# If electrode potential of 0.01M Cu^2+/Cu is 0.32V then what will be its E°_RP?

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#### Explanation

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Michael Share
Mar 15, 2018

+0.38 V

#### Explanation:

The half cell is:

$\textsf{C {u}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s C u}$

The potential of a metal / metal ion half cell is given by:

$\textsf{E = {E}^{\circ} - \frac{R T}{z F} \ln \textcolor{w h i t e}{x} \left(\left[\text{reduced form"])/(["oxidised form}\right]\right)}$

At $\textsf{{25}^{\circ} C}$ this simplifies to:

$\textsf{E = {E}^{\circ} + \frac{0.0591}{z} \log \left[C {u}^{2 +}\right]}$

$\textsf{z}$ is the number of moles of electrons transferred which, in this case = 2.

$\therefore$$\textsf{+ 0.32 = {E}^{\circ} + \frac{0.0591}{2} \times - 2}$

$\textsf{{E}^{\circ} = + 0.38 \textcolor{w h i t e}{x} V}$

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