Question

what is the ecell for the electrolysis of molten mni₂(l)? half-reaction | e°(v) mn²⁺ + 2e⁻ → mn | -1.18 i₂ + 2e⁻ → 2i⁻ | +0.54 -0.64v -1.72v 0.64v 1.72v

Explanation:

Step1: Identify reactions (oxidation/reduction)

In electrolysis of molten \( \text{MnI}_2 \), \( \text{Mn}^{2+} \) will be reduced (gain electrons) and \( \text{I}^- \) will be oxidized (lose electrons, reverse of \( \text{I}_2 + 2e^-
ightarrow 2\text{I}^- \)).

Step2: Find \( E^\circ \) for each reaction

• Reduction: \( \text{Mn}^{2+} + 2e^-

ightarrow \text{Mn} \), \( E^\circ_{\text{red}} = -1.18 \, \text{V} \)

• Oxidation: Reverse of \( \text{I}_2 + 2e^-

ightarrow 2\text{I}^- \), so \( 2\text{I}^-
ightarrow \text{I}_2 + 2e^- \), \( E^\circ_{\text{ox}} = -E^\circ_{\text{red (reverse)}} = -0.54 \, \text{V} \) (since oxidation is reverse of reduction, \( E^\circ_{\text{ox}} = -E^\circ_{\text{red}} \) for the reverse reaction).

Step3: Calculate \( E^\circ_{\text{cell}} \)

\( E^\circ_{\text{cell}} = E^\circ_{\text{red (cathode)}} + E^\circ_{\text{ox (anode)}} \)
Substitute values: \( E^\circ_{\text{cell}} = (-1.18) + (-0.54) = -1.72 \, \text{V} \)

Answer:

-1.72 V (corresponding to the option: -1.72 V)