# Voltaic cell and Nernst equation

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}

$$\ce{Cu(s)}~|~\ce{CuSO4(aq)}~(10^{-3}~\mathrm{M})~||~\ce{CH3COONa~(aq)}~(10^{-2}~\mathrm{M})~|~\ce{Pt(s)}~(\ce{H2(g)})~(1~ \mathrm{atm})$$
In your case, notice that the ionic salt $\ce{CuSO4}$ is assumed to be a $100\%$ dissociated ionic salt, hence, its aqueous solution will produce $[\ce{Cu^{2+}}]=[\ce{SO4^{2-}}]=10^{-3}\mathrm{M}$. That should answer your final query.
Apart from that, your expression for $\mathrm{Q_c}$ is wrong. It should be $\displaystyle\mathrm{Q}=\frac{p_{\ce{H2}}}{[\ce{Cu^{+2}}]}$. Since $\ce{H2}$ is present in gaseous form, use its pressure value in the expression for $\mathrm{Q}$ (and not molarity which doesn't exist)
PS: The concentration ($\pu{10^−3M}$) that's given to you is the concentration of copper ions after equilibrium has been achieved for the interconversion of $\ce{Cu+2(aq)}$ and $\ce{Cu(s)}$.