Question

When a solution of $AgNO_3$ (1 M) is electrolyzed using platinum anode and copper cathode. What are the occurring at two electrodes?

Given: $E^o_{Cu^{2+}|Cu} = +0.34$ volt; $E^o_{O_2,H^+|H_2O} = +1.23$ volt; $E^o_{H^+|H_2} = +0.0$ volt; $E^o_{Ag^+|Ag} = +0.8$volt

• A. $Cu \longrightarrow Cu^{2+} + 2e^-$ at anode; $Ag^+ + e^- \longrightarrow Ag$ at cathode
• B. $\frac{1}{2}H_2O \longrightarrow \frac{1}{2}O_2 + 2H^+ + 2e^-$ at anode; $Cu^{2+} + 2e^- \longrightarrow Cu$ at cathode
• C. $H_2O \longrightarrow \frac{1}{2}O_2 + 2H^+ + 2e^-$ at anode; $Ag^+ + e^- \longrightarrow Ag$ at cathode
• D. $e^- + 2H^+ + NO_3^- \longrightarrow NO_2 + H_2O$ at anode; $Ag^+ + e^- \longrightarrow Ag$ at cathode

Answer 1

Answer: (C)

(C)

Answer 2

Solution Explanation:
Since AgNO₃ is aqueous, at the inert platinum anode water is oxidized:

$$\text{Anode: } H_2O \rightarrow \tfrac{1}{2}O_2 + 2H^+ + 2e^-$$

At the copper cathode, Ag⁺ is reduced to Ag:

$$\text{Cathode: } Ag^+ + e^- \rightarrow Ag$$

Thus, the correct option is (C).