Q24 / 160 · TS EAMCET 2012 · Chemistry

The emf (in V) of a Daniell cell containing $0.1 \mathrm{MZnSO}_4$ and $0.01 \mathrm{M} \mathrm{CuSO}_4$ solutions at their respective electrodes is $\left(E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0.34 \mathrm{~V} ; E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}\right)$

  • A
    $1.10$
  • B
    $1.16$
  • C
    $1.13$
  • Correct answer
    D
    $1.07$

Solution

$\begin{aligned} E_{\text {cell }}^{\circ} & =E_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{-}}^{\circ}-E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ} \\ & =+0.34-(-0.76) \mathrm{V} \\ & =1.1 \mathrm{~V}\end{aligned}$ Further $E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \frac{\text { [products] }}{\text { [reactants] }}$ For the reaction, $\mathrm{CuSO}_4+\mathrm{Zn} \longrightarrow \mathrm{ZnSO}_4+\mathrm{Cu}$ $\mathrm{Cu}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Cu}$ $\begin{aligned} E_{\text {cell }} & =E_{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]} \\ & =1.1-\frac{0.059}{2} \log \frac{0.1}{0.01} \\ & =1.1-\frac{0.059}{2} \log 10 \\ & =1.1-0.0295 \times 1 \quad \quad[\because \log 10=1] \\ & =1.07 \mathrm{~V}\end{aligned}$

Hence, the correct answer is option: 'D'.

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