability of both electrons to donate during oxidation and empty orbitals to occupy during reduction.

For MnO₄^2-, the disproportionation reaction is,

the oxidation state is +6, the maximum oxidation state Mn can exhibit is +7, so it has the tendency to loose one more electron thus can undergo oxidation. And as the Mn nucleus is electron deficient with +6 ON, it can definitely undergo reduction.

In case of MnO₄^-, the oxidation state is +7, so it can no longer donate any electron and hence can no longer undergo oxidation.

∴MnO₄^2- undergoes disproportionation reaction in acidic medium but MnO^-₄ does not.

say first reaction is (i) ad second as (ii)

Primary reason (1)

The electronic configuration of central atom plays an important role in showing multiple oxidation states. Pb has an EC as: [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p². The outermost orbital can give 4 electrons, so can exhibit multiple oxidation states. In PbO the Oxidation number is +2 and that in PbO₂ is +4.

There is a difference in energy required and steps in both reactions (i) and (ii). For reaction

The ionization of PbO will result as

This gives easy substitution by Cl atoms and the reaction proceeds at faster rate. Although for PbO₂,

Dissociates as,

This ion is unstable and needs to first get reduced to Pb²⁺, also thus resulting in oxidation of Cl^- ion produced to Cl₂ due availability of enough energy. Thus the difference in reactivity is justified.

Reason (2)

In order to substitute oxygen in both the compounds, the bond between Pb and O needs to be broken. The driving force needed to break these bonds is the condition that it can attain a lower energy state and a more stable compound by electron transfer or by substitution. The one bond can be broken easily in PbO as compared to PbO₂ which has two bonds to replace with Cl. Thus the substitution reaction in (ii) moves slowly as compared to reaction (i).

Also when in an aqueous solution, the more number of moles of PbO further allows more of substitution reaction to occur, thus giving more amount of substituted product. Thus even with same reactant a given compound can give different amounts of product based on the extensive nature of reactant.

Actually Nitric oxide reacts with both PbO and PbO₂

Pb has an EC as: [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p². The outermost orbital can give 4 electrons, so can exhibit multiple oxidation states. In PbO the Oxidation number is +2 and that in PbO₂ is +4. As seen by the EC, Pb can only donate 4 outermost electrons to gain an oxidation state of +4. Oxidation by nitric acid of PbO takes place as:

Oxidation of PbO₂ takes place as:

As seen by the reactions of PbO₂ we require drastic conditions and excess of nitric acid in order for the reaction to occur spontaneously. Also if we consider whether HNO₃ can oxidize PbO or not, under non – reactive conditions, can be discussed. As PbO has ON as +2, it can undergo oxidation by giving two electrons to nitric acid to attain Pb⁴⁺ state. As there are no more valence electrons in the outer most EC of Pb it won’t undergo oxidation further, thus nitric acid cannot oxidize PbO₂. Also the ionization energy for Pb to give 4 electrons is high as compared to Sn which shows ON +4. Thus oxidation is not possible for a Pb⁴⁺ ionic state.

Option (i) Permanganate ion (MnO₄^- ) reacts with sulphur dioxide gas in acidic medium to produce Mn²⁺ and hydrogensulphate ion. (Balance by ion electron method)

Equation balance :-

Step (i) Generate unbalanced equation so this reaction can be written as:

Here Mn undergoes reduction from +7 to +2 and sulphur ion undergoes oxidation from +4 to +6. By using ion electron or hal-reaction method,

Step (ii) The half reduction reaction is:

The half oxidation reaction is:

Step (iii) we balance the number of atoms other than O and H, as here they are already balanced we then balance the number of oxygen and hydrogen atoms.

Step (iv) We assume the reaction is undertaking under aqueous acidic conditions and thus balance by adding H₂O on RHS of the equation to balance number of oxygen, and H⁺ on LHS to balance the hydrogen atoms.

Reduction:

Oxidation:

Step (v) in order to balance charges you need to add electrons on the unbalanced side, here reduction reaction has 1 electron less in LHS and 5 electron less in oxidation reaction in LHS(if needed you can multiply the equation by coefficients.)

Reduction:

Oxidation:

Step (vi) we add the Half-cell reactions to attain the overall reaction.

Step (vii) make electron gain and lost during oxidation and reduction equal, thus multiplying reduction reaction by 2 and oxidation reaction by 5.

Reduction:

Oxidation:

The overall reaction is met by adding thus half-reactions:

(ii) Reaction of liquid hydrazine (N₂H₄ ) with chlorate ion (ClO^-₃ ) in basic medium produces nitric oxide gas and chloride ion in gaseous state. (Balance by oxidation number method)

Step (i) we generate the unbalanced skeleton:

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples. Since its aready balanced, reactions are same as previous steps.

Oxidation:

Reduction:

Step (iv) For basic solutions, balance the charge by adding OH^- on required side of the equation.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 6 oxygen atoms in LHS of oxidation reaction, so add water malecule to RHS of the equation to balance oxygen and 3 molecules to LHS of reduction reaction.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction. As they are same no need for any changes.

Oxidation:

Reduction:

Step (vii) Combine these two equation by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

Check the charge on both sides whether is balanced or not for verification. As the charge is -1 on both sides reaction is balanced.

(iii) Dichlorine heptaoxide (Cl₂O₇ ) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion (ClO₂^- ) and oxygen gas.

(Balance by ion electron method)

Step (i) we generate the unbalanced skeleton:

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H⁺ on required side of the equation.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 7 oxygen atoms in reduction reaction, so add needed water molecule to RHS of the equation to balance oxygen.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction. Multiply oxidation reaction by 3.

Oxidation:

Reduction:

Step (vii) Combine these two equations by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

The charge on both sides is 0, thus reaction is balanced.

We calculate oxidation number based on the rules of oxidation. For the unknown oxidation number we assign variable x, and for all the known values for alkali metals, oxygen and halogens, we use the fact that the algebraic sum of all the oxidation states in the compound is equal to the charge on the compound. The oxidation number for hydrogen is +1, Sodium (Na) is +1 and for oxygen is -2.

HPO^2-₃ è x= +3

Na₂S₄O₆ è x= 5/2

Na₂SO₃ è x= +4

Na₂SO₄ è x= +6

The fractional ON of 5/2 can be explained by its structure, where it contains S-S bond, thus the charge being divided by two atoms.

(i)

Step (i) we generate the unbalanced skeleton:

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefullymake equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H⁺ on required side of the equation.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 7 oxygen atoms in reduction reaction, so add water molecule to RHS of the equation to balance oxygen.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction.

Oxidation:

Reduction:

Step (vii) Combine these two equation by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

Verify whether all charges are balanced.

LHS = 6×+2 + 1× -2 + 14 = 24

RHS = 6×+3 + 2×+3 + 7×0 = 24

Thus the sum total is same the solve reaction is correct.

Option (ii) given is wrong in sense of molecules given there, the correct equation is:

Step (i) we generate the unbalanced skeleton:

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H⁺ on required side of the equation. On the RHS of oxidation reaction the charge is -12 so we balance it by adding 12 H⁺ ions. And for reduction reaction we add 2 H⁺ ions on LHS.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 6 oxygen atoms in oxidation reaction, so add 3 water molecule to RHS of the equation to balance oxygen and 1 oxygen atom in reduction reaction.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction.

Oxidation:

Reduction:

Step (vii) Combine these two equation by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

Finally always check the balance of charge for equation

LHS = 2×0 + 10×-1 + 8 = -2

RHS = 2×-1 + 10×0 + 4×0 = -2

Thus the sum total is same the solved reaction is correct.

Given Option (iii) is wrong equation, corrected equation is stated as

I₂ + 2S₂ O₃^2-→2I^- + S₄ O^2-₆

Step (i) we generate the unbalanced skeleton:

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H⁺ on required side of the equation. Since the charges are balanced we can proceed to next step.

Step (v) Balance the oxygen atoms. The oxygen atoms are balanced in both the reactions so we don’t need to add water molecule to adjust oxygen.

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction. Since the electrons gained and lost is 2 in both reactions we don’t need to multiply by any coefficients.

Step (vii) Combine these two equation by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

The charges on both sides of equation are -4 so the reaction is balanced.

Option (iv) MnO₂ + C₂ O^2-₄→ Mn²⁺ + CO₂

Step (i) we generate the unbalanced skeleton:

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples. Here Mn undergoes reduction, and C undergoes oxidation

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H⁺ on required side of the equation. On the RHS of oxidation reaction the charge is balanced, for reduction reaction we add 4 H⁺ ions on LHS.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 4 oxygen atoms in oxidation reaction, which are balanced and we add 2 water molecules in RHS of reduction reaction.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction. They are same so we can directly add the equations.

Step (vii) Combine these two equation by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

The charge on LHS is +2 which matches with the charge on the RHS, thus the equation is balanced.

Option (i)

Cl oxidizes from -1 to 0 and Nitrogen reduced from +5 to +3. Thus as Cl oxidizes it acts as reducing agent and nitric acid acts as an oxidizing agent.

Reducing agent: HCl

Oxidizing agent: HNO₃

Option (ii)

This reaction is an example of displacement reaction, and none of the components undergo oxidation or reduction, as the oxidation states oh Hg, Cl, K, & I are same in both reactants and products.

Option (iii)

Oxidising agent: Fe₂O₃

Reducing agent: CO

Option (iv)

Oxidising agent: O₂

Reducing agent: NH₃

Option (i) Cr₂O₇^2- + H⁺ + I^-→ Cr³⁺ + I₂ + H₂O

Step (i) we generate the unbalanced skeleton:

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H⁺ on required side of the equation.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 7 oxygen atoms in reduction reaction, so add water molecule to RHS of the equation to balance oxygen.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction. Multiply oxidation reaction by 3.

Oxidation:

Reduction:

Step (vii) Combine these two equations by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

The charge on the LHS is +6 which matches with the RHS of the equation thus the reaction is balanced.

Option (ii)

Step (i) we generate the unbalanced skeleton:

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H⁺ on required side of the equation.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 7 oxygen atoms in reduction reaction, so add water molecule to RHS of the equation to balance oxygen.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction.

Oxidation:

Reduction:

Step (vii) Combine these two equation by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

Verify whether all charges are balanced.

LHS = 6×+2 + 1× -2 + 14 = 24

RHS = 6×+3 + 2×+3 + 7×0 = 24

Thus the sum total is same the solve reaction is correct.

Option (iii)

Step (i) we generate the unbalanced skeleton:

Here Mn undergoes reduction and S oxidation.

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples. Since the atoms are balanced the reaction is same as previous step.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H⁺ on required side of the equation.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 4 oxygen atoms in reduction reaction, so add water molecule to RHS of the equation to balance oxygen and 1 H₂O molecule on LHS of oxidation reaction.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction.

Oxidation:

Reduction:

Step (vii) Combine these two equations by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

LHS = 2×-1+6+5×-2= -6

RHS = 2×+2+3× 0+5×-2 = -6

Equal charges on both sides imply balanced equation.

Option (iv)

Step (i) we generate the unbalanced skeleton:

Step (ii) Identify which reactants are oxidized and reduced, and write the half-cell reaction for them along with the electron transfers carefully make equal number of atoms of reduced and oxidized redox couples.

Oxidation:

Reduction:

Step (iii) Balance the number of atoms of reduced and oxidized redox couples. Since the atoms are balanced the reaction is same as previous step.

Oxidation:

Reduction:

Step (iv) For acidic solutions, balance the charge by adding H⁺ on required side of the equation.

Oxidation:

Reduction:

Step (v) Balance the oxygen atoms. There are 4 oxygen atoms in reduction reaction, so add water molecule to RHS of the equation to balance oxygen.

Oxidation:

Reduction:

Step (vi) Make the electron lost and gain equal in both oxidation and reduction reaction.

Oxidation:

Reduction:

Step (vii) Combine these two equations by adding them so that all reactants and products remain in the same side.

Step (viii) Simplify the equation by eliminating similar terms on both sides of equation.

The charge on LHS and RHS is equal.

Option (i) Cr₂O₇^2-→ (d) +6

By oxidation rule we can calculate the unknown oxidation number.

Cr₂O₇^2-

Option (ii) MnO^-₄→ (e) +7

MnO^-₄

Option (iii) VO^-₃→ (c) +5

VO^-₃

Option (iv) FeF^3-₆→ (a) +3

FeF^3-₆

Option (i) Ions having positive charge → (e) Cation

Cations are positive charge entities in ionic compounds.

Option (ii) the sum of oxidation numberof all atoms in a neutral molecule → (d) 0

The algebraic sum of all oxidation numbers of a neutral compound is equal to 0. For ionic compounds

Option (iii) Oxidation number of hydrogen ion → (c) +1

Hydrogen atom is +1 charge for bonding with no9n-metals, whereas its oxidation number is -1 with metals is -1, for example metal hydrides.

Option (iv) Oxidation number of F in NaF → (b) -1

For all compounds of fluorine the oxidation state is -1 as it is most electronegative element in the entire periodic table.

Option (v) Ions having negative charge → (f) anion

By definition the ions having negative charge are called anion.

option (i)

Both statements are correct and (A) is a correct explanation of (B).

Fluorine has +2.87 V of reduction potential for a reduction reaction:

As the reduction potential is highest in the electrochemical series there is no other element that can oxidize fluorine. And thus fluorine acts as the best oxidizing agent.

Size of fluorine is small for which the amount of electron attraction is very large. This results into very strong nucleus electron attraction making it difficult to release any electrons. This also impacts during compound formation, such the nuclear attraction is very strong thus resulting into high electronegativity and a very high intermolecular attraction for electrons. Due to this reason the electronegativity of F is 3.98, which is highest in the periodic table, thus causing good oxidant properties.

Option (iii) A is true but R is false.

The above shows the reaction is a redox reaction, where Mn undergoes reduction from +7 to +4 and Iodine undergoes oxidation from -1 to 0. The maximum oxidation state that Mn can reach according to electronic configuration is +7. Thus as it cannot undergo oxidation it acts as oxidising agent and itself undergoes reduction, whereas Iodine undergoes oxidation from -1 to 0.

Oxidation state of Mn changes from +7 to +4 an thus (B) is false.

Option (ii) Both A and R are true but R is not the correct explanation of A.

The decomposition of hydrogen peroxide gives water and oxygen molecule as a product. The oxidation number of oxygen in peroxide is -1 and that of hydrogen is +1. Thus oxygen under consideration is reduced from -1 to -2 in H₂O molecule whereas it is oxidized to zero in oxygen molecule.

Thus both statements are true, but the main driving force is reason that peroxide undergoes disproportionation reaction is due to the availability of more stable oxygen molecule and water form for the peroxide. Thus the statement two is not the correct explanation for the (A).

(iii) A is true but R is false.

A redox couple is a reducing species and its corresponding oxidizing form written in a particular representation.

Fe³⁺|Fe²⁺ and Cu²⁺|Cu are the correct representations of a redox couple, implying that Fe³⁺ is an oxidized state of Fe and same for Cu.

Initially it was thought that the oxidation & reduction are just related to addition of oxygen molecule and removal of hydrogen atoms. As the process developed study showed that, its mostly related to transfer of electrons from one species to another resulting in more stable ions.

Thus in a redox reaction if one species loses electrons its considered to be undergoing oxidation reaction and acts as oxidizing agent or oxidant, and for species who accepts electrons is said to undergo reduction and behave as reductant.

This transfer is mainly based on the relative electronegativity of difference between the two interacting species. The more electronegative element attracts electrons and the electropositive element loses electrons.

For example, consider a redox reaction between Zn and HCl

Here the zinc loses electrons to a more electronegative atom Cl with the reaction for oxidation and reduction as follows:

Oxidation:

Reduction:

Thus transfer of electrons causes the redox reaction to occur.

The electrochemical series gives a relative strength arrangement of standard electrode reduction potentials of common ions and elements. It is arranged in increasing order of Reduction Potential from bottom to top with F being at the top and Li being at the bottom.

Now whether a reaction should takes place or not would depend on the net cell EMF of the cell, given by equation

Option (i)Cu + Zn²⁺→ Cu²⁺ + Zn

Here Cu undergoes oxidation so it acts as anode and Zn acts as cathode. So from the table

For cathode E°_cathode = -0.76 V

For anode E°_anode = 0.52 V

As the EMF of cell is negative the given reaction will not occur spontaneously if they were to form a cell placed as electrodes.

Option (ii) Mg + Fe²⁺→ Mg²⁺ + Fe

Similarly, we can say that Mg undergoes oxidation and Fe undergoes reduction.

E°_cathode = -0.44 V

E°_anode = -2.36 V

Positive EMF implies that the reaction will give out energy and attain stability, thus it will occur spontaneously. So the given redox reaction will occur.

Option (iii) Br₂ + 2Cl^-→ Cl₂ + 2Br^-

Here Br undergoes reduction thus acting as cathode and Cl acting as anode.

For cathode E°_cathode = 1.09 V

For anode E°_anode = 1.36 V

The negative potential prevents easy reaction, so the redox reaction will not occur.

Option (iv) Fe + Cd²⁺→ Cd + Fe²⁺

Fe is the cathode and Cd is the anode

For cathode E°_cathode = -0.44 V

For anode E°_anode = -0.40 V

The negative potential prevents easy reaction, so the redox reaction will not occur.

Electronegativity (EN) is the ability of an atom or molecule to attract electron towards its nucleus, which is relative to the atom or molecule to which it is bonded or is attracted to. Fluorine has the highest electronegativity in the entire periodic table with EN value of 3.98; this means that out of 4 bonded electrons 3.98 fraction of it is shared with fluorine. Thus the ability to attract electron from other elements is more pronounced than the atom with which it is bonded. And as fluorine has the highest reduction potential (E°_cell =2.87) in the spectrochemical series, it cannot undergo oxidation itself. Thus cannot display disproportionation reactions.

Redox couple is a reducing element along with its oxidizing form. So for the given example,

(i) Cu²⁺ /Cu and Zn²⁺/Zn

(ii) Mg⁺/Mg and Fe²⁺/Fe

(iii) Br^-/Br and Cl^-/Cl

(iv) Fe²⁺ /Fe and Cd²⁺/Cd

Oxidation number is calculated by assuming the unknown ON as x and applying the rules of oxidation number for the rest of the atom/molecules bonded with the unknown atom.

For all the chlorine atoms we assume its oxidation state as x, and for atoms Na, K = +1, O = -2. Thus by thumb rule of oxidation, the algebraic sum of oxidation numbers of all the atoms/molecules in a compound is equal to zero or to the charge on that compound for an ion.

The Cl₂O₇ structure is interesting to learn, as there are two atoms of chlorine atom with different kinds of O ─ O, Cl ─ Cl. Its structure is given as:

Ascending order of compounds w.r.t their oxidation number is:

NaCl (-1), Cl₂(0), Cl₂O(+1), KClO₂(+3), ClO₂(+4), NaClO₃(+5), ClO₃(+6), Cl₂O₇=NaClO₄(+7).

Among the above compounds the oxidation number +2 is not present.

Strength of a reductant (reducing agent) or oxididant (oxidising agent) can be found out by measuring the relative electrode potential when it’s connected in a solution using a cell. Element under consideration can be made electrode of a standard cell whose electrode potential is already known. For example Fe³⁺/Fe is the element we want to test with Standard Hydrogen electrode (SHE). The half-cell reaction for Fe and H are given below.

When any element needs to be evaluated it is placed as an electrode with SHE. The amount of emf it generates in the cell can be considered as the potential of the element.

The tendency of Fe³⁺ ion to undergo reduction is more as compared to hydrogen. Hence the above assumed configuration of Fe being anode can be reversed and hence strength Fe as a reductant can be established. Hence the strength of an oxidant can be determined.