Question

Consider a $70\%$ efficient hydrogen-oxygen fuel cell working under standard conditions at $1\text{bar}$ and $298K$. Its cell reaction is

$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right)$
The work derived from the cell on the consumption of $1.0\times {10}^{-3}\text{mol}$ of $H_2\left(g\right)$ is used to compress $1.00\text{mol}$ 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.
$O_2\left(g\right)+4H^{+}\left(aq\right)+4e^{-}\to 2H_{2}O\left(l\right),E^o=1.23V2H^{+}\left(aq\right)+2e^{-}\to H_2\left(g\right),E^o=0.00V$
Use $F=96500Cmo{l}^{-1},R=8.314Jmo{l}^{-1}{K}^{-1}$

• A. 13.32
• B. 13.324
• C. 13.3

Answer 1

Answer: (A), (B), (C)

$E_{cell}^o=1.23\text{Volt}\Delta G^o=-nF{E}_{cell}^o\Delta G^o=-2\times 96500\times 1.23\text{J}$
Since, work derived from this fuel cell =$70/100\times (-\Delta G_{cell}^o)\times 1.0\times {10}^{-3}=x\text{J}$
Since, insulated vessel, hence $q=0$
From equation, for monoatomic gas,
$w=\Delta U\Rightarrow x=n{C}_V,m\Delta T(C_V,m=\frac{3R}{2})70/100\times 2\times 96500\times 1.23\times {10}^{-3}=1\times 3/2\times 8.314\times \Delta T$
$\Delta T=13.32$