(a) $E_{cell} = E_{cell}^{{^\circ}} - \frac{0.059}{n}\log\frac{\left\lbrack {Zn}^{2 +} \right\rbrack}{\left\lbrack {Cu}^{2 +} \right\rbrack}$ $= 1.10 - \frac{0.059}{2}\log\frac{0.1}{0.1}$ $= 1.10\ V$
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