Question

Consider a $70 \%$ efficient hydrogen-oxygen fuel cell working under standard conditions at $1$ bar and $298 \,K$. Its cell reaction is
$H _{2}( g )+\frac{1}{2} O _{2}( g ) \rightarrow H _{2} O ( l )$
The work derived from the cell on the consumption of $1.0 \times 10^{-3} mol$ of $H _{2}( g )$ is used to compress $1.00\, mol$ of a monoatomic ideal gas in a thermally insulated container.
What is the change in the temperature (in $K$ ) of the ideal gas?
The standard reduction potentials for the two half-cells are given below.
$O _{2}(g)+4 H ^{+}(a q)+4 e^{-} \rightarrow 2 H _{2} O ( I ), E^{\circ}=1.23 \,V$
$2 H ^{+}(a q)+2 e^{-} \rightarrow H _{2}(g), E^{\circ}=0.00 \,V$
Use $F =96500 \,C\,mol ^{-1}, R=8.314\, J\,mol ^{-1} K ^{-1}$

Answer 1

$H_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow H_{2} O(l)$
$E_{\text {cell }}^{\circ}=1.23\, V$ (from given data)
$\because \Delta G^{\circ}=-n F E_{\text {cell }}^{\circ}$
$=-2 \times 96500 \times 1.23\, J / mol$
Work derived using $70 \%$ efficiency and on consumption of
$1.0 \times 10^{-3}$ mol of $H_{2}(g)$
$W=2 \times 96500 \times 1.23 \times 0.7 \times 1 \times 10^{-3}$
$=166.17\, J$
This work done $=$ Change in internal energy of monoatomic gas.
$\Rightarrow 166.17=n C_{v, m} \Delta T$
$\Rightarrow \Delta T=\frac{166.17 \times 2}{1 \times 3 R}$
$\Delta T=13.32\, K$