Chemistry - Electrochemistry Question with Solution | TestHub

We know:

$\begin{array}{rlr}2H_2(g)⟶4H^{+}+4e^{-}& E=0.0& \end{array}$

$O_2(g)+4H^{+}(aq)+4e^{-}⟶2H_{2}O(l)E=1.23V$

$\begin{array}{rlrlr}& 2H_2(g)+O_2(g)⟶2H_{2}O(l)E_{\text{cell }}^{\circ }=1.23V& & & \end{array}$

$\Delta G_{298}^{\circ }=-nF{E}^{\circ }=-4\times 96500\times 1.23=-474.78kJ$

$\Delta G_{298}^{\circ }=\Delta H_{298}^{\circ }-T\cdot \Delta S_{298}^o$

$\Delta S_{298}^{\circ }=\frac{2\times (-285.5)-(-475)}{298}=-0.322kJ/K]$