First slide

Electrochemical cells

Question

${Cu}^{2 +} + 2 e ⟶ Cu; \log [{Cu}^{2 +}]$ vs . E_red graph is of the type as shown in figure where OA = 0.34V, then electrode potential of the half-cell of Cu ;Cu²⁺ (0.1 M) will be:

Moderate

A

$- 0.34 + \frac{0.059}{2} V$
B

0.34 +0.059V
C

0.34V
D

None of these

Solution

$E_{Cu / {Cu}^{2 +}} = E_{Cu / {Cu}^{2 +}}^{\circ} - \frac{0 .059}{2} \log [{Cu}^{2 +}]$
If $\log [{Cu}^{2 +}] = 0 i.e. [{Cu}^{2 +}] = 1$
then $E_{Cu / {Cu}^{2 +}} = {E^{0}}_{Cu / {Cu}^{2 +}}$
or $O A = E_{Cu / {Cu}^{+ 2}}^{o} = - E_{{Cu}^{+ 2} / Cu}^{\circ} = - 0.34$
Now, $E_{Cu / {Cu}^{2 \cdot}} = - 0 .34 - \frac{0 .059}{2} \log 0 .1$
$= - 0.34 + \frac{0.059}{2}$

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