Ncert Exercises & Solutions

8.27 : Predict the products of electrolysis in each of the following:
(i) An aqueous solution of ${AgNO}_{3}$ with silver electrodes
(ii) An aqueous solution ${AgNO}_{3}$ with platinum electrodes
(iii) A dilute solution of $H_{2} {SO}_{4}$ with platinum electrodes
(iv) An aqueous solution of ${CuCl}_{2}$ with platinum electrodes.

NEETprep Answer:

(i) AgNO3 ionizes in aqueous solutions to form

{Ag}^{+}

and

{NO}^{-}_{3}

ions.
On electrolysis, either

{Ag}^{+}

ions or

H_{2} O

molecules can be reduced at the cathode. But the reduction potential of

{Ag}^{+}

ions is higher than that of

H_{2} O

{Ag}^{+}_{(aq)} + e^{-} \to {Ag}_{(s)}; E ° = + 0.80 V

2 H_{2} O_{(l)} + 2 e^{-} \to {H_{2}}_{(g)} + 2 {OH}^{-}_{(aq)}; E ° = - 0.83 V

Hence,

{Ag}^{+}

ions are reduced at the cathode. Similarly, Ag metal or

H_{2} O

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

H_{2} O

molecules.

{Ag}_{(s)} + e^{-} \to {Ag}^{+}_{(a q)} + e^{-}; E ° = - 0.80 V

2 H_{2} O_{(l)} \to O_{2 (g)} + 4 {H^{+}}_{(aq)} + 4 e^{-}; E ° = - 1.23 V

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

O_{2}

. At the cathode,

{Ag}^{+}

ions are reduced and get deposited.
(iii)

H_{2} {SO}_{4}

ionizes in aqueous solutions to give

H^{+}

and

{SO}_{4}^{2 -}

ions.

H_{2} {SO}_{4 (aq)} \to 2 {H^{+}}_{(aq)} + {SO}_{4}^{2 -}_{(aq)}

On electrolysis, either of

H^{+}

ions or

H_{2} O

molecules can get reduced at the cathode. But the reduction potential of

H^{+}

ions is higher than that of

H_{2} O

molecules.

2 {H^{+}}_{(aq)} + 2 e^{-} \to H_{2 (g)}; E ° = 0.0 V

2 H_{2} O_{(aq)} + 2 e^{-} \to {H_{2}}_{(g)} + 2 {OH}^{-}_{(aq)}; E ° = - 0.83 V

Hence, at the cathode,

H^{+}

ions are reduced to liberate

H_{2}

gas.

On the other hand, at the anode, either

{SO}_{4}^{2 -}

of ions or

H_{2} O

molecules can get oxidized.
But the oxidation of

{SO}_{4}^{2 -}

involves breaking of more bonds than that of

H_{2} O

molecules.
Hence,

{SO}_{4}^{2 -}

ions have a lower oxidation potential than

H_{2} O

. Thus,

H_{2} O

is oxidized at the anode to liberate

O_{2}

molecules.
(iv) In aqueous solutions,

{CuCl}_{2}

ionizes to give

{Cu}^{2 +} and {Cl}^{-}

ions as:

{CuCl}_{2 (aq)} \to {Cu}^{2 +}_{(aq)} + 2 {Cl}^{-}_{(aq)}

On electrolysis, either of

{Cu}^{2 +}

ions or

H_{2} O

molecules can get reduced at the cathode. But the reduction potential of

{Cu}^{2 +}

is more than that of

H_{2} O

molecules.

{Cu}^{2 +}_{(aq)} + 2 e^{-} \to {Cu}_{(aq)}; E ° = + 0.34 V

H_{2} O_{(l)} + 2 e^{-} \to H_{2 (g)} + 2 {OH}^{-}; E ° = - 0.83 V

Hence,

{Cu}^{2 +}

ions are reduced at the cathode and get deposited.
Similarly, at the anode, either of

{Cl}^{-}

H_{2} O

is oxidized. The oxidation potential of

H_{2} O

is higher than that of

{Cl}^{-}

2 {Cl}^{-}_{(aq)} \to {Cl}_{2 (g)} + 2 e^{-}; E ° = - 1.36 V

2 H_{2} O_{(l)} \to O_{2 (g)} + 4 {H^{+}}_{(aq)} + 4 e^{-}; E ° = - 1.23 V

But oxidation of $H_{2} O$ molecules occurs at a lower electrode potential than that of ${Cl}^{-}$ ions because of over-voltage (extra voltage required to liberate gas). As a result, ${Cl}^{-}$ ions are oxidized at the anode to liberate ${Cl}_{2}$ gas.

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