Question

For hydrogen-oxygen fuel cell, the cell reaction is:
$2 H _{2}( g )+ O _{2}( g ) \rightarrow 2 H _{2} O ( l )$
If $\Delta G _{ f }^{ o }\left( H _{2} O \right)=-237.2\, kJ\,mol ^{-1}$, then emf of this cell is:

• A. + 2.46 V
• B. - 2.46 V
• C. + 1.23 V
• D. - 1.23 V

Answer 1

Answer: (C)

Gibbs free energy, $\Delta G ^{\circ}=- nFE ^{\circ}$
$\therefore E ^{\circ}=\frac{\Delta G ^{\circ}}{ nF }$
$=\frac{237.2 \times 1000 J }{2 \times 96500}$
$=1.23 V [\because n =2]$