Q43 / 225 · JEE Main 2003 · Chemistry

For the redox reaction $\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.1 \mathrm{M}) \rightarrow \mathrm{Zn}^{2+}(1 \mathrm{M})+\mathrm{Cu}(\mathrm{s})$ taking place in a cell, $\mathrm{E}_{\text {cell }}^0$ is $1.10$ volt. $\mathrm{E}^{\circ}$ for the cell will be $\left(2.303 \frac{\mathrm{RT}}{\mathrm{F}}=0.0591\right)$

  • A
    $1.80$ volt
  • Correct answer
    B
    $1.07$ volt
  • C
    $0.82$ volt
  • D
    $2.14$ volt

Solution

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^0+\frac{0.059}{\mathrm{n}} \log \left[\frac{\mathrm{Cu}^{+2}}{\mathrm{Zn}^{+2}}\right]$ $=1.10+\frac{0.059}{2} \log [0.1]=1.10-0.0295=1.07 \mathrm{~V}$

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

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