Question

calculate the electrochemical cell potential.
$2\ce{al} + 3\ce{sn^{+2}} \
ightarrow 3\ce{sn} + 2\ce{al^{+3}}$

half-reaction	$\text{e}^\circ$(v)

| $\ce{al^{3+}} + 3\ce{e^-} \
ightarrow \ce{al}$ | -1.66 |
| $\ce{sn^{2+}} + 2\ce{e^-} \
ightarrow \ce{sn}$ | -0.14 |
$\text{e}^\circ_{\text{cell}} = ?$ v

• 0.64 v
• -1.94 v
• 1.52 v
• 1.06 v

Explanation:

Step1: Identify Oxidation and Reduction

Oxidation: \( \text{Al}
ightarrow \text{Al}^{3+} + 3e^- \) (reverse of \( \text{Al}^{3+} + 3e^-
ightarrow \text{Al} \)), so \( E^{\circ}_{\text{ox}} = 1.66 \, \text{V} \) (reverse the sign of reduction potential).
Reduction: \( \text{Sn}^{2+} + 2e^-
ightarrow \text{Sn} \), so \( E^{\circ}_{\text{red}} = -0.14 \, \text{V} \).

Step2: Calculate Cell Potential

\( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{red (cathode)}} + E^{\circ}_{\text{ox (anode)}} \)
Substitute values: \( E^{\circ}_{\text{cell}} = -0.14 + 1.66 \)
\( E^{\circ}_{\text{cell}} = 1.52 \, \text{V} \)

Answer:

1.52 V