Question

The standard reduction potentials are: \[ \mathrm{Cu^{2+} + 2e^- \rightarrow Cu} \quad E^\circ = +0.34\,V \] \[ \mathrm{Zn^{2+} + 2e^- \rightarrow Zn} \quad E^\circ = -0.76\,V \] Calculate the standard cell potential for the reaction: \[ \mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu} \]

• A. 1.10 V
• B. 0.42 V
• C. -1.10 V
• D. -0.42 V

Hint:

Cell potential is calculated as \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\), remembering to reverse the sign for the oxidation half-reaction.

Answer 1

The Correct Option is A

The given half reactions and their reduction potentials are: \[ \mathrm{Cu^{2+} + 2e^- \rightarrow Cu} \quad E^\circ_{\text{red}} = +0.34\,V \] \[ \mathrm{Zn^{2+} + 2e^- \rightarrow Zn} \quad E^\circ_{\text{red}} = -0.76\,V \] In the cell reaction: \[ \mathrm{Zn} \rightarrow \mathrm{Zn^{2+}} + 2e^- \quad \text{(oxidation)} \] \[ \mathrm{Cu^{2+}} + 2e^- \rightarrow \mathrm{Cu} \quad \text{(reduction)} \] Oxidation potential of Zn is the negative of its reduction potential: \[ E^\circ_{\text{ox}}(\mathrm{Zn}) = -(-0.76) = +0.76\,V \] Standard cell potential is: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = E^\circ_{\text{red (Cu)}} - E^\circ_{\text{red (Zn)}} \] \[ E^\circ_{\text{cell}} = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10\, V \] Hence, the standard cell potential is \(1.10\,V\).