Question

Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of AgNO3 with platinum electrodes.
(iii) A dilute solution of H2SO4 with platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.

Answer 1

(i) At cathode:
The following reduction reactions compete to take place at the cathode.
Ag+(aq) + e- $\rightarrow$ Ag(s) ; $E^0$ = 0.80 V
H+(aq) + e- $\rightarrow$ $\frac{1}{2}$ H2(g) ;$E^0$= 0.00 V
The reaction with a higher value of $E^0$ takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
The Ag anode is attacked by NO-3 ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag+ .
(ii) At cathode:
The following reduction reactions compete to take place at the cathode
Ag+(aq) +e- $\rightarrow$ Ag(s) ;$E^0$ = 0.80 V
H+(aq) + e- \rightarrow$$\frac{1}{2} H2(g) ; $E^0$ = 0.00 V
The reaction with a higher value of $E^0$ takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
Since Pt electrodes are inert, the anode is not attacked by NO-3 ions. Therefore, OH- or NO-3 ions can be oxidized at the anode. But OH- ions having a lower discharge potential and get preference and decompose to liberate O2.
OH-- $\rightarrow$ OH+e-
4OH- $\rightarrow$ 2H2O+O2
(iii) At the cathode, the following reduction reaction occurs to produce H2 gas.
H+(aq)+e- $\rightarrow$ $\frac{1}{2}$ H2(g)
At the anode, the following processes are possible.
2H2O(l) $\rightarrow$ O2(g) +4H+(aq)+4e- ; $E^0$ = +1.23 v (i)
2SO2-4(aq) $\rightarrow$ S2O2-6(aq) + 2e- ; $E^0$ = +1.96 V (ii)
For dilute sulphuric acid, reaction (i) is preferred to produce O2 gas. But for concentrated sulphuric acid, reaction (ii) occurs.
(iv) At cathode:
The following reduction reactions compete to take place at the cathode.
Cu2+(aq) +2e- → Cu(g) ; $E^0$ = 0.34 V
H+(aq) +e- $\rightarrow$ $\frac{1}{2}$ H2(g) ; $E^0$ = 0.00 V
The reaction with a higher value of $E^0$ takes place at the cathode. Therefore, deposition of copper will take place at the cathode. At anode:
The following oxidation reactions are possible at the anode
Cl-(aq) $\rightarrow$ $\frac{1}{2}$ Cl2(g)+e-1 ; $E^0$ = 1.36 V
2H2O(l) $\rightarrow$ O2(g)+4H+(aq) + 4e- ; $E^0$= +1.23 V
At the anode, the reaction with a lower value of $E^0$ is preferred. But due to the over-potential of oxygen, Cl- gets oxidized at the anode to produce Cl2 gas.