Use arrows ↓ and ↑ !
Calculate the standard oxidation-reduction potential of the following system: $CuI(s)$ +1$e^-$ $\leftrightarrows$ $Cu(s)+I^{-}(aq)$ Given: $Cu^+$//$Cu$ $E^o_2$= $0.52 \;V$ $L(CuI)$ = $K_{sp}(CuI)$ = $1.1\cdot 10^{-12}\;\frac{mol^2}{L^2}$ (Solubility product)
The following oxidation-reduction reaction is considered:
$CuI+e^-$ $\leftrightarrows$ $Cu+I^{-}$ $E^o_1$= $0.52 \;V$
$Cu-e^-$ $\leftrightarrows$ $Cu^{+}$ $E^o_2$= $0.52 \;V$
$CuI$ $\leftrightarrows$ $Cu^{+}+I^{-}$
For this last reaction we can calculate $ E^o $ (see → potential )! $E^o$ $ =$ $ \frac{0.059}{1}log\;K$ = $ \frac{0.059}{1}log\;(1.1\cdot 10^{-12})$ = $E^o_1-E^o_2$ then $E^o_1$ = $ 0.52 + \frac{0.059}{1}log\;(1.1\cdot 10^{-12})$ = $-0.19\; V$