The useful work done during the reaction $\mathrm{Ag}_{(\mathrm{s})}+\frac{1}{2}…

The useful work done during the reaction $\mathrm{Ag}_{(\mathrm{s})}+\frac{1}{2} \mathrm{Cl}_{2(\mathrm{~g})} ightarrow \mathrm{AgCl}_{(\mathrm{s})}$ Would be Given $\mathrm{E}_{\mathrm{Cl}_{2} ~|~ \mathrm{Cl}}^{0}=1.36 \mathrm{~V}, \mathrm{E}_{\mathrm{AgCl~|~Ag~|~} \mathrm{Cl}^{-}}^{0}=0.220 \mathrm{~V}$ $P_{C l_{2}}=1 a t m$, and $T=298 K$
  1. $110 \mathrm{KJ} \mathrm{mol}^{-1}$
  2. $220 \mathrm{KJ} \mathrm{mol}^{-1}$
  3. $55 \mathrm{KJ} \mathrm{mol}^{-1}$
  4. $100 \mathrm{KJ} \mathrm{mol}^{-1}$

Solution

For the cell reaction
$\mathrm{Ag}_{(\mathrm{s})}+\frac{1}{2} \mathrm{Cl}_{2(\mathrm{~g})} ightarrow \mathrm{AgCl}_{(\mathrm{s})}$
$\mathrm{E}^{0}=-1.14 \mathrm{~V}$
or $\mathrm{E}=\mathrm{E}^{0}-\frac{0.0592}{1} \log \mathrm{P}_{\mathrm{Cl}_{2}}^{\mathrm{1} / 2}$
Under standard conditions, $\mathrm{P}_{\mathrm{Cl}_{2}}=0$
$\therefore \log \mathrm{P}_{\mathrm{Cl}_{2}}^{1 / 2}=0$
$\therefore$ Useful work $=-\mathbf{W}_{\max }=-\mathrm{nFE}$
$=(-1) \times(-1.14) \times 96500 \times 10^{-3} \mathrm{KJ}$
$=110 \mathrm{KJ} \mathrm{mol}^{-1}$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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