NCERT Exemplar Class 11 Chemistry Solutions Chapter 8 Redox Reactions

Question 1. Which of the following is not an example of a redox reaction?
$(i)$$CuO + H_{2}\rightarrow Cu + H_2O$
$(ii) Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_{2}$
$(iii) 2K + F_{2} \rightarrow 2KF$
$(iv) BaCl_{2} + H_{2}SO_{4} \rightarrow BaSO_{4} + 2HCl$

Answer:

The answer is the option $(iv) BaCl_{2} + H_{2}SO_{4} \rightarrow BaSO_{4} + 2HCl$
Explanation: The correct answer is the option (iv) because only its oxidation no. changes while that of all elements remains unchanged.

Question 2. The more positive the value of $E^{}\Theta$ , the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent.
$E^{}\Theta$ values: $Fe^{3+}/Fe^{2+}= +0.77; I_{2}(s)/I^{-}=+0.54;$
$Cu^{2+}/Cu= +0.34; Ag^{+}/Ag=+0.80V$
$(i) Fe^{3+}$
$(ii) I_{2} (s)$
$(iii) Cu^{2+}$
$(iv) Ag^{+}$

Answer:

The answer is the option $(iv) Ag^{+}$
Explanation: Reduction potential of $Ag^{+}/Ag$ is the highest.

Question 3. $E^{}\Theta$ values of some redox couples are given below. On the basis of these values choose the correct option.
$E^{}\Theta$ values: $Br_{2}/Br^{-}=+1.90; Ag^{+}/Ag(s)=+0.80$
$Cu^{2+}/Cu(s)=+0.34; I_{2}(s)/I^{-}=+0.54$
(i) Cu will reduce $Br^{-}$
(ii) Cu will reduce Ag
(iii) Cu will reduce $I^{-}$
(iv) Cu will reduce $Br_{2}$
Answer:

The answer is the option (iv) Cu will reduce Br₂
Explanation: Compared to Cu, Br₂ is a better reducing agent.

Question 4. Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
$E^{}\Theta$ values: $Fe^{3+}/Fe^{2+} =0.77;I_{2}/I^{-}=+0.54;$
$Cu^{2+}/Cu=0.34;Ag^{+}/Ag=+0.80V$

$(i) Fe^{3+} and \;I^{-}$
$(ii) Ag^{+} and \;Cu$
$(iii) Fe^{3+}$ $and \;Cu$
$(iv) Ag \;and \;Fe^{3+}$
Answer:

The answer is the option $(iv) Ag \;and \;Fe^{3+}$
Explanation: To make the reaction feasible, $E^{0}$cell need to be positive for the pair Ag and $Fe^{3+}$, but it is negative. Hence the reaction is not feasible.

Question 5. Thiosulphate reacts differently with iodine and bromine in the reactions given below:
$2S_{2}O^{2-}_{3}+I_{2}\rightarrow S_{4}O_{6}^{2-} +2I^{-}$
$S_{2}O^{2-}_{3}+2Br_{2} +5H_{2}O\rightarrow 2SO_{4}^{2-} +2Br^{-}+10H^{+}$
Which of the following statements justifies the above dual behaviour of thiosulphate?
(i) Bromine is a stronger oxidant than iodine.
(ii) Bromine is a weaker oxidant than iodine.
(iii) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions.
(iv) Bromine undergoes oxidation and iodine undergoes reduction in these reactions.
Answer:

Ans. The answer is the option (i) Bromine is a stronger oxidant than iodine.
Explanation: To be a better oxidant, the standard reduction potential of the particular element should be higher than the other, but, here, the standard reduction potential of bromine is higher than iodine. Hence, bromine is a stronger oxidant than iodin

Question 6. The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?
(i) The oxidation number of hydrogen is always +1.
(ii) The algebraic sum of all the oxidation numbers in a compound is zero.
(iii) An element in the free or the uncombined state bears oxidation number zero.
(iv) In all its compounds, the oxidation number of fluorine is -1.
Answer:

The answer is the option (i) The oxidation no. of hydrogen is always +1.
Explanation: In ionic hydrides, hydrogen exists in -1 oxidation state because the hydrogen acquires negative charge in the presence of its companion.

Question 7. In which of the following compounds, an element exhibits two different oxidation states.
$(i) NH_{2}OH$
$(ii) NH_{4}NO_{3}$
$(iii) N_{2}H_{4}$
$(iv) N_{3}H$

Answer:

The answer is the option $(ii) NH_{4}NO_{3}$
Explanation: Here, Nitrogen exists in two different oxidation states, i.e., +5 and -3.

Question 8. Which of the following arrangements represent increasing oxidation number of the central atom?
(i) $CrO_{2}^{-}, ClO_{3}^{-},CrO_{4}^{2-},MnO_{4}^{-}$
(ii) $ClO_{3}^{-},CrO_{4}^{2-},MnO_{4}^{-},CrO_{2}^{-}$
(iii) $CrO_{2}^{-}, ClO_{3}^{-},MnO_{4}^{-},CrO_{4}^{2-}$
(iv) $CrO_{4}^{2-}, MnO_{4}^{-},CrO_{2}^{-}, ClO_{3}^{-}$

Answer:

The answer is the option (i) $CrO_{2}^{-}, ClO_{3}^{-},CrO_{4}^{2-},MnO_{4}^{-}$
Explanation: the above arrangement represents increasing oxidation no. of the central atom because here the oxidation no. of the central element increases as +3, +5, +6, +7, respectively.

Question 9. The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?
$(i) 3d^1 4s^2$
$(ii) 3d^3 4s^2$
$(iii) 3d^5 4s^1$
$(iv) 3d^5 4s^2$
Answer:

The answer is the option $(iv) 3d^5 4s^2$
Explanation: the above arrangement will exhibit the largest oxidation number since total no. of electrons present in d and s subshell = 7.

Question 10. Identify disproportionation reaction
$(i) CH_4 + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$
$(ii) CH_{4} + 4Cl_{2}\rightarrow CCl_{4} + 4HCl$
$(iii) 2F_{2} + 2OH^{-}\rightarrow 2F^{-} + OF_{2} + H_{2}O$
$(iv) 2NO_{2} + 2OH^{-} \rightarrow NO_{2}^{-} + NO_{3}^{-} + H_2O$
Answer:

The answer is the option $(iv) 2NO_{2} + 2OH^{-} \rightarrow NO_{2}^{-} + NO_{3}^{-} + H_2O$
Explanation: Oxidation no. of nitrogen decreases by 1 from $NO_{2}$ to $NO_{2}^{-}$ and increase by +1 from $NO_{2}$ to $NO_{3}^{-}$

Question 11. Which of the following elements does not show disproportionation tendency?
(i) Cl
(ii) Br
(iii) F
(iv) I
Answer:

The answer is the option (iii) F
Explanation: Fluorine is the most electronegative element; therefore, it does not show a disproportionation tendency.

Question 13. Identify the correct statement (s) in relation to the following reaction:
$Zn + 2HCl \rightarrow ZnCl_{2} + H_{2}$
(i) Zinc is acting as an oxidant
(ii) Chlorine is acting as a reductant
(iii) Hydrogen ion is acting as an oxidant
(iv) Zinc is acting as a reductant
Answer:

The answer is the option (iii) and (iv)
Explanation: The above statement is correct because you according to the reaction zinc is being oxidised and hydrogen is reduced.

Question 14. The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its compounds.
$(i) 3s^{1}$
$(ii) 3d^{1}4s^{2}$
$(iii) 3d^{2}4s^{2}$
$(iv) 3s^{2} 3p^{3}$
Answer:

The answer is the option (iii) and (iv)
Explanation: Atoms having electronic configurations such as in Option (iii) and (iv) will exhibit more than one oxidation state in the compound because in option (iii) electron can be removed from 4s as well as 3d. Similarly, in option (iv) electron can be removed from 3p and 3s both.

Question 15. Identify the correct statements with reference to the given reaction
$P_{4} + 3OH^{-} + 3H_{2}O \rightarrow PH_{3} + 3H_{2}PO_{2}^{-}$
(i) Phosphorus is undergoing reduction only.
(ii) Phosphorus is undergoing oxidation only.
(iii) Phosphorus is undergoing oxidation as well as reduction.
(iv) Hydrogen is undergoing neither oxidation nor reduction.
Answer:

The answer is the option (iii) and (iv)
Explanation: This is a kind of disproportionation reaction in which phosphorous is being reduced as well as oxidised, as given in opt (iii); whereas hydrogen remains same in +1 oxidation state as in opt (iv). Therefore, option (iii) and (iv) are correct.

Question 16. Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode?
$(i)Al/Al^{3+} , E^{\Theta }=-1.66$
$(ii)Fe/Fe^{2+} , E^{\Theta }=-0.44$
$(iii)Cu/Cu^{2+} , E^{\Theta }=+0.34$
$(iv)F_{2}(g)/2F^{-}(aq) , E^{\Theta }=+2.87$

Answer:

The answer is the option (i) and (ii)
Explanation: They will act as anodes because both of the electrodes has a negative value of standard reduction potential

Question 18. $MnO_{4}^{2-}$undergoes disproportionation reaction in acidic medium but $MnO_{4}^{-}$ does not. Give reason.
Answer:

In $MnO_{4}^{-}$, Mn has the highest oxidation state (+7); therefore, it cannot undergo oxidation which is required for disproportionation here. Hence $MnO_{4}^{-}$ doesn’t undergo disproportionation reaction. Whereas in $MnO_{4}^{2-}$ manganese is in +6 (i.e., less than that of in $MnO_{4}^{-}$) oxidation state which can be oxidised as well as reduced, hence it undergoes disproportionation.

Question 19. PbO and PbO₂ react with HCl according to following chemical equations :
$2PbO + 4HCl \rightarrow 2PbCl_{2} + 2H_{2}O$
$PbO_{2} + 4HCl \rightarrow PbCl_{2} + Cl_{2} + 2H_{2}O$
Why do these compounds differ in their reactivity?
Answer:

Lead is present in +4 oxidation state and the stable oxidation state of lead in PbO is +2. $PbO_{2}$ thus it can act as an oxidising agent. Therefore, it can oxidise $Cl^{-}$ ions of HCl into Chlorine.

Question 20. Nitric acid is an oxidising agent and reacts with PbO but it does not react with $PbO_{2}$. Explain why?
Answer:

We know that HNO₃ is and oxidising agent and $PbO_2$ is a basic oxide. Therefore, it is unlikely that any reaction will occur between them, however, the acid base reaction occurs between these two compounds, as shown below:
$2PbO + 4HNO_{3} \rightarrow 2Pb(NO_{3})_{2} + 2H_{2}O$

Question 21. Write balanced chemical equation for the following reactions:
(i) Permanganate ion $(MnO_{4}^{-})$ reacts with sulphur dioxide gas in acidic medium to produce $Mn^{2+}$ and hydrogensulphate ion.
(Balance by ion electron method)
(ii) Reaction of liquid hydrazine $(N_{2}H_{4})$ with chlorate ion $(ClO_{3}^{-})$ in basic medium produces nitric oxide gas and chloride ion in gaseous state.
(Balance by oxidation number method)
(iii) Dichlorine heptaoxide $(Cl_{2}O_{7})$ in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion $(ClO_{2}^{-})$ and oxygen gas.
(Balance by ion electron method)
Answer:

$(i) 2MnO_{4}^{-} + 5SO_{2}+ 2H_{2}O + H^{+} \rightarrow 2Mn^{+2}+5HSO_{4}^{-}$

$(ii) N_{2}H_{4} + ClO_{3}^{-} \rightarrow Cl^{-} + N_{2}O + 2H_{2}O$

$(iii) Cl_{2}O_{7} + 4H_{2}O_{2} \rightarrow 2ClO_{2}^{-} + 3H_{2}O + 4O_{2}+ 2H^{+}$

Question 22. Calculate the oxidation number of phosphorus in the following species.
$(a) HPO_{3}^{2-}$ and
$(b) PO_{4}^{3-}$
Answer:

$(a) HPO_{3}^{2-}$

Let us consider the oxidation number of phosphorus to be x.

$+1 + x + (-2) \times 3 = -2$
$+1 + x - 6 = -2$
$X -5 = -2$
$X = -2 +5$
$X = +3$
Thus, O.S. of phosphorous is +3.

$(b) PO_{4}^{3-}$
$x + (-2) \times 4 = -3$
$x-8 = -3$
$x = -3 +8$
$x = +5$
Thus, O.S. of phosphorous in this ion is +5.

Question 23. Calculate the oxidation number of each sulphur atom in the following compounds:
$(a) Na_{2}S_{2}O_{3}$
$(b) Na_{2}S_{4}O_{6}$
$(c) Na_{2}SO_{3}$
$(c) Na_{2}SO_{4}$
Answer:

(a) Consider oxidation number of sulphur atom x.

$2 \times (+1) + 2x + (-2) \times 3 = 0$
$2 + 2x -6 = 0$
$2x -4 = 0$
$2x = 4$
$x = +2$

$(b) Na_{2}S_{4}O_{6}$
$2 \times (+1) + 2x + (-2) \times 6 = 0$
$2 + 2x -12 = 0$
$2x -10 = 0$
$X = +5$

$(c) Na_{2}SO_{3}$
$2 \times (+1) + x + (-2) X = 0$
$2 + x - 6 = 0$
$x = +4$

$(d) Na_{2}SO_{4}$
$2 \times (+1) + x + (-2) \times 4 = 0$
$2 + x - 8 = 0$
$x = +6$

Question 24. Balance the following equations by the oxidation number method.
$(i) Fe^{2+}+H^{+}+Cr_{2}O_{7}^{2-}\rightarrow Cr^{3+}+Fe^{3+}+H_{2}O$
$(ii) I_{2}+NO_{3}^{-}\rightarrow NO_{2}+IO_{3}^{-}$
$(iii) I_{2}+S_{2}O_{3}^{2-}\rightarrow I^{-}+S_{4}O_{6}^{2-}$
$(iv) MnO_{2}+C_{2}O_{4}^{2-}\rightarrow Mn^{2+}+CO_{2}$

Answer:

$(i) Fe^{2+}+H^{+}+Cr_{2}O_{7}^{2-}\rightarrow Cr^{3+}+Fe^{3+}+H_{2}O$
$Fe^{2+}+H^{+}+Cr_{2}O_{7}^{2-}\rightarrow Cr^{3+}+Fe^{3+}+H_{2}O$
(a) Balance the inceases and decreases in O.N.
$6Fe^{2+}+H^{+}+Cr_{2}O_{7}^{2-}\rightarrow 2Cr^{3+}+6Fe^{3+}+H_{2}O$
(b) Balancing H and O atoms by adding $H^{+} and H_{2}O$ molecules
$6Fe^{2+}+14H^{+}+Cr_{2}O_{7}^{2-}\rightarrow 2Cr^{3+}+6Fe^{3+}+7H_{2}O$
$(ii) I_{2}+NO_{3}^{-}\rightarrow NO_{2}+IO_{3}^{-}$

$I_{2}+NO_{3}^{-}\rightarrow NO_{2}+IO_{3}^{-}$

O.N. increases by +5
Now, we know that the total increase in O.N. = $5\times2= 10$ and the reduction in O.N. = 1
In order to equalize O.N., we will multiply $NO_{3}^{-}$ with 10, which will give us,
$I_{2} + 10NO_{3} \rightarrow 10NO_{2} + IO_{3}^{-}$
Now, we will be balancing the atoms other than O and H, to give us
$I_{2} + 10NO_{3} \rightarrow 10NO_{2} + 2IO_{3}^{-}$
Now, on balancing O and H we will get,
$I_{2} + 10NO_{3} + 8H^{+}\rightarrow 10NO_{2} + 2IO_{3}^{-}+4H_{2}O$
$(iii) I_{2}+S_{2}O_{3}^{2-}\rightarrow I^{-}+S_{4}O_{6}^{2-}$
$I_{2}+S_{2}O_{3}^{2-}\rightarrow I^{-}+S_{4}O_{6}^{2-}$
Total increase in O.N. $=0.5 \times 4 = 2$
Total decrease in O.N. $=1 \times 2 = 2$
To equalise O.N. multiply $S_{2}O_{3}^{2-}$ and $I^{-}$by 2.
$I_{2}+S_{2}O_{3}^{2-}\rightarrow 2I^{-}+S_{4}O_{6}^{2-}$
$(iv) MnO_{2}+C_{2}O_{4}^{2-}\rightarrow Mn^{2+}+CO_{2}$
$MnO_{2}+C_{2}O_{4}^{2-}\rightarrow Mn^{2+}+CO_{2}$
Total increase in O.N. $=1 \times 2 = 2$
Total decrease in O.N. $=2$
Now, in order to equalize O.N., we will multiply $CO_{2}$ with 2 to give us,
$MnO_2 + C_{2}O^{2-}_{4} \rightarrow Mn^{2+} + 2CO_{2}$
Now, we will balance H and O by adding 2H₂O on the right side, and 4H⁺ on the left side of equation.
$MnO_2 + C_{2}O^{2-}_{4} +4H^{+}\rightarrow Mn^{2+} + 2CO_{2}+2H_{2}O$

Question 25. Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.
$(i)3HCl(aq) +HNO_{3} (aq)\rightarrow Cl_{2}(g)+NOCl(g)+2H_{2}O(l)$
$(ii)HgCl_{2}(aq) +2KI (aq)\rightarrow HgI_{2}(s)+2KCl(aq)$
$(iii)Fe_{2}O_{3}(s)+3CO(g)\overset{\Delta }{\rightarrow} 2Fe(s)+3CO_{2}(g)$
$(iv)PCl_{3}(l)+3H_{2}O(l)\rightarrow 3HCl(aq)+H_{3}PO_{3}(aq)$
$(v)4NH_{3}+3O_{2}(g)\rightarrow 2N_{2}(g)+6H_{2}O(g)$
Answer:

$HCl(aq) +HNO_{3} (aq)\rightarrow Cl_{2}(g)+NOCl(g)+2H_{2}O(l)$

In the above reaction, it can be seen that Cl oxidizes from -1 to 0 and Nitrogen reduces from +5 to +3. Thus, Cl oxidizes to act as a reducing agent and the nitric acid works as the oxidizing agent.

Therefore, in the above reaction, the reducing agent is HCl and the oxidizing agent is $HNO_{3}$
$(ii)HgCl_{2}(aq) +2KI (aq)\rightarrow HgI_{2}(s)+2KCl(aq)$
The above mentioned reaction is the perfect example of a displacement reaction. As a result, none of the components undergo the process of reduction or oxidation. This is because the oxidation states of Hg, Cl, K, & I are unchanged in both the products and reactants.
$(iii)Fe_{2}O_{3}(s)+3CO(g)\overset{\Delta }{\rightarrow} 2Fe(s)+3CO_{2}(g)$


In the above reaction, the oxidising agent is $Fe_{2}O_{3}$and the reducing agent is CO.
$(iv)PCl_{3}(l)+3H_{2}O(l)\rightarrow 3HCl(aq)+H_{3}PO_{3}(aq)$

In the above reaction, oxidising agent is $O_2$and the reducing agent is $NH_3$.

Question 26. Balance the following ionic equations

$(i) Cr_{2}O_{7}^{2-} + H^{+} + I^{-}\rightarrow Cr^{3+} + I_2 + H_{2}O$
$(ii) Cr_{2}O^{2-}_{7} + Fe^{2+} + H^{+}\rightarrow Cr^{3+} + Fe^{3+} + H_{2}O$

$(iii) MnO^{-}_{4} + SO^{2-}_{3} + H^{+}\rightarrow Mn^{2+} + S O^{2-}_{4} + H_2O$
$(iv) MnO_{4}^{-} + H^{+} + Br^{-}\rightarrow Mn^{2+} + Br_{2} + H_{2}O$

Answer:

$(i) Cr_{2}O_{7}^{2-} + H^{+} + I^{-}\rightarrow Cr^{3+} + I_2 + H_{2}O$
As step (i), we will be generating the unbalanced skeleton i.e.,
$(i) Cr_{2}O_{7}^{2-} + H^{+} + I^{-}\rightarrow Cr^{3+} + I_2 + H_{2}O$
As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.

Oxidation is $2I^{-}\rightarrow I_{2}+2e^{-}$and Reduction is $Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow Cr^{3+}$

As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples.

Oxidation is $2I^{-}\rightarrow I_{2}+2e^{-}$ and Reduction is $Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow Cr^{3+}$

Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.

The oxidation is $2I^{-}\rightarrow I_{2}+2e^{-}$ and reduction is $Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}$

As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.

Hence, oxidation is $2I^{-}\rightarrow I_{2}+2e^{-}$and

reduction is $Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O$

Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction. We will now multiply the oxidation reaction by 3, which will give us,

The oxidation as $6I^{-}\rightarrow 3I_{2}+6e^{-}$ and
the reduction as $Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O$

Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.

$6I^{-}+Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O+3I_{2}+6e^{-}$

Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.
$6I^{-}+Cr_{2}O_{7}^{2-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O+3I_{2}$

Now, we know that the charge on the LHS is +6 which is same as the RHS of the equation, thereby enduring that the reaction is balanced.

$(ii) Cr_{2}O^{2-}_{7} + Fe^{2+} + H^{+}\rightarrow Cr^{3+} + Fe^{3+} + H_{2}O$

As step (i), we will be generating the unbalanced skeleton i.e.,

$Cr_{2}O^{2-}_{7} + Fe^{2+} + H^{+}\rightarrow Cr^{3+} + Fe^{3+} + H_{2}O$


As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.

Thus, oxidation is $Fe^{2+}\rightarrow Fe^{3+}+e^{-}$ and the reduction is $Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow Cr^{3+}$

As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples.

Thus, oxidation is $Fe^{2+}\rightarrow Fe^{3+}+e^{-}$ and the reduction is $Cr_{2}O_{7}^{2-}+6e^{-}\rightarrow Cr^{3+}$

Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.

Thus, oxidation $Fe^{2+}\rightarrow Fe^{3+}+e^{-}$is and
reduction is $Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}$

As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.

Thus, oxidation is$Fe^{2+}\rightarrow Fe^{3+}+e^{-}$ and
reduction is $Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O$

Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.

Thus, oxidation is$6Fe^{2+}\rightarrow 6Fe^{3+}+6e^{-}$ and
reduction is $Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 2Cr^{3+}+7H_{2}O$

Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.

$6Fe^{2+}+Cr_{2}O_{7}^{2-}+6e^{-}+14H^{+}\rightarrow 6Fe^{3+}+6e^{-}+2Cr^{3+}+7H_{2}O$
Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.

$6Fe^{2+}+Cr_{2}O_{7}^{2-}+14H^{+}\rightarrow 6Fe^{3+}+2Cr^{3+}+7H_{2}O$
We will now verify whether all the charges are balanced.

So, the LHS = $6\times +2 + 1\times -2 + 14 = 24$ and the RHS = $6\times +3 + 2\times +3 + 7\times 0 = 24$

Thus the sum total is same on both the sides, therefore the solve reaction is right.


$(iii) MnO^{-}_{4} + SO^{2-}_{3} + H+\rightarrow Mn^{2+} + S O^{2-}_{4} + H_2O$

As step (i), we will be generating the unbalanced skeleton i.e.,

$MnO^{-}_{4} + SO^{2-}_{3} + H+\rightarrow Mn^{2+} + S O^{2-}_{4} + H_2O$

Here in Mn undergoes reduction and S undergoes oxidation.

As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.

Thus, oxidation is $SO^{2-}_{3} \rightarrow SO^{2-}_{4}+2e^{-}$and reduction is $MnO_{4}^{-}+5e^{-}\rightarrow Mn^{2+}$


As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples. But as we can see that the atoms are balanced, so the reaction will be the same as the previous step.

Thus, oxidation is $SO^{2-}_{3} \rightarrow SO^{2-}_{4}+2e^{-}$ and reduction is $MnO_{4}^{-}+5e^{-}\rightarrow Mn^{2+}$


Now, as step (iv) for acidic solutions, we will balance the charge by adding H+ on the required side of the given equation.

Thus, oxidation is $SO^{2-}_{3} \rightarrow SO^{2-}_{4}+2e^{-}+2H^{+}$ and
reduction is $MnO_{4}^{-}+5e^{-}+8H^{+}\rightarrow Mn^{2+}$


As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.

Thus, oxidation is $SO^{2-}_{3}+H_{2}O \rightarrow SO^{2-}_{4}+2e^{-}+2H^{+}$and
reduction is $MnO_{4}^{-}+5e^{-}+8H^{+}\rightarrow Mn^{2+}+4H_{2}O$


Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.

Thus, oxidation is $5SO^{2-}_{3}+5H_{2}O \rightarrow 5SO^{2-}_{4}+10e^{-}+10H^{+}$and
reduction is $2MnO_{4}^{-}+10e^{-}+16H^{+}\rightarrow 2Mn^{2+}+8H_{2}O$


Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.

$2MnO_{4}^{-}+10e^{-}+16H^{+}+5SO_{3}^{2-} +5H_{2}O\rightarrow 2Mn^{2+}+8H_{2}O+5SO_{4}^{2-}+10e^{-}+10H^{+}$

Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.

$2MnO_{4}^{-}+6H^{+}+5SO_{3}^{2-} \rightarrow 2Mn^{2+}+3H_{2}O+5SO_{4}^{2-}$

We will now verify whether all the charges are balanced.

So, LHS = $2\times-1+6+5\times-2= -6$and the RHS = $2\times+2+3\times 0+5\times-2 = -6$
Equal charges on both sides imply towards a balanced equation.


$(iv) MnO_{4}^{-} + H^{+} + Br^{-}\rightarrow Mn_{2+} + Br_{2} + H_{2}O$

As step (i), we will be generating the unbalanced skeleton i.e.,


$(iv) MnO_{4}^{-} + H^{+} + Br^{-}\rightarrow Mn^{2+} + Br_{2} + H_{2}O$

As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.


Thus, oxidation is $2Br^{-}\rightarrow Br_{2}+2e^{-}$ and reduction is $MnO_{4}^{-}+5e^{-}\rightarrow Mn^{2+}$

As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples.
As we can see that the atoms are balanced, therefore the reaction is going to be the same as the previous step.

Thus, oxidation is $2Br^{-}\rightarrow Br_{2}+2e^{-}$ and reduction is $MnO_{4}^{-}+5e^{-}\rightarrow Mn^{2+}$

Now, as step (iv) for acidic solutions, we will balance the charge by adding H⁺ on the required side of the given equation.


Thus, oxidation is $2Br^{-}\rightarrow Br_{2}+2e^{-}$and
reduction is $MnO_{4}^{-}+5e^{-}+8H^{+}\rightarrow Mn^{2+}$


As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.

Thus, the oxidation is $2Br^{-}\rightarrow Br_{2}+2e^{-}$ and
reduction is $MnO_{4}^{-}+5e^{-}+8H^{+}\rightarrow Mn^{2+}+4H_{2}O$

Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.

Thus, oxidation is $10Br^{-}\rightarrow 5Br_{2}+10e^{-}$and
reduction is $2MnO_{4}^{-}+10e^{-}+16H^{+}\rightarrow 2Mn^{2+}+8H_{2}O$


Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.

$2MnO_{4}^{-}+5Br^{-}+10e^{-}+16H^{+}\rightarrow 2Mn^{2+}+8H_{2}O+5Br_{2}+10e^{+}$

Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.

$2MnO_{4}^{-}+10Br^{-}+16H^{+}\rightarrow 2Mn^{2+}+8H_{2}O+5Br_{2}$

Thus, the charge on LHS and RHS is equal.

Question 28.Match the items in Column I with relevant items in Column II.

Answer:

(i) → (e); (ii) → (d); (iii) → (c); (iv) → (b); (v) → (f)

(i) Ions having a positive charge are called cations.

(ii) The sum of the oxidation numbers of all atoms in a neutral molecule is 0.

(iii) Oxidation number of hydrogen ion $(H^{+})$ is +1

(iv) Oxidation number of fluorine in NaF

Let the oxidation state of F is x

+1 + (x) = 0

x= -1

(v) Ions having a negative charge are called anions.

Question 30. In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): In the reaction between potassium permanganate and potassium iodide, permanganate ions act as oxidising agent.
Reason (R) : Oxidation state of manganese changes from +2 to +7 during the reaction.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Answer:

The answer is the option (iii) A is true, but R is false.Explanation: Due to the change of permanganate ion changes to $MnO_{2}$

As per this equation, the maximum oxidation state that Mn is capable of reaching as per its electronic configuration is +7. Therefore, being an oxidising agent, it cannot undergo oxidation and hence undergoes reduction. On the other hand, Iodine undergoes oxidation from -1 to 0.

Question 31. In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : The decomposition of hydrogen peroxide to form water and oxygen is an example of disproportionation reaction.
Reason (R) : The oxygen of peroxide is in –1 oxidation state and it is converted to zero oxidation state in $O_{2}$ and –2 oxidation state in $H_{2}O$.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Answer:

The answer is the option (i) Both A and R are true, and R is the correct explanation of A.
Explanation: $2H_{2}O_{2}(aq) \rightarrow 2H_{2}O(l) + O_{2�}(g)$
Here the oxygen of peroxide is present in -1 state, and it is converted to zero oxidation state in $O_2$ undergoing oxidation and decreases to -2 oxidation state in $H_{2}O$undergoing reduction.

Question 32. In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : Redox couple is the combination of oxidised and reduced form of a substance involved in an oxidation or reduction half cell.
Reason (R) : In the representation $E^{\Theta }_{Fe^{3+}/Fe^{2+}}$ and $E^{\Theta }_{Cu^{2+}/Cu}, Fe^{3+}/Fe^{2+}$ and $Cu^{2+}/Cu$ are redox couples.
(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.
Answer:

The answer is the option (ii) Both A and R are true, but R is not the correct explanation of A.
Explanation: A redox couple is defined as a pair of compounds or elements having together the oxidised and reduced forms of it and taking part in an oxidation or reduction half-reaction.

Question 34. On the basis of standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for $E^{\Theta }$ value).

$(i) Cu + Zn^{2+}\rightarrow Cu^{2+} + Zn$
$(ii) Mg + Fe^{2+}\rightarrow Mg^{2+} + Fe$
$(iii) Br_{2} + 2Cl^{-}\rightarrow Cl_{2} + 2Br^{-}$
$(iv) Fe + Cd^{2+}\rightarrow Cd + Fe^{2+}$
Answer:

If a reaction might take place or not is dependent on the net cell EMF of the cell. The equation for the same is
$E^{^{\circ}}_{cell}=E^{^{\circ}}_{cathode}-E^{^{\circ}}_{anode}$
In option (ii) it is clearly visible that, the reaction can take place as Mg has a more negative value of $E^{^{\circ}}_{cell}$. Hence, Mg is oxidised by losing electron and iron is reduced by gaining an electron.
$Mg + Fe^{2+}\rightarrow Mg^{2+} + Fe$
We can say that Fe undergoes reduction and Mg undergoes oxidation,

$E^{^{\circ}}_{cathode}=-0.44V$
$E^{^{\circ}}_{anode}=-2.36V$
$E^{^{\circ}}_{cell}=-0.44- \left (-2.36 \right )V$
$E^{^{\circ}}_{cell}=+1.92V$

Question 35. Why does fluorine not show disproportionation reaction?
Answer:

We already know that; fluorine has the highest electronegativity in the entire periodic table with EN value of 3.98. Therefore, it is the most electronegative element in its period as well as in the group. It shows only -1 oxidation state and is smallest in size of all the halogens. We know that fluorine has the greatest reduction potential $(E^{^{\circ}}cell =2.87)$therefore, it is not able to undergo oxidation by itself. Hence, fluorine shows a disproportionation reaction.

Question 36. Write redox couples involved in the reactions (i) to (iv) given in question 34.
Answer:

$(i) Cu/Cu^{2+}\; and\; Zn^{2+} / Zn$
$(ii) Mg/Mg^{+2} \;and \;Fe^{2+}/ Fe$
$(iii) Br_{2}/Br^{-} \;and\;Cl^{-}/ Cl_{2}$
$(iv) Fe/Fe^{2+}\; and\; Cd^{2+}/ Cd$

Question 37. Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine.
$NaClO_{4}, NaClO_{3}, NaClO, KClO_{2}, Cl_{2}O_{7}, ClO_{3}, Cl_{2}O, NaCl, Cl_{2}, ClO_{2}.$
Which oxidation state is not present in any of the above compounds?
Answer:

For all the chlorine atoms, we will assume the oxidation state to be x, and for atoms Na, K = +1, O = -2. Now, we know that, the sum of oxidation numbers of all the atoms in a compound is the same as the charge on that compound for an ion or zero.

Hence,

$NaClO_{4}\rightarrow +1+x+4\times -2 = 0\rightarrow x = +7$
$NaClO_{3}\rightarrow +1+x+4\times -2 = 0\rightarrow x = +5$


In the $Cl_{2}O_{7}$ structure there are two atoms of chlorine with varying types of O ? O, Cl ? Cl. The structure is explained as follows: -


$Cl_{2}O_{7}\rightarrow 2\times x+7\times -2 = 0\rightarrow x = +7$
$ClO_{3}\rightarrow x+2\times -2 = 0\rightarrow x = +6$
$Cl_{2}O\rightarrow 2 \times x -2 = 0\rightarrow x = +1$
$NaCl\rightarrow +1+ x = 0\rightarrow x = -1$
$Cl_{2}\rightarrow$ As chlorine is in its naturalthe oxidation number is zero.
$ClO_{2}\rightarrow x +2\times -2= 0\rightarrow x = +4$
Therefore the ascending order is,
$NaCl (-1), Cl_{2}(0), Cl_{2}O(+1), KClO_{2}(+3), ClO_{2}(+4), NaClO_{3}(+5), ClO_{3}(+6), Cl_{2}O_{7}=NaClO_{4}(+7).$

Question 38. Which method can be used to find out strength of reductant/oxidant in a solution? Explain with an example.
Answer:

In order to determine the strength of a reductant/oxidant, we can use the titration method; it can be performed with the help of a redox-sensitive indicator.
The usage indicators in redox titration are as shown below:

In one scenario, the reagent itself is intensely coloured, e.g., permanganate ion, $MnO_{4}^{-}$. In here, $MnO_{4}^{-}$ acts as a self-indicator. In this case, the visible endpoint is attained after the last of the reductant$(Fe^{2+} or C_{2}O_{4}^{2-})$is oxidised. Now, the first lasting tinge of pink colour appears as $MnO_{4}^{-}$ concentration as low as $10^{-5}$mol $(10^{-6} mol L^{-1})$. This make sure of a minimal ‘overshoot’ in colour beyond the equivalence point, which is the point from where the reductant and the oxidant are equal in terms of their mole stoichiometry.