Question

For the fuel cell reaction,
$ 2{{H}_{2}}+{{O}_{2}}\xrightarrow[{}]{{}}2{{H}_{2}}O $
$ \Delta G=-475\,kJ, $ Hence, $ E_{cell}^{o} $ is

• A. 1.23 V
• B. 2.46 V
• C. 0.615 V
• D. 0.31 V

Answer 1

Answer: (A)

$ 2{{H}_{2}}+{{O}_{2}}\xrightarrow[{}]{{}}2{{H}_{2}}O $
Total electrons involved = 4
$ \Delta G{}^\circ =-nFE{}^\circ $
$ -475\times 1000=-4\times 96500\text{ }E{}^\circ $
$ E{}^\circ =1.23V $