# Question 867f3

May 16, 2017

$\textsf{{E}_{c e l l} = + 0.58 \textcolor{w h i t e}{x} V}$

#### Explanation:

$\textsf{N {i}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s N i \text{ } {E}^{\circ} = - 0.25 \textcolor{w h i t e}{x} V}$

$\textsf{C {u}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s C u \text{ } {E}^{\circ} = + 0.34 \textcolor{w h i t e}{x} V}$

The nickel 1/2 cell has the more -ve electrode potential so will push out electrons and shift right to left.

The copper 1/2 cell will take in these electrons and shift left to right.

This gives the overall cell reaction:

$\textsf{N i + C {u}^{2 +} \rightarrow N {i}^{2 +} + C u}$

To find $\textsf{{E}_{c e l l}^{\circ}}$ subtract the least +ve value from the most +ve:

$\textsf{{E}_{c e l l}^{\circ} = + 0.34 - \left(- 0.25\right) = + 0.59 \textcolor{w h i t e}{x} V}$

Now we need to use the Nernst Equation:

$\textsf{{E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{R T}{z F} \ln Q}$

At 298K this can be written:

$\textsf{{E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{0.0591}{z} \log Q}$

$\textsf{z}$ is the no. moles of electrons transferred which, in this case = 2.

Q is the reaction quotient and from the cell equation is given by:

sf(Q=a_(Ni^(2+))/(a_(Cu^(2+)))#

Putting in the numbers:

$\textsf{{E}_{c e l l} = + 0.59 - \frac{0.0591}{2} \log \left[\frac{0.05}{0.025}\right]}$

$\textsf{{E}_{c e l l} = + 0.59 - 0.008895 \textcolor{w h i t e}{x} V}$

$\textsf{{E}_{c e l l} = + 0.58 \textcolor{w h i t e}{x} V}$