Question

calculate the electrochemical cell potential. pb + cu²⁺ → pb²⁺ + cu half-reaction e° (v) pb²⁺ + 2e⁻ → pb -0.13 cu²⁺ + 2e⁻ → cu +0.34 e°_cell = ? v 1.64v 0.47v 1.06v -0.26v

Explanation:

Step1: Identify oxidation and reduction

Oxidation: \( \text{Pb}
ightarrow \text{Pb}^{2+} + 2e^- \) (reverse of \( \text{Pb}^{2+} + 2e^-
ightarrow \text{Pb} \)), so \( E^{\circ}_{\text{oxidation}} = -(-0.13) = +0.13 \, \text{V} \) (since oxidation is reverse of reduction, flip the sign).
Reduction: \( \text{Cu}^{2+} + 2e^-
ightarrow \text{Cu} \), \( E^{\circ}_{\text{reduction}} = +0.34 \, \text{V} \).

Step2: Calculate cell potential

\( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{reduction}} + E^{\circ}_{\text{oxidation}} \)
Substitute values: \( E^{\circ}_{\text{cell}} = 0.34 + 0.13 = 0.47 \, \text{V} \).

Answer:

0.47V