Question

The standard reduction potentials are: \( E^\circ_{\text{Ag}^+/\text{Ag}} = +0.80\, \text{V} \) and \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\, \text{V} \). What will happen when a copper rod is placed in a solution of silver nitrate?

• A. Copper dissolves, silver precipitates
• B. No reaction occurs
• C. Silver dissolves, copper precipitates
• D. Both dissolve

Hint:

To predict redox reaction spontaneity, compare standard reduction potentials. The metal with lower reduction potential gets oxidized, and the ion with higher potential gets reduced.

Answer 1

The Correct Option is A

- The two half-reactions with their standard reduction potentials are: \[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad E^\circ = +0.80\, V \] \[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad E^\circ = +0.34\, V \] - When a copper rod is placed in a solution of silver nitrate (\(\text{AgNO}_3\)), the reaction will be a redox reaction between copper metal and silver ions. - To analyze if the reaction is spontaneous, we write the oxidation and reduction half-reactions: - Oxidation (anode): Copper metal is oxidized to copper ions: \[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \quad E^\circ_{\text{ox}} = -0.34\, V \] - Reduction (cathode): Silver ions are reduced to silver metal: \[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad E^\circ_{\text{red}} = +0.80\, V \] - The overall cell reaction is: \[ \text{Cu} (s) + 2 \text{Ag}^+ (aq) \rightarrow \text{Cu}^{2+} (aq) + 2 \text{Ag} (s) \] - Calculate the standard cell potential: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.80\, V - 0.34\, V = +0.46\, V \] - Since \( E^\circ_{\text{cell}}>0 \), the reaction is spontaneous. - Conclusion: - Copper dissolves (oxidized to \( \text{Cu}^{2+} \)) - Silver ions get reduced and precipitate as silver metal on the copper rod.