Q25 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Chemistry

The maximum work that can be obtained from the following cells is $X\left|X^{2+}(a q) \| Y^{+}(a q)\right| Y$ Given, $E_{X^{2+} / X}^{\circ}=-1.7 \mathrm{~V}, E_{Y^{2+} / Y}^{\circ}=0.8 \mathrm{~V}$

  • A
    $579 \mathrm{~kJ} / \mathrm{mol}$
  • Correct answer
    B
    $482.5 \mathrm{~kJ} / \mathrm{mol}$
  • C
    $289.5 \mathrm{~kJ} / \mathrm{mol}$
  • D
    $301.8 \mathrm{~kJ} / \mathrm{mol}$

Solution

$\because \Delta G=(-) W=n F E_{\text {cell }}^{\circ}$ where, $\Delta G=$ free energy, $W=$ maximum work done $F=96500 \mathrm{C}, \quad E=$ cell potential and $E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}$ $n=$ number of electrons envolved Given, $\quad E_{\text {cathode }}^{\circ}=0.8 \mathrm{~V}$ $E_{\text {anode }}^{\circ}=(-) 1.7 \mathrm{~V}$ $\therefore \quad E_{\text {cell }}^{\circ}=0.8-[(-) 1.7]=2.5 \mathrm{~V}$ $\therefore \quad W_{\text {maximum }}=\frac{2 \times 96500 \times 2.5}{1000}(\mathrm{in} \mathrm{kJ} / \mathrm{mol})$ $=482.5 \mathrm{~kJ} / \mathrm{mol}$

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

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