A) 0.40 V
B) 0.81 V
C) 1.23 V
D) \[-0.40\text{ }V\]
Correct Answer: A
Solution :
[a] Cell reaction cathode: \[\begin{align} & \,\,\,\,\,\,{{H}_{2}}O(l)+\frac{1}{2}{{O}_{2}}(g)+2{{e}^{-}}\xrightarrow{{}}2O{{H}^{-}}(aq) \\ & \,\,\,\,\,\text{anode}:{{H}_{2}}(g)\xrightarrow{{}}2{{H}^{+}}(aq)+2{{e}^{-}} \\ & \overline{{{H}_{2}}O(l)+\frac{1}{2}{{O}_{2}}(g)+{{H}_{2}}(g)\xrightarrow{{}}2{{H}^{+}}(aq)+2O{{H}^{-}}(aq)} \\ \end{align}\] Also we have \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}O(l);\] \[\Delta G_{f}^{{}^\circ }=-237.2\text{ }kJ/mole\] \[{{H}_{2}}O(l)\xrightarrow{{}}{{H}^{+}}(aq)+O{{H}^{-}}(aq);\] \[\Delta G{}^\circ =80kJ/mol\] Hence for cell reaction \[\Delta G{}^\circ =-237.2+\left( 2\times 80 \right)=-77.20\,kJ/mol\] \[\therefore E{}^\circ =-\frac{\Delta G{}^\circ }{nF}=\frac{77200}{2\times 96500}=-0.40V\]You need to login to perform this action.
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