Question
The $$emf$$ of a particular voltaic cell with the cell reaction $$Hg_2^{2 + } + {H_2} \rightleftharpoons 2Hg + 2{H^ + }$$ is $$0.65\,V.$$ The maximum electrical work of this cell when $$0.5\,g$$ of $${H_2}$$ is consumed.
A.
$$ - 3.12 \times {10^4}J$$
B.
$$ - 1.25 \times {10^5}J$$
C.
$$25.0 \times {10^6}J$$
D.
$${\text{None of these}}$$
Answer :
$$ - 3.12 \times {10^4}J$$
Solution :
$$\eqalign{
& {W_{\max }} = - n.FE; \cr
& {W_{\max }} = - 2 \times 96500 \times 0.65 = - 1.25 \times {10^5}J \cr
& 0.5g\,{H_2} = 0.25\,mole \cr
& {\text{Hence}}\,\,{W_{\max }} \cr
& = - 1.25 \times {10^5} \times 0.25 \cr
& = - 3.12 \times {10^4}J \cr} $$