NCERT · Class 11 · Chemistry

Chapter 7: Redox Reactions

Practice each question, then open the answer or worked solution.

Questions

30 questions

Q21Written answer

The

\(\begin{array}{l}Mn^{ 3+ }\end{array} \)

ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂, and H⁺ ion. Write a balanced ionic equation for the reaction.

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Worked solution

The reaction is as given below:

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; \rightarrow \; Mn^{ 2+ }_{ (aq) } \; + \; MnO_{ 2 \; (s) } \; + \; H^{ + }_{ (aq) }\end{array} \)

The oxidation half-reaction:

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; \rightarrow \; MnO_{ 2 \; (s) }\end{array} \)

Add 1 electron to balance the oxidation no. :

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; \rightarrow \; MnO_{ 2 \; (s) } \;+ \; e^{ – }\end{array} \)

Add

\(\begin{array}{l}4 \; H^{ + }\end{array} \)

ions to balance the charge:

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; \rightarrow \; MnO_{ 2 \; (s) } \;+ \; e^{ – }\; + \; 4 \; H^{ + }_{ (aq) }\end{array} \)

Add 2

\(\begin{array}{l}H_{ 2 }O\end{array} \)

to balance O atoms and

\(\begin{array}{l}H^{ + }\end{array} \)

ions:

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) } \; \rightarrow \; MnO_{ 2 \; (s) } \;+ \; e^{ – }\; + \; 4 \; H^{ + }_{ (aq) }\end{array} \)

———— (1)

The reduction half-reaction:

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; \rightarrow \; Mn^{ 2+ }_{ (aq) }\end{array} \)

Add 1 electron to balance oxidation no. :

\(\begin{array}{l}Mn^{ 3+ }_{ (aq) } \; + \; e^{ – } \; \rightarrow \; Mn^{ 2+ }_{ (aq) }\end{array} \)

——— (2)

Add equations (1) and (2) to get the balanced chemical equation:

\(\begin{array}{l}2 \; Mn^{ 3+ }_{ (aq) } \; + \; 2 \; H_{ 2 }O_{ (l) } \; \rightarrow \; MnO_{ 2 \;(s) } \; + \; 2 \; Mn^{ 2+ }_{ (aq) } \; + \; 4 \; H^{ + }_{ (aq) }\end{array} \)

Q22Written answer

Consider the elements:

Cs, Ne, I and F

(a) Identify the element that exhibits only negative oxidation.

(b) Identify the element that exhibits only positive oxidation.

(c) Identify the element that exhibits both negative and positive oxidation states.

(d) Identify the element that exhibits neither negative nor positive oxidation state?

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Worked solution

(a) F exhibits only negative oxidation no. That is -1.

(b) Cs exhibits only positive oxidation no. That is +1.

(d) Ne exhibits neither negative nor positive oxidation no. That is 0.

Q23Written answer

Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

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Worked solution

The redox reaction is as given below:

\(\begin{array}{l}Cl_{ 2 \;(s)} \; + \; SO_{ 2\;(aq)} \; + \; H_{ 2 }O_{(l)} \; \rightarrow \; Cl^{ – }_{(aq)} \; + \; SO^{ 2- }_{ 4 \; (aq) }\end{array} \)

The oxidation half-reaction:

\(\begin{array}{l}SO_{ 2 \;(aq)} \;\rightarrow \; SO^{ 2- }_{ 4 \; (aq) }\end{array} \)

Add 2 electrons to balance the oxidation no. :

\(\begin{array}{l}SO_{ 2 \;(aq)} \;\rightarrow \; SO^{ 2- }_{ 4 \; (aq) } \; + \; 2 \; e^{ – }\end{array} \)

Add

\(\begin{array}{l}4 \; H^{ + }\end{array} \)

ions to balance the charge:

\(\begin{array}{l}SO_{ 2 \;(aq)} \;\rightarrow \; SO^{ 2- }_{ 4 \; (aq) } \; + \;4 \; H^{ + }_{ (aq) } \;+ \; 2 \; e^{ – }\end{array} \)

Add 2

\(\begin{array}{l}H_{ 2 }O\end{array} \)

to balance O atoms and

\(\begin{array}{l}H^{ + }\end{array} \)

ions:

\(\begin{array}{l}SO_{ 2 \;(aq)} \; + \; 2 \; H_{ 2 }O \; \rightarrow \; SO^{ 2- }_{ 4 \; (aq) } \; + \;4 \; H^{ + }_{ (aq) } \;+ \; 2 \; e^{ – }\end{array} \)

——— (1)

The reduction half-reaction:

\(\begin{array}{l}Cl_{2\;(s)} \; \rightarrow Cl^{-}_{(aq)}\end{array} \)

Balance Cl atoms:

\(\begin{array}{l}Cl_{ 2 \; (s) } \; \rightarrow \; 2 \; Cl^{ – }_{ (aq) }\end{array} \)

Add 2 electrons to balance the oxidation no. :

\(\begin{array}{l}Cl_{ 2 \; (s) } \; + \; 2 \; e^{ – } \rightarrow \; 2 \; Cl^{ – }_{ (aq) }\end{array} \)

——— (2)

Add equations (1) and (2) to get the balanced chemical equation:

\(\begin{array}{l}Cl_{ 2 \; (s) }\; + \; SO_{ 2 \;(aq)} \; + \; 2 \; H_{ 2 }O_{ (l) } \; \rightarrow \; 2 \; Cl^{ – }_{ (aq) } \; + \; SO^{ 2- }_{ 4 \; (aq) } \; + \; 4 \; H^{ + }_{ (aq) }\end{array} \)

Q24Written answer

Refer to the periodic table given in your book and now answer the following questions:

(a) Select the possible non–metals that can show disproportionation reaction?

(b) Select three metals that show disproportionation reaction?

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One of the reacting elements always has an element that can exist in at least 3 oxidation numbers.

(i) The non–metals which can show disproportionation reactions are P, Cl and S.

(ii) The three metals which can show disproportionation reactions are Mn, Ga and Cu.

Q25Written answer

In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?

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Worked solution

The balanced reaction is as given below:

\(\begin{array}{l}4 \; NH_{ 3 \; (g) } \; + \; 5 \; O_{ 2 \; (g) } \; \rightarrow \; 4 \; NO_{ (g) } \; + \; 6 \; H_{ 2 }O_{ (g) }\end{array} \)

\(\begin{array}{l}4 \; NH_{ 3 }\end{array} \)

= 4 × 17 g = 68 g

\(\begin{array}{l}5 \; O_{ 2 }\end{array} \)

= 5 × 32 g = 160 g

\(\begin{array}{l}4 \; NO\end{array} \)

= 4 × 30 g = 120 g

\(\begin{array}{l}6 \; H_{ 2 }O\end{array} \)

= 6 × 18 g = 108 g

Thus,

\(\begin{array}{l}NH_{ 3 }\end{array} \)

(68 g) reacts with

\(\begin{array}{l}O_{ 2 }\end{array} \)

( 20 g )

Therefore, 10 g of

\(\begin{array}{l}NH_{ 3 }\end{array} \)

reacts with

\(\begin{array}{l}\frac{ 160 \; \times \; 10 }{ 68 }\end{array} \)

g = 23.53 g of

\(\begin{array}{l}O_{ 2 }\end{array} \)

But only 20 g of

\(\begin{array}{l}O_{ 2 }\end{array} \)

is available.

Hence,

\(\begin{array}{l}O_{ 2 }\end{array} \)

is a limiting reagent.

Now, 160 g of

\(\begin{array}{l}O_{ 2 }\end{array} \)

gives

\(\begin{array}{l}\frac{ 120 \; \times \; 20}{ 160 }\end{array} \)

g of N = 15 g of NO.

Therefore, max of 15 g of nitric oxide can be obtained.

Q26Written answer

Using the standard electrode potentials given in Table 8.1, predict if the reaction between the following is feasible:

(a)

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

and
\(\begin{array}{l}I^{ – }_{ (aq) }\end{array} \)

(b)

\(\begin{array}{l}Ag^{ + }_{ (aq) }\end{array} \)

and
\(\begin{array}{l}Cu^{ }_{ (s) }\end{array} \)

(c)

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

and
\(\begin{array}{l}Cu^{ }_{ (s) }\end{array} \)

(d)

\(\begin{array}{l}Ag^{ }_{ (s) }\end{array} \)

and
\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

(e)

\(\begin{array}{l}Br_{ 2 \; (aq) }\end{array} \)

and
\(\begin{array}{l}Fe^{ 2+ }_{ (aq) }\end{array} \)

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Worked solution

(a)

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

and

\(\begin{array}{l}I^{ – }_{ (aq) }\end{array} \)

\(\begin{array}{l}2 \; Fe_{ (aq) }^{ 3+ } \; + \; 2 \; I^{ – }_{ (aq) } \; \rightarrow \; 2 \; Fe_{ (aq) }^{ 2+ } \; + \; I_{ 2 \; (s) }\end{array} \)

Oxidation half reaction:

\(\begin{array}{l}2 \; I^{ – }_{ (aq) } \; \rightarrow \; I_{ 2 \; (s) } \; + \; 2 \; e^{ – }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.54V\end{array} \)

Reduction half reaction:

\(\begin{array}{l}[Fe_{ (aq) }^{ 3+ } \; + \; e^{ – } \; \rightarrow \; Fe_{ (aq) }^{ 2+ }] \; \times \; 2;\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.77V\end{array} \)

\(\begin{array}{l}2\; Fe_{ (aq) }^{ 3+ } \; + \;2 \; I^{ – } \; \rightarrow \; 2 \; Fe_{ (aq) }^{ 2+ } \; + \; I_{ 2 \; (s) };\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.23V\end{array} \)

\(\begin{array}{l}E^{\circ}\end{array} \)

for the overall reaction is positive. Therefore, the reaction between

\(\begin{array}{l}Fe_{ (aq) }^{ 3+ }\end{array} \)

and

\(\begin{array}{l}I^{ – }_{ (aq) }\end{array} \)

is feasible.

(b)

\(\begin{array}{l}Ag^{ + }_{ (aq) }\end{array} \)

and

\(\begin{array}{l}Cu^{ }_{ (s) }\end{array} \)

\(\begin{array}{l}2 \; Ag^{ + }_{ (aq) } \; + \; Cu_{ (s) } \; \rightarrow \; 2 \; Ag_{ (s) } \; + \; Cu_{ (aq) }^{ 2+ }\end{array} \)

Oxidation half reaction:

\(\begin{array}{l}Cu_{ (s) } \; \rightarrow \; Cu^{ 2+ }_{ (aq) } \; + \; 2 \; e^{ – }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.34V\end{array} \)

Reduction half reaction:

\(\begin{array}{l}[Ag^{ + }_{ (aq) } \; + \; e^{ – } \; \rightarrow \; Ag_{ (s) }] \; \times \; 2\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.80V\end{array} \)

\(\begin{array}{l}2 \; Ag^{ + }_{ (aq) } \; + \; Cu_{ (s) } \; \rightarrow \; 2 \; Ag_{ (s) } \; + \; Cu^{ 2+ }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.46V\end{array} \)

\(\begin{array}{l}E^{\circ}\end{array} \)

for the overall reaction is positive. Therefore, the reaction between

\(\begin{array}{l}Ag_{ (aq) }^{ + }\end{array} \)

and

\(\begin{array}{l}Cu_{ (s) }\end{array} \)

is feasible.

(c)

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

and

\(\begin{array}{l}Cu^{ }_{ (s) }\end{array} \)

\(\begin{array}{l}2 \; Fe^{ 3+ }_{ (aq) } \; + \; Cu_{ (s) } \; \rightarrow \; 2 \; Fe^{ 2+ }_{ (s) } \; + \; Cu^{ 2+ }_{ (aq) }\end{array} \)

Oxidation half reaction:

\(\begin{array}{l}Cu_{ (s) } \; \rightarrow \; Cu^{ 2+ }_{ (aq) } \; + \; 2 \; e^{ – }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.34V\end{array} \)

Reduction half reaction:

\(\begin{array}{l}[Fe_{ (aq) }^{ 3+ }\; + \; e^{ – } \; \rightarrow \; Fe_{ (s) }^{ 2+ }] \; \times \; 2\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.77V\end{array} \)

\(\begin{array}{l}2 \; Fe_{ (aq) }^{ 3+ } \; + \; Cu_{ (s) } \; \rightarrow \; 2 \; Fe_{ (s) }^{ 2+ } \; + \; Cu_{ (aq) }^{ 2+ }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.43V\end{array} \)

\(\begin{array}{l}E^{\circ}\end{array} \)

for the overall reaction is positive. Therefore, the reaction between

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

and

\(\begin{array}{l}Cu^{ }_{ (s) }\end{array} \)

is feasible.

(d)

\(\begin{array}{l}Ag^{ }_{ (s) }\end{array} \)

and

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

\(\begin{array}{l}Ag_{ (s) } \; + \; 2 \; Fe_{ (aq) }^{ 3+ } \; \rightarrow \; Ag_{ (aq) }^{ + } \; + \; Fe_{ (aq) }^{ 2+ }\end{array} \)

Oxidation half reaction:

\(\begin{array}{l}Ag_{ (s) } \; \rightarrow \; Ag_{ (aq) }^{ + } \; + \; e^{ – }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.80V\end{array} \)

Reduction half reaction:

\(\begin{array}{l}Fe_{ (aq) }^{ 3+ } \; + \; e^{ – } \; \rightarrow \; Fe_{ (aq) }^{ 2+ }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.77V\end{array} \)

\(\begin{array}{l}Ag_{ (s) } \; + \; Fe_{ (aq) }^{ 3+ } \; \rightarrow \; Ag_{ (aq) }^{ + } \; + \; Fe_{ (aq) }^{ 2+ }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.03V\end{array} \)

\(\begin{array}{l}E^{\circ}\end{array} \)

for the overall reaction is positive. Therefore, the reaction between

\(\begin{array}{l}Ag^{ }_{ (s) }\end{array} \)

and

\(\begin{array}{l}Fe^{ 3+ }_{ (aq) }\end{array} \)

is feasible.

(e)

\(\begin{array}{l}Br_{ 2 \; (aq) }\end{array} \)

and

\(\begin{array}{l}Fe^{ 2+ }_{ (aq) }\end{array} \)

\(\begin{array}{l}Br_{ 2 \; (s) } \; + \; 2 \; Fe_{ (aq) }^{ 2+ } \; \rightarrow \; 2 \; Br_{ (aq) }^{ – } \; + \; 2 \; Fe_{ (aq) }^{ 3+ }\end{array} \)

Oxidation half reaction:

\(\begin{array}{l}[Fe_{ (aq) }^{ 2+ } \; \rightarrow \; Fe_{ (aq) }^{ 3+ } \; + \; e^{ – } ] \; \times \; 2\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.77V\end{array} \)

Reduction half reaction:

\(\begin{array}{l}Br_{2\; (aq) } \; + \; 2 \; e^{ – } \; \rightarrow \; 2 \; Br_{ (aq) }^{ – }\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +1.09V\end{array} \)

\(\begin{array}{l}Br_{2\; (s) } \; + \; 2 \; Fe_{ (aq) }^{ 2+ } \; \rightarrow \; 2 \; Br_{(aq)}^{ – } \; + \; 2 \; Fe_{ (aq) }^{3+}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.32V\end{array} \)

\(\begin{array}{l}E^{\circ}\end{array} \)

for the overall reaction is positive. Therefore, the reaction between

\(\begin{array}{l}Br_{2\; (aq) }\end{array} \)

and

\(\begin{array}{l}Fe^{2+}_{ (aq) }\end{array} \)

is feasible.

Q27Written answer

Predict the products of electrolysis in each of the following:

(i) An aqueous solution of

\(\begin{array}{l}AgNO_{3}\end{array} \)

with silver electrodes

(ii) An aqueous solution

\(\begin{array}{l}AgNO_{3}\end{array} \)

with platinum electrodes

(iii) A dilute solution of

\(\begin{array}{l}H_{2}SO_{4}\end{array} \)

with platinum electrodes

(iv) An aqueous solution of

\(\begin{array}{l}CuCl_{2}\end{array} \)

with platinum electrodes.

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Worked solution

(i)

\(\begin{array}{l}AgNO_{3}\end{array} \)

ionizes in aqueous solution to form

\(\begin{array}{l}Ag^{+}\end{array} \)

and

\(\begin{array}{l}NO^{-}_{3}\end{array} \)

ions.

On electrolysis, either

\(\begin{array}{l}Ag^{+}\end{array} \)

ion or

\(\begin{array}{l}H_{2}O\end{array} \)

molecule can be decreased at cathode. But the reduction potential of

\(\begin{array}{l}Ag^{+}\end{array} \)

ions is higher than that of

\(\begin{array}{l}H_{2}O\end{array} \)

\(\begin{array}{l}Ag^{+}_{(aq)} \; + \; e^{-} \; \rightarrow Ag_{\left ( s \right )}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.80V\end{array} \)

\(\begin{array}{l}2\; H_{2}O_{\left ( l \right )} \; +\; 2 \;e^{-} \; \rightarrow \;H_{2\;\left ( g \right )} \;+ \; 2 \;OH^{-}_{\left ( aq \right )}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.83V\end{array} \)

Therefore,

\(\begin{array}{l}Ag^{+}\end{array} \)

ions are decreased at the cathode. Same way, Ag metal or

\(\begin{array}{l}H_{2}O\end{array} \)

molecules can be oxidized at the anode. But the oxidation potential of Ag is greater than that of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules.

\(\begin{array}{l}Ag_{\left ( s \right )} \; \rightarrow \;Ag^{+}_{\left ( aq \right )} \;+ \;e^{-}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.80V\end{array} \)

\(\begin{array}{l}2\; H_{2}O_{\left ( l \right )}\; \rightarrow \;O_{2\;\left ( g \right )}\; +\; 4 \;H^{+}_{\left ( aq \right )}\; +\; 4\; e^{-}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -1.23V\end{array} \)

Hence, Ag metal gets oxidized at the anode.

(ii) Pt cannot be oxidized very easily. Therefore, at the anode, oxidation of water occurs to liberate

\(\begin{array}{l}O_{2}\end{array} \)

. At the cathode,

\(\begin{array}{l}Ag^{+}\end{array} \)

ions are decreased and get deposited.

(iii)

\(\begin{array}{l}H_{2}SO_{4}\end{array} \)

ionizes in aqueous solutions to give

\(\begin{array}{l}H^{+}\end{array} \)

and

\(\begin{array}{l}SO_{4}^{2-}\end{array} \)

ions.

\(\begin{array}{l}H_{2}SO_{4\;\left ( aq \right )} \;\rightarrow \; 2\; H^{+}_{\left ( aq \right )} \; +\; SO^{2-}_{4 \left ( aq \right )}\end{array} \)

On electrolysis, either of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules or

\(\begin{array}{l}H^{+}\end{array} \)

ions can get decreased at cathode. But the decreased potential of

\(\begin{array}{l}H^{+}\end{array} \)

ions is higher than that of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules.

\(\begin{array}{l}2 \;H^{+}_{\left ( aq \right )} \;+ \;2 \;e^{-} \;\rightarrow \;H_{2\;\left ( g \right )}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; 0.0V\end{array} \)

\(\begin{array}{l}2 \;H_{2}O_{\left ( aq \right )} \;+ \;2 \;e^{-} \;\rightarrow \;H_{2\;\left ( g \right )} \;+ \;2 \;OH^{-}_{\left ( aq \right )}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.83V\end{array} \)

Therefore, at cathode,

\(\begin{array}{l}H^{+}\end{array} \)

ions are decreased to free

\(\begin{array}{l}H_{2}\end{array} \)

gas.

On the other hand, at the anode, either of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules or

\(\begin{array}{l}SO_{4}^{2-}\end{array} \)

ions can be oxidized. But the oxidation of

\(\begin{array}{l}SO_{4}^{2-}\end{array} \)

involves breaking of more bonds than that of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules. Therefore,

\(\begin{array}{l}SO_{4}^{2-}\end{array} \)

ions have lower oxidation potential than

\(\begin{array}{l}H_{2}O\end{array} \)

. Hence,

\(\begin{array}{l}H_{2}O\end{array} \)

is oxidized at anode to free

\(\begin{array}{l}O_{2}\end{array} \)

molecules.

(iv) In aqueous solutions,

\(\begin{array}{l}CuCl_{2}\end{array} \)

ionizes to give

\(\begin{array}{l}Cu^{2+}\end{array} \)

and

\(\begin{array}{l}Cl^{-}\end{array} \)

ions as:

\(\begin{array}{l}CuCl_{2\; \left ( aq \right )} \;\rightarrow \;Cu^{2+}_{\left ( aq \right )} \;+ \;2 \;Cl_{\left ( aq \right )}^{-}\end{array} \)

\(\begin{array}{l}CuCl_{2\;\left ( aq \right )} \;\rightarrow \;Cu^{2+}_{\left ( aq \right )} \;+ \;2 \;Cl_{\left ( aq \right )}^{-}\end{array} \)

On electrolysis, either of

\(\begin{array}{l}Cu^{2+}\end{array} \)

ions or

\(\begin{array}{l}H_{2}O\end{array} \)

molecules can get decreased at cathode. But the decreased potential of

\(\begin{array}{l}Cu^{2+}\end{array} \)

is more than that of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules.

\(\begin{array}{l}Cu^{2+}_{\left ( aq \right )} \;+ \;2 \;e^{-} \;\rightarrow \;Cu_{\left ( aq \right )}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.34V\end{array} \)

\(\begin{array}{l}H_{2}O_{\left ( l \right )} \;+ \;2 \;e^{-} \;\rightarrow \;H_{2\;\left ( g \right )} \;+ \; 2 \;OH^{-}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -0.83V\end{array} \)

Therefore,

\(\begin{array}{l}Cu^{2+}\end{array} \)

ions are decreased at the cathode and get deposited. In the same way, at the anode, either of

\(\begin{array}{l}Cl^{-}\end{array} \)

\(\begin{array}{l}H_{2}O\end{array} \)

is oxidized. The oxidation potential of

\(\begin{array}{l}H_{2}O\end{array} \)

is higher than that of

\(\begin{array}{l}Cl^{-}\end{array} \)

\(\begin{array}{l}2 \;Cl^{-}_{\left ( aq \right )} \;\rightarrow \;Cl_{2\;\left ( g \right )} \;+ \;2 \;e^{-}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; +0.34V\end{array} \)

\(\begin{array}{l}2\; H_{2}O_{\left ( l \right )} \;\rightarrow \;O_{2\;\left ( g \right )} \;+ \;4 \;H^{+}_{\left ( aq \right )} \;+ \;4 \;e^{-}\end{array} \)

;

\(\begin{array}{l}E^{\circ} \; = \; -1.23V\end{array} \)

But oxidation of

\(\begin{array}{l}H_{2}O\end{array} \)

molecules occurs at a lower electrode potential compared to that of

\(\begin{array}{l}Cl^{-}\end{array} \)

ions because of over-voltage (extra voltage required to liberate gas). As a result,

\(\begin{array}{l}Cl^{-}\end{array} \)

ions are oxidized at the anode to liberate

\(\begin{array}{l}Cl_{2}\end{array} \)

gas.

Q28Written answer

Arrange the given metals in the order in which they displace each other from the solution of their salts.

Al, Fe, Cu, Zn, Mg

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Worked solution

A metal with stronger reducing power displaces another metal with weaker reducing power from its solution of salt.

The order of the increasing reducing power of the given metals is as given below:

Cu < Fe < Zn < Al < Mg

Therefore, Mg can displace Al from its salt solution, but Al cannot displace Mg. Thus, the order in which the given metals displace each other from the solution of their salts is as given below: Mg >Al>Zn> Fe >Cu

Q29Written answer

Given the standard electrode potentials,

\(\begin{array}{l}K^{+}\end{array} \)

/K = –2.93V

\(\begin{array}{l}Ag^{+}\end{array} \)

/Ag = 0.80V

\(\begin{array}{l}Hg^{2+}\end{array} \)

/Hg = 0.79V

\(\begin{array}{l}Mg^{2+}\end{array} \)

/Mg = –2.37V

\(\begin{array}{l}Cr^{3+}\end{array} \)

/Cr = –0.74V

Arrange these metals in their increasing order of reducing power.

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Worked solution

The reducing agent is stronger as the electrode potential decreases. Hence, the increasing order of the reducing power of the given metals is as given below:

Ag < Hg < Cr < Mg < K

Q30Written answer

Depict the galvanic cell in which the reaction is:

\(\begin{array}{l}Zn_{\left ( s \right )} \;+ \;2 \;Ag_{\left ( aq \right )}^{+} \;\rightarrow \;Zn^{2+}_{\left ( aq \right )} \;+ \;2 \;Ag_{\left ( s \right )}\end{array} \)

Further show:

(i) which of the electrode is negatively charged?

(ii) the carriers of the current in the cell.

(iii) individual reaction at each electrode.

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Worked solution

The galvanic cell corresponding to the given redox reaction can be shown as:

Zn|

\(\begin{array}{l}Zn^{2+}_{\left ( aq \right )}\end{array} \)

\(\begin{array}{l}Ag^{+}_{\left ( aq \right )}\end{array} \)

|Ag

(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to

\(\begin{array}{l}Zn^{2+}\end{array} \)

and the leaving electrons accumulate on this electrode.

(ii) The carriers of current are ions in the cell.

(iii) Reaction at Zn electrode is shown as:

\(\begin{array}{l}Zn_{\left ( s \right )} \;\rightarrow \;Zn^{2+}_{\left ( aq \right )} \;+ \;2 \;e^{-}\end{array} \)

Reaction at Ag electrode is shown as:

\(\begin{array}{l}Ag^{+}_{\left ( aq \right )} \;+ \;e^{-}\;\rightarrow \;Ag_{\left ( s \right )}\end{array} \)

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Chapter 7: Redox Reactions

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