Excess zinc dust is stirred into 0.050 mol/L copper(II) sulphate until the reaction Zn(s) +…

Excess zinc dust is stirred into $0.050\ \text{mol/L}$ copper(II) sulphate until the reaction $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$ is at equilibrium. The standard emf is $1.10\ \text{V}$ and $n = 2$. Find $\log K$ at $298\ \text{K}$. Then find the concentration of $\text{Cu}^{2+}$ left, and give it as $-\log[\text{Cu}^{2+}]$, in the way that pH is given. Take $[\text{Zn}^{2+}]$ to be $0.050\ \text{mol/L}$, since almost all the copper ion has reacted.