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\to Cu+H_{2}O$
$(ii)F{e}_2{O}_3+3CO\to 2Fe+3C{O}_2$
$(iii)2K+F_2\to 2KF$
$(iv)BaC{l}_2+H_{2}S{O}_4\to BaS{O}_4+2HCl$

Answer:

The answer is the option $(iv)BaC{l}_2+H_{2}S{O}_4\to BaS{O}_4+2HCl$
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 $E^{}\mathrm{\Theta }$ , 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.
$E^{}\mathrm{\Theta }$ values: $F{e}^{3+}/F{e}^{2+}=+0.77;I_2(s)/I^{-}=+0.54;$
$C{u}^{2+}/Cu=+0.34;A{g}^{+}/Ag=+0.80V$
$(i)F{e}^{3+}$
$(ii)I_2(s)$
$(iii)C{u}^{2+}$
$(iv)A{g}^{+}$

Answer:

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

Question 3. $E^{}\mathrm{\Theta }$ values of some redox couples are given below. On the basis of these values choose the correct option.
$E^{}\mathrm{\Theta }$ values: $B{r}_2/B{r}^{-}=+1.90;A{g}^{+}/Ag(s)=+0.80$
$C{u}^{2+}/Cu(s)=+0.34;I_2(s)/I^{-}=+0.54$
(i) Cu will reduce $B{r}^{-}$
(ii) Cu will reduce Ag
(iii) Cu will reduce $I^{-}$
(iv) Cu will reduce $B{r}_2$
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.
$E^{}\mathrm{\Theta }$ values: $F{e}^{3+}/F{e}^{2+}=0.77;I_2/I^{-}=+0.54;$
$C{u}^{2+}/Cu=0.34;A{g}^{+}/Ag=+0.80V$

$(i)F{e}^{3+}and{I}^{-}$
$(ii)A{g}^{+}andCu$
$(iii)F{e}^{3+}$ $andCu$
$(iv)AgandF{e}^{3+}$
Answer:

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

Question 5. Thiosulphate reacts differently with iodine and bromine in the reactions given below:
$2S_2{O}_3^{2-}+I_2\to S_4{O}_6^{2-}+2I^{-}$
$S_2{O}_3^{2-}+2B{r}_2+5H_{2}O\to 2S{O}_4^{2-}+2B{r}^{-}+10H^{+}$
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.
$(i)N{H}_{2}OH$
$(ii)N{H}_{4}N{O}_3$
$(iii)N_2{H}_4$
$(iv)N_{3}H$

Answer:

The answer is the option $(ii)N{H}_{4}N{O}_3$
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) $Cr{O}_2^{-},Cl{O}_3^{-},Cr{O}_4^{2-},Mn{O}_4^{-}$
(ii) $Cl{O}_3^{-},Cr{O}_4^{2-},Mn{O}_4^{-},Cr{O}_2^{-}$
(iii) $Cr{O}_2^{-},Cl{O}_3^{-},Mn{O}_4^{-},Cr{O}_4^{2-}$
(iv) $Cr{O}_4^{2-},Mn{O}_4^{-},Cr{O}_2^{-},Cl{O}_3^{-}$

Answer:

The answer is the option (i) $Cr{O}_2^{-},Cl{O}_3^{-},Cr{O}_4^{2-},Mn{O}_4^{-}$
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
$(i)C{H}_4+2O_2\to C{O}_2+2H_{2}O$
$(ii)C{H}_4+4C{l}_2\to CC{l}_4+4HCl$
$(iii)2F_2+2O{H}^{-}\to 2F^{-}+O{F}_2+H_{2}O$
$(iv)2N{O}_2+2O{H}^{-}\to N{O}_2^{-}+N{O}_3^{-}+H_{2}O$
Answer:

The answer is the option $(iv)2N{O}_2+2O{H}^{-}\to N{O}_2^{-}+N{O}_3^{-}+H_{2}O$
Explanation: Oxidation no. of nitrogen decreases by 1 from $N{O}_2$ to $N{O}_2^{-}$ and increase by +1 from $N{O}_2$ to $N{O}_3^{-}$

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.

NCERT Exemplar Solutions Class 11 Chemistry Chapter 8: MCQ (Type 2)

All the MCQ (type 2) questions with solutions are given below:

Question 12. Which of the following statement(s) is/are not true about the following decomposition reaction.
$2KCl{O}_3\to 2KCl+3O_2$
(i) Potassium is undergoing oxidation
(ii) Chlorine is undergoing oxidation
(iii) Oxygen is reduced
(iv) None of the species are undergoing oxidation or reduction
Answer:

The answer is the option (i) and (iv)
Explanation: The statements above are not true because it is evident in the reaction that, potassium remains in the same oxidation state, and oxygen is being oxidised

Question 13. Identify the correct statement (s) in relation to the following reaction:
$Zn+2HCl\to ZnC{l}_2+H_2$
(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
$P_4+3O{H}^{-}+3H_{2}O\to P{H}_3+3H_{2}P{O}_2^{-}$
(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?
$(i)Al/A{l}^{3+},E^{\mathrm{\Theta }}=-1.66$
$(ii)Fe/F{e}^{2+},E^{\mathrm{\Theta }}=-0.44$
$(iii)Cu/C{u}^{2+},E^{\mathrm{\Theta }}=+0.34$
$(iv)F_2(g)/2F^{-}(aq),E^{\mathrm{\Theta }}=+2.87$

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

NCERT Exemplar Solutions Class 11 Chemistry Chapter 8: Short Answer Type

All the short-answer type questions with solutions are given below:

Question 17. The reaction
$C{l}_2(g)+2O{H}^{-}(aq)\to Cl{O}^{-}(aq)+C{l}^{-}(aq)+H_{2}O(l)$
represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action.
Answer:

The hypochlorite ion $(Cl{O}^{-})$ formed in the reaction oxidises the colour-bearing stains of the substances to colourless compounds; therefore, in the above reaction, the hypochlorite ion acts as a bleach

Question 18. $Mn{O}_4^{2-}$undergoes disproportionation reaction in acidic medium but $Mn{O}_4^{-}$ does not. Give reason.
Answer:

In $Mn{O}_4^{-}$, Mn has the highest oxidation state (+7); therefore, it cannot undergo oxidation which is required for disproportionation here. Hence $Mn{O}_4^{-}$ doesn’t undergo disproportionation reaction. Whereas in $Mn{O}_4^{2-}$ manganese is in +6 (i.e., less than that of in $Mn{O}_4^{-}$) 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 :
$2PbO+4HCl\to 2PbC{l}_2+2H_{2}O$
$Pb{O}_2+4HCl\to PbC{l}_2+C{l}_2+2H_{2}O$
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. $Pb{O}_2$ thus it can act as an oxidising agent. Therefore, it can oxidise $C{l}^{-}$ ions of HCl into Chlorine.

Question 20. Nitric acid is an oxidising agent and reacts with PbO but it does not react with $Pb{O}_2$. Explain why?
Answer:

We know that HNO3 is and oxidising agent and $Pb{O}_2$ 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:
$2PbO+4HN{O}_3\to 2Pb(N{O}_3{)}_2+2H_{2}O$

Question 21. Write balanced chemical equation for the following reactions:
(i) Permanganate ion $(Mn{O}_4^{-})$ reacts with sulphur dioxide gas in acidic medium to produce $M{n}^{2+}$ and hydrogensulphate ion.
(Balance by ion electron method)
(ii) Reaction of liquid hydrazine $(N_2{H}_4)$ with chlorate ion $(Cl{O}_3^{-})$ in basic medium produces nitric oxide gas and chloride ion in gaseous state.
(Balance by oxidation number method)
(iii) Dichlorine heptaoxide $(C{l}_2{O}_7)$ in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion $(Cl{O}_2^{-})$ and oxygen gas.
(Balance by ion electron method)
Answer:

$(i)2Mn{O}_4^{-}+5S{O}_2+2H_{2}O+H^{+}\to 2M{n}^{+2}+5HS{O}_4^{-}$

$(ii)N_2{H}_4+Cl{O}_3^{-}\to C{l}^{-}+N_{2}O+2H_{2}O$

$(iii)C{l}_2{O}_7+4H_2{O}_2\to 2Cl{O}_2^{-}+3H_{2}O+4O_2+2H^{+}$

Question 22. Calculate the oxidation number of phosphorus in the following species.
$(a)HP{O}_3^{2-}$ and
$(b)P{O}_4^{3-}$
Answer:

$(a)HP{O}_3^{2-}$

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.

$(b)P{O}_4^{3-}$
$x+(-2)\times 4=-3$
$x-8=-3$
$x=-3+8$
$x=+5$
Thus, O.S. of phosphorous in this ion is +5.

Question 23. Calculate the oxidation number of each sulphur atom in the following compounds:
$(a)N{a}_2{S}_2{O}_3$
$(b)N{a}_2{S}_4{O}_6$
$(c)N{a}_{2}S{O}_3$
$(c)N{a}_{2}S{O}_4$
Answer:

(a) Consider oxidation number of sulphur atom x.

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

$(b)N{a}_2{S}_4{O}_6$
$2\times (+1)+2x+(-2)\times 6=0$
$2+2x-12=0$
$2x-10=0$
$X=+5$

$(c)N{a}_{2}S{O}_3$
$2\times (+1)+x+(-2)X=0$
$2+x-6=0$
$x=+4$

$(d)N{a}_{2}S{O}_4$
$2\times (+1)+x+(-2)\times 4=0$
$2+x-8=0$
$x=+6$

Question 24. Balance the following equations by the oxidation number method.
$(i)F{e}^{2+}+H^{+}+C{r}_2{O}_7^{2-}\to C{r}^{3+}+F{e}^{3+}+H_{2}O$
$(ii)I_2+N{O}_3^{-}\to N{O}_2+I{O}_3^{-}$
$(iii)I_2+S_2{O}_3^{2-}\to I^{-}+S_4{O}_6^{2-}$
$(iv)Mn{O}_2+C_2{O}_4^{2-}\to M{n}^{2+}+C{O}_2$

Answer:

$(i)F{e}^{2+}+H^{+}+C{r}_2{O}_7^{2-}\to C{r}^{3+}+F{e}^{3+}+H_{2}O$
2⁺ 6⁺ 2- +3 +3
$F{e}^{2+}+H^{+}+C{r}_2{O}_7^{2-}\to C{r}^{3+}+F{e}^{3+}+H_{2}O$
(a) Balance the inceases and decreases in O.N.
2⁺ 6⁺ 2- +3 +3
$6F{e}^{2+}+H^{+}+C{r}_2{O}_7^{2-}\to 2C{r}^{3+}+6F{e}^{3+}+H_{2}O$
(b) Balancing H and O atoms by adding $H^{+}and{H}_{2}O$ molecules
$6F{e}^{2+}+14H^{+}+C{r}_2{O}_7^{2-}\to 2C{r}^{3+}+6F{e}^{3+}+7H_{2}O$
$(ii)I_2+N{O}_3^{-}\to N{O}_2+I{O}_3^{-}$

$I_2+N{O}_3^{-}\to N{O}_2+I{O}_3^{-}$

O.N. increases by +5
Now, we know that the total increase in O.N. = $5\times 2=10$ and the reduction in O.N. = 1
In order to equalize O.N., we will multiply $N{O}_3^{-}$ with 10, which will give us,
$I_2+10N{O}_3\to 10N{O}_2+I{O}_3^{-}$
Now, we will be balancing the atoms other than O and H, to give us
$I_2+10N{O}_3\to 10N{O}_2+2I{O}_3^{-}$
Now, on balancing O and H we will get,
$I_2+10N{O}_3+8H^{+}\to 10N{O}_2+2I{O}_3^{-}+4H_{2}O$
$(iii)I_2+S_2{O}_3^{2-}\to I^{-}+S_4{O}_6^{2-}$
0 +2 -1 +2.5
$I_2+S_2{O}_3^{2-}\to I^{-}+S_4{O}_6^{2-}$
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.
$I_2+S_2{O}_3^{2-}\to 2I^{-}+S_4{O}_6^{2-}$
$(iv)Mn{O}_2+C_2{O}_4^{2-}\to M{n}^{2+}+C{O}_2$
+4 +3 +2 +4
$Mn{O}_2+C_2{O}_4^{2-}\to M{n}^{2+}+C{O}_2$
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 $C{O}_2$ with 2 to give us,
$Mn{O}_2+C_2{O}_4^{2-}\to M{n}^{2+}+2C{O}_2$
Now, we will balance H and O by adding 2H₂O on the right side, and 4H⁺ on the left side of equation.
$Mn{O}_2+C_2{O}_4^{2-}+4H^{+}\to M{n}^{2+}+2C{O}_2+2H_{2}O$

Question 25. Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.
$(i)3HCl(aq)+HN{O}_3(aq)\to C{l}_2(g)+NOCl(g)+2H_{2}O(l)$
$(ii)HgC{l}_2(aq)+2KI(aq)\to Hg{I}_2(s)+2KCl(aq)$
$(iii)F{e}_2{O}_3(s)+3CO(g)\stackrel{\mathrm{\Delta }}{\to }2Fe(s)+3C{O}_2(g)$
$(iv)PC{l}_3(l)+3H_{2}O(l)\to 3HCl(aq)+H_{3}P{O}_3(aq)$
$(v)4N{H}_3+3O_2(g)\to 2N_2(g)+6H_{2}O(g)$
Answer:

$HCl(aq)+HN{O}_3(aq)\to C{l}_2(g)+NOCl(g)+2H_{2}O(l)$

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 $HN{O}_3$
$(ii)HgC{l}_2(aq)+2KI(aq)\to Hg{I}_2(s)+2KCl(aq)$
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.
$(iii)F{e}_2{O}_3(s)+3CO(g)\stackrel{\mathrm{\Delta }}{\to }2Fe(s)+3C{O}_2(g)$


In the above reaction, the oxidising agent is $F{e}_2{O}_3$and the reducing agent is CO.
$(iv)PC{l}_3(l)+3H_{2}O(l)\to 3HCl(aq)+H_{3}P{O}_3(aq)$

In the above reaction, oxidising agent is $O_2$and the reducing agent is $N{H}_3$.

Question 26. Balance the following ionic equations

$(i)C{r}_2{O}_7^{2-}+H^{+}+I^{-}\to C{r}^{3+}+I_2+H_{2}O$
$(ii)C{r}_2{O}_7^{2-}+F{e}^{2+}+H^{+}\to C{r}^{3+}+F{e}^{3+}+H_{2}O$

$(iii)Mn{O}_4^{-}+S{O}_3^{2-}+H^{+}\to M{n}^{2+}+S{O}_4^{2-}+H_{2}O$
$(iv)Mn{O}_4^{-}+H^{+}+B{r}^{-}\to M{n}^{2+}+B{r}_2+H_{2}O$

Answer:

$(i)C{r}_2{O}_7^{2-}+H^{+}+I^{-}\to C{r}^{3+}+I_2+H_{2}O$
As step (i), we will be generating the unbalanced skeleton i.e.,
$(i)C{r}_2{O}_7^{2-}+H^{+}+I^{-}\to C{r}^{3+}+I_2+H_{2}O$
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 $2I^{-}\to I_2+2e^{-}$and Reduction is $C{r}_2{O}_7^{2-}+6e^{-}\to C{r}^{3+}$

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

Oxidation is $2I^{-}\to I_2+2e^{-}$ and Reduction is $C{r}_2{O}_7^{2-}+6e^{-}\to C{r}^{3+}$

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 $2I^{-}\to I_2+2e^{-}$ and reduction is $C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}$

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 $2I^{-}\to I_2+2e^{-}$and

reduction is $C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$

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 $6I^{-}\to 3I_2+6e^{-}$ and
the reduction as $C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$

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.

$6I^{-}+C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O+3I_2+6e^{-}$

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.
$6I^{-}+C{r}_2{O}_7^{2-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O+3I_2$

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.

$(ii)C{r}_2{O}_7^{2-}+F{e}^{2+}+H^{+}\to C{r}^{3+}+F{e}^{3+}+H_{2}O$

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

$C{r}_2{O}_7^{2-}+F{e}^{2+}+H^{+}\to C{r}^{3+}+F{e}^{3+}+H_{2}O$


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 $F{e}^{2+}\to F{e}^{3+}+e^{-}$ and the reduction is $C{r}_2{O}_7^{2-}+6e^{-}\to C{r}^{3+}$

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

Thus, oxidation is $F{e}^{2+}\to F{e}^{3+}+e^{-}$ and the reduction is $C{r}_2{O}_7^{2-}+6e^{-}\to C{r}^{3+}$

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 $F{e}^{2+}\to F{e}^{3+}+e^{-}$is and
reduction is $C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}$

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$F{e}^{2+}\to F{e}^{3+}+e^{-}$ and
reduction is $C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$

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$6F{e}^{2+}\to 6F{e}^{3+}+6e^{-}$ and
reduction is $C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$

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.

$6F{e}^{2+}+C{r}_2{O}_7^{2-}+6e^{-}+14H^{+}\to 6F{e}^{3+}+6e^{-}+2C{r}^{3+}+7H_{2}O$
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.

$6F{e}^{2+}+C{r}_2{O}_7^{2-}+14H^{+}\to 6F{e}^{3+}+2C{r}^{3+}+7H_{2}O$
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.


$(iii)Mn{O}_4^{-}+S{O}_3^{2-}+H+\to M{n}^{2+}+S{O}_4^{2-}+H_{2}O$

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

$Mn{O}_4^{-}+S{O}_3^{2-}+H+\to M{n}^{2+}+S{O}_4^{2-}+H_{2}O$

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 $S{O}_3^{2-}\to S{O}_4^{2-}+2e^{-}$and reduction is $Mn{O}_4^{-}+5e^{-}\to M{n}^{2+}$


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 $S{O}_3^{2-}\to S{O}_4^{2-}+2e^{-}$ and reduction is $Mn{O}_4^{-}+5e^{-}\to M{n}^{2+}$


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 $S{O}_3^{2-}\to S{O}_4^{2-}+2e^{-}+2H^{+}$ and
reduction is $Mn{O}_4^{-}+5e^{-}+8H^{+}\to M{n}^{2+}$


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 $S{O}_3^{2-}+H_{2}O\to S{O}_4^{2-}+2e^{-}+2H^{+}$and
reduction is $Mn{O}_4^{-}+5e^{-}+8H^{+}\to M{n}^{2+}+4H_{2}O$


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 $5S{O}_3^{2-}+5H_{2}O\to 5S{O}_4^{2-}+10e^{-}+10H^{+}$and
reduction is $2Mn{O}_4^{-}+10e^{-}+16H^{+}\to 2M{n}^{2+}+8H_{2}O$


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.

$2Mn{O}_4^{-}+10e^{-}+16H^{+}+5S{O}_3^{2-}+5H_{2}O\to 2M{n}^{2+}+8H_{2}O+5S{O}_4^{2-}+10e^{-}+10H^{+}$

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.

$2Mn{O}_4^{-}+6H^{+}+5S{O}_3^{2-}\to 2M{n}^{2+}+3H_{2}O+5S{O}_4^{2-}$

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.


$(iv)Mn{O}_4^{-}+H^{+}+B{r}^{-}\to M{n}_{2+}+B{r}_2+H_{2}O$

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


$(iv)Mn{O}_4^{-}+H^{+}+B{r}^{-}\to M{n}^{2+}+B{r}_2+H_{2}O$

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 $2B{r}^{-}\to B{r}_2+2e^{-}$ and reduction is $Mn{O}_4^{-}+5e^{-}\to M{n}^{2+}$

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 $2B{r}^{-}\to B{r}_2+2e^{-}$ and reduction is $Mn{O}_4^{-}+5e^{-}\to M{n}^{2+}$

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 $2B{r}^{-}\to B{r}_2+2e^{-}$and
reduction is $Mn{O}_4^{-}+5e^{-}+8H^{+}\to M{n}^{2+}$


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 $2B{r}^{-}\to B{r}_2+2e^{-}$ and
reduction is $Mn{O}_4^{-}+5e^{-}+8H^{+}\to M{n}^{2+}+4H_{2}O$

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 $10B{r}^{-}\to 5B{r}_2+10e^{-}$and
reduction is $2Mn{O}_4^{-}+10e^{-}+16H^{+}\to 2M{n}^{2+}+8H_{2}O$


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.

$2Mn{O}_4^{-}+5B{r}^{-}+10e^{-}+16H^{+}\to 2M{n}^{2+}+8H_{2}O+5B{r}_2+10e^{+}$

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.

$2Mn{O}_4^{-}+10B{r}^{-}+16H^{+}\to 2M{n}^{2+}+8H_{2}O+5B{r}_2$

Thus, the charge on LHS and RHS is equal.

NCERT Exemplar Solutions Class 11 Chemistry Chapter 8: Matching Type

All the matching type questions with solutions are given below:

Question 27. Match Column I with Column II for the oxidation states of the central atoms.