Question

The emf of a particular voltaic cell with the cell reaction $Hg_{2}^{2+}+H_{2} {\rightleftharpoons} 2Hg+2H^{+}$ is 0.65 V. The maximum electrical work of this cell when 0.5 g of $H_{2}$ is consumed.

• A. $-3.12\times10^{4} J$
• B. $-1.25\times10^{5} J$
• C. $25.0\times10^{6}J$
• D. None

Answer 1

Answer: (A)

$-3.12\times10^{4} J$

Answer 2

$W_{max} =-n. FE;$ $W_{max} =-2\times96500\times0.65=-1.25\times10^{5}J$ $0.5g H_{2}=0.25 mole.$ Hence, $W_{max}=1.25\times10^{5}\times0.25=-3.12\times10^{4}J$