The emf of a particular voltaic cell with the cell reaction $\mathrm{Hg}_{2}^{2+}+\mathrm{H}_{2}…
V. The maximum electrical work of this cell when $0.5 \mathrm{~g}$ of $\mathrm{H}_{2}$ is consumed.
- $-3.12 \times 10^{4} \mathrm{~J}$
- $-1.25 \times 10^{5} \mathrm{~J}$
- $\quad 25.0 \times 10^{6} \mathrm{~J}$
- None of these
Solution
$W_{\max }=-2 \times 96500 \times 0.65=-1.25 \times 10^{5} \mathrm{~J}$
$0.5 \mathrm{~g} \mathrm{H}_{2}=0.25 \mathrm{~mole}$
Hence $W_{\max }$
$=-1.25 \times 10^{5} \times 0.25=-3.12 \times 10^{4} \mathrm{~J}$
Asked in: JEE-TOPICTESTS-CHEMISTRY