It is given voltaic cell $\ce{Cu}|\ce{CuSO4}(\pu{10^-3 M})||\ce{CH3COONa}(10^{-2})|\ce{Pt}(\ce{H2})(\pu{1 atm})$. Find the potential at $\pu{50^\circ C}$.
Given $K_\mathrm{a}$ for $H= 1.8 \times 10^{-3}$. $E_{\ce{Cu^2+/Cu}} = \pu{0.336 V}$
When voltaic cell starts working: $\ce{H2}$ becomes $\ce{2H+}$. In the other hand $\ce{CH3COO-}$ hydrolizes forming $\ce{CH3COOH}$ and $\ce{OH-}$. We know $K_\mathrm{a}$ and we can find $K_\mathrm{b}$. And from that we can find the concentration of $\ce{OH-}$ and after that using the formula $\mathrm{pH}=14-\mathrm{pOH}$ we can find the concentration of $\ce{H+}$.
Now how we can find the concentration of $\ce{Cu^2+}$ in order to replace it into Nernst equation? \begin{align} E &=E^\circ - \frac{RT}{nF} \cdot \ln\{\text{Qc}\} \text{, where}& \text{Qc} &= \frac{[\ce{H+}]}{[\ce{Cu^2+}]}. \end{align}