Question

Predict the products of electrolysis in each of the following:

1. An aqueous solution of \(AgNO_3\) with silver electrodes
2. An aqueous solution \(AgNO_3\) with platinum electrodes 
3. A dilute solution of \(H_2SO_4\) with platinum electrodes 
4. An aqueous solution of \(CuCl_2\) with platinum electrodes.

Answer 1

(i) \(AgNO_3\) ionizes in aqueous solutions to form \(Ag ^+\) and \(NO_3^-\) ions.
On electrolysis, either \(Ag ^+\) ions or \(H_2O\) molecules can be reduced at the cathode. But the reduction potential of \(Ag ^+\) ions is higher than that of \(H_2O\).

\(Ag^+_{(aq)} +e^−\rightarrow Ag_{(s)};E^0=+0.80 V\)

\(2H_2O_{(l)}+2e^−\rightarrow H_2 (g)+2OH^−_{(aq)};E^0=−0.83 V\)
Hence, \(Ag ^+\) ions are reduced at the cathode. 
Similarly, \(Ag\) metal or \(H_2O\) molecules can be oxidized at the anode. But the oxidation potential of \(Ag\) is higher than that of \(H_2O\) molecules.

\(Ag_{(s)}\rightarrow Ag^+_{(aq)}+e^−;E^0=−0.80 V\)

\(2H_2O_{(l)}\rightarrow O_2(g)+4H^+_{(aq)}+4e^-;E^0=-1.23 V\)

Therefore, \(Ag\) metal gets oxidized at the anode.

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(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.

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(iii) \(H_2SO_4\) ionizes in aqueous solutions to give \(H ^+\) and \(SO_4^{2-}\) ions
\(H_2SO_4(aq) \rightarrow 2H^+_{(aq)}+SO^{2−}_4(aq)\)
On electrolysis, either of \(H ^+\) ions or \(H_2O\) molecules can get reduced at the cathode. But the reduction potential of \(H ^+\) ions is higher than that of \(H_2O\) molecules.

\(2H^+_{(aq)}+2e^−\rightarrow H_2( g);E^0=0.0 V\)

\(2H_2O_{(aq)}+2e^−\rightarrow H_2( g)+2OH^−_{(aq)};E^0=0.83 V\)

Hence, at the cathode, \(H ^+\) ions are reduced to liberate \(H_2\) gas. 
On the other hand, at the anode, either of \(SO_4^{2-}\)ions or \(H_2O\) molecules can get oxidized. But the oxidation of \(SO_4^{2-}\) involves breaking of more bonds than that of \(H_2O\) molecules.

Hence, \(SO_4^{2-}\) ions have a lower oxidation potential than \(H_2O\). 
Thus, \(H_2O\) is oxidized at the anode to liberate \(O_2\) molecules

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(iv) In aqueous solutions, \(CuCl_2\) ionizes to give \(Cu ^{2+}\) and \(Cl ^-\) ions as:
\(CuCl_2(aq)\rightarrow Cu^{2+}_{(aq)}+2Cl^−(aq)\)
On electrolysis, either of \(Cu ^{2+}\) ions or \(H_2O\) molecules can get reduced at the cathode. But the reduction potential of \(Cu ^{2+}\)is more than that of \(H_2O\) molecules.

\(Cu^ 2+_{(aq)}+2e^- \rightarrow Cu_{(aq)};E^0=+0.34 V\)
\(H_2O_{(l)}+2e^- \rightarrow H_2(g) + 2OH^- ;E^0=−0.83V\)

Hence, \(Cu ^{2+}\) ions are reduced at the cathode and get deposited.
Similarly, at the anode, either of \(Cl ^-\) or \(H_2O\) is oxidized. The oxidation potential of \(H_2O\) is higher than that of \(Cl^-\)  .

\(2Cl^-(aq)\rightarrow Cl_2(g)+2e^-;E^0=-1.36V\)

\(2H_2O_{(l)}\rightarrow O_2(g)+4H^+_{(aq)}+4e^−;E^0=−1.23 V\)
But oxidation of \(H_2O\) 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.