Iodometric titration of copper – standard electrode potentials
Let's consider the following redox couples:

$\ce{Cu^{2+} + e^- -&gt;Cu^+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{ E^o_1}(\ce{Cu^{2+}/Cu^+}=0.17\, \ce{V})$

$\ce{Cu^{2+} + e^- + I^- -&gt;CuI\,\,\,\,\,\,\,\,\,\,\,\,}{ E^o_2}(\ce{Cu^{2+}/CuI}=?)$

As the chemical species $\ce{Cu^{2+}}$, $\ce{Cu^+}$ and $\ce{CuI}$ are together in the aqueous solution, the two redox couples  $\ce{Cu^{2+}/CuI}$ and $\ce{Cu^{2+}/Cu^+}$  are in equilibrium and that means: $$ E_1(\ce{Cu^{2+}/Cu^+})=E_2(\ce{Cu^{2+}/CuI})$$ Using Nernst equation for each couple:$$ E_1(\ce{Cu^{2+}/Cu^+})=E^0_1(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{[\ce{Cu^2+}]}{[\ce{Cu^+}]}}$$ 
$$ E_2(\ce{Cu^{2+}/CuI})=E^0_2(\ce{Cu^{2+}/CuI})+\frac{RT}{F}\ln{[\ce{I^-}][\ce{Cu^2+}]}$$
We can write then:
$$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{[\ce{Cu^2+}]}{[\ce{Cu^+}][\ce{Cu^2+}][\ce{I^-}]}}$$ We rewrite the above equation after simplification:$$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{1}{[\ce{Cu^+}][\ce{I^-}]}}$$
$$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{1}{K_{sp}(\ce{CuI})}}$$

$$E^0_2(\ce{Cu^{2+}/CuI})= 0.17+\frac{8.314 \times 298}{96500}\ln{\frac{1}{10^{-12}}}=0.88 \ce{V}$$