# Question #ffc6a

Jan 21, 2016

(a) Decrease.
(b) Ditto
(c) n = 2

#### Explanation:

(a)

It helps to use $r i g h t \le f t h a r p \infty n s$ instead:

$Z {n}_{\left(s\right)} + 2 A {g}_{\left(a q\right)}^{+} r i g h t \le f t h a r p \infty n s Z {n}_{\left(a q\right)}^{2 +} + 2 A {g}_{\left(s\right)}$

Just by using Le Chatelier's Principle you can predict that by increasing $\left[Z {n}^{2 +}\right]$ the position of equilibrium will shift to the left thus decreasing $\Delta {E}_{\text{cell}}$.

(c)

$n$ is the number of moles of electrons transferred which you can see must be 2 since we have :

$Z n \rightarrow Z {n}^{2 +} + 2 e$

I'll attempt a fuller explanation:

The equation in the question refers to the electrode potential of the the 1/2 cell ${E}_{\text{red}}$, not ${E}_{\text{cell}}$. It should read:

${E}_{\text{cell")=E_("cell}}^{\circ} - \frac{R T}{n F} \ln Q$

At 298K this simplifies to:

${E}_{\text{cell")=E_("cell}}^{\circ} - \frac{0.05916}{2} \log Q$

$Q$ is the reaction quotient and is given by:

$Q = \frac{\left[Z {n}^{2 +}\right]}{{\left[A {g}^{+}\right]}^{2}}$

We can work out ${E}_{\text{cell}}^{\circ}$ from standard electrode potentials:

$Z {n}^{2 +} \text{/"Zn " } {E}^{\circ} = - 0.76 V$

$A {g}^{+} \text{/"Ag" } {E}^{\circ} = + 0.8 V$

To find ${E}_{\text{cell}}^{\circ}$ subtract the least +ve ${E}^{\circ}$ from the most +ve $\Rightarrow$

${E}_{\text{cell}}^{\circ} = 0.8 - \left(- 0.76\right) = + 1.56 V$

If we had standard conditions i.e unit concentrations etc then $Q = 1$ so $\text{log} Q = 0$ and $E = {E}_{\text{cell}} - 0 = + 1.56 V$.

What happens if we increase $\left[Z {n}^{2 +}\right]$ to 2M as might happen in the question?

${E}_{\text{cell")=E_("cell}}^{\circ} - \frac{0.05916}{2} \log \left(\frac{2}{1}\right)$

$\therefore {E}_{\text{cell}} = + 1.56 - 0.0089 = + 1.551 V$

So you can see that there is a slight decrease as predicted.

I note the equation has now been corrected. The same reasoning applies. If $\left[Z {n}^{2 +}\right]$ is increased you can see that ${E}^{\circ}$ is made less -ve so ${E}_{\text{cell}}$ is reduced.