Question

The value of $\Delta S^{\ominus }$ for the fuel cell reaction at 298 K is:
$2H_2+O_2\to 2H_{2}O$
$E_{cell}^{\ominus }=1.23V$
${\Delta }_f{H}^{\ominus }\left(H_{2}O\right)=-285.8Kjmo{l}^{-1}$

• A. $-324.9J{K}^{-1}$
• B. $-214.9J{K}^{-1}$
• C. $-154.9J{K}^{-1}$
• D. $Noneofthese$

Answer 1

The correct option is A $-324.9J{K}^{-1}$

$2H_{2}O+O_2\to 2H_{2}O$,
$E_{cell}^{\ominus }=1.23V$,
${\Delta }_f{H}_{\left(H_{2}O\right)}^{\ominus }=-285.8kJmo{l}^{-1}$
Anode : $H_2\to 2H^{\oplus }+2e^{-}$
Cathode : $O_2+4H^{\oplus }+4e^{-}\to 2H_{2}O$
Number of electrons transferred = 4
$\Rightarrow G^{\ominus }=-4\times 96500\times (+1.23)=-474.78kJ$
$\Delta G^{\ominus }=\Delta H^{\ominus }-T\Delta S^{\ominus }$
$\Rightarrow \Delta S^{\ominus }\frac{2\times (-285.8)-(-474.78)}{298}kJ{K}^{-1}=-324.9J{K}^{-1}$