Q25 / 160 · TS EAMCET 2021 (06 Aug Shift 2) · Chemistry

For the cell reaction, $\mathrm{Cu} / \mathrm{Cu}^{2+}(0.1 \mathrm{M}) \| \mathrm{Cu}^{2+}$ $(1.0 \mathrm{M}) / \mathrm{Cu}$, the emf of the cell at $25^{\circ} \mathrm{C}$ is $\left[E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}=0.34 \mathrm{~V}\right]$

  • A
    0.059 V
  • B
    0.311 V
  • C
    0.369 V
  • Correct answer
    D
    0.029 V

Solution

Given, $$ \begin{aligned} & E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\mathrm{o}}=0.34 \mathrm{~V} \\ & E_{\mathrm{Cu} / \mathrm{Cu}^{2+}}^{\circ}=-0.34 \mathrm{~V} \end{aligned} $$ At anode $: \mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 e^{-}$ At cathode : $\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu}(s)$ Overall reaction: $\overline{\mathrm{Cu}(s)+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}(s)}$ Using formula, $E_{\text {cell }}^o=E_{\text {red. }}^o-E_{\text {oxi. }}^o$ Using Nernst equation, $$ \begin{aligned} E_{\text {cell }} & =E_{\text {cell }}^{\circ}-\frac{0.0591}{n} \log \frac{C_1}{C_2} \\ E_{\text {cell }}^{\circ} & =(0.34-0.34)-\frac{0.059}{2} \log \left[\frac{0.1}{1}\right] \\ & =-0.029 \times \log 10^{-1} \quad(\because \log 10=1) \\ & =-0.029 \times(-\log 10)=-0.029 \times(-1)=0.029 \mathrm{~V} \end{aligned} $$

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

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