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^- \(\rightarrow\) \(\frac{1}{2}\) H_2(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^- \(\rightarrow\)\(\frac{1}{2}\) H_2(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^- \(\rightarrow\) \(\frac{1}{2}\) H_2(g)
At the anode, the following processes are possible. 
2H₂O_(l) \(\rightarrow\) O₂(g) +4H⁺_(aq)+4e^-   ; \(E^0\) = +1.23 v        (i)
2SO^2-_4(aq) \(\rightarrow\) S₂O^2-_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.
Cu^2+_(aq) +2e^- → Cu_(g)    ; \(E^0\) = 0.34 V
H⁺_(aq) +e^- \(\rightarrow\) \(\frac{1}{2}\) H_2(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.