1) The emf of a particular voltaic cell with the cell reaction $Hg_{2}^{2}+ +H_{2} \rightleftharpoons 2Hg+H^{+}$ is 0.65 V. The maximum electrical work of this cell when 0.5 g of H2 is consumed. A) $-3.12\times10^{4}J$ B) $-1.25\times10^{5}J$ C) $25.0\times10^{6}J$ D) None Answer: Option AExplanation:Wmax = -n.FE Wmax = -2 x 96500 x 0.65 = -1.25 x 105J 0.5g H2 = 0.25 mole. Hence, Wmax = 1.25 x 105 x 0.25 = -3.12 x 104 J
