This is the problem:
We can determine the solubility equilibrium for silver bromide using cell:
Ag (s) | AgNO3 (aq) || KBr (aq) | AgBr (s) | Ag (s)
We know that:
$\text{AgBr} + \text{e}^- \rightarrow \text{Ag} + \text{Br}^- \ \ \ \text{E}^0=0.095 \text{V}$
$\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}\ \ \ \text{E}^0=0.799 \text{V}$
First of all isn't that cell diagram incorrect? As far as I understand it is the silver reducing from silver nitrate to metallic silver thus being cathode. And vice versa for the silver bromide that is being produced more as silver is oxidating. Shouldn't the cell diagram be like this instead:
KBr (aq) | AgBr (s) | Ag (s) || Ag (s) | AgNO3 (aq)
Anyways, if we consider the Nernst equation:
$$E = E_0-\frac{RT}{nF}\ln{k}$$
and assume that the galvanic cell is in equilibrium when there is no voltage (i.e. $E=0$). We can then write:
$$\ln{k}=\frac{E^0nF}{RT} \\ \leftrightarrow k=e^{\frac{E^0nF}{RT}}$$
Then plug in the numbers:
$$\ln{k}=\frac{(-0.095+0.799)\ \text{V}\cdot 1 \ \text{mol}\cdot 96485.31 \ \frac{\text{C}}{\text{mol}}}{8.31451\ \frac{\text{J}}{\text{mol}\cdot \text{K}}\cdot 293.15 \ \text{K}} \\ \leftrightarrow k= 1.26754\cdot 10^{12}$$
But the right answer is:
$$1.26 \cdot 10^{-12}\ \text{mol}^2/\text{dm}^6$$
Any ideas what am I doing wrong?