Question

Consider a $70\%$ efficient hydrogen-oxygen fuel cell working under standard conditions at $1$ bar and $298K$. Its cell reaction is
$H_2(g)+\frac{1}{2}{O}_2(g)\to H_{2}O(l)$
The work derived from the cell on the consumption of $1.0\times 10^{-3}mol$ of $H_2(g)$ is used to compress $1.00mol$ 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.
$O_2(g)+4H^{+}(aq)+4e^{-}\to 2H_{2}O(I),E^{\circ }=1.23V$
$2H^{+}(aq)+2e^{-}\to H_2(g),E^{\circ }=0.00V$
Use $F=96500Cmo{l}^{-1},R=8.314Jmo{l}^{-1}{K}^{-1}$

Answer 1

Answer: 13.32

$H_2(g)+\frac{1}{2}{O}_2(g)\to H_{2}O(l)$
$E_{\text{cell }}^{\circ }=1.23V$ (from given data)
$\because \Delta G^{\circ }=-nF{E}_{\text{cell }}^{\circ }$
$=-2\times 96500\times 1.23J/mol$
Work derived using $70\%$ efficiency and on consumption of
$1.0\times 10^{-3}$ mol of $H_2(g)$
$W=2\times 96500\times 1.23\times 0.7\times 1\times 10^{-3}$
$=166.17J$
This work done $=$ Change in internal energy of monoatomic gas.
$\Rightarrow 166.17=n{C}_{v,m}\Delta T$
$\Rightarrow \Delta T=\frac{166.17\times 2}{1\times 3R}$
$\Delta T=13.32K$