KCET · Chemistry · Thermodynamics (C)
For hydrogen - oxygen fuel cell at one atm and \( 298 \mathrm{~K} \)
\( \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{H}_{2} \mathrm{O}(l) ; \Delta \mathrm{G}^{\circ}=-240 \mathrm{~kJ} \)
\( \mathrm{E}^{\circ} \) for the cell is approximately,
(Given \( \mathrm{F}=96,500 \mathrm{C} \) )
- A \( 1.24 \mathrm{~V} \)
- B \( 1.26 \mathrm{~V} \)
- C \( 2.48 \mathrm{~V} \)
- D \( 2.5 \mathrm{~V} \)
Answer & Solution
Correct Answer
(A) \( 1.24 \mathrm{~V} \)
Step-by-step Solution
Detailed explanation
At Anode : \(2 \mathrm{H}_{2}(g)+4 \mathrm{OH}^{-}(a q) \rightarrow 4 \mathrm{H}_{2} \mathrm{O}(l)+4 e^{-}\)
At Cathode : \(\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)+4 e^{-} \rightarrow 4 \mathrm{OH}^{-}(a q)\)
\(\quad\) Overall : \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(1) ; \Delta G^{\circ}\)
\(\quad-240 \times 2=-480 \mathrm{~kJ}=-480000 \mathrm{~J}\)
Now, \(\Delta G^{\circ}=-n F E_{\text {cell }}{ }^{\circ}\)
\(\Rightarrow-480 \mathrm{~kJ}=-4 \times 96500 E_{\text {cell }}\)
\(E_{\text {cell }}^{\circ}=\frac{-480000}{-4 \times 96500}=1.24 \mathrm{~V}\)
At Cathode : \(\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)+4 e^{-} \rightarrow 4 \mathrm{OH}^{-}(a q)\)
\(\quad\) Overall : \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(1) ; \Delta G^{\circ}\)
\(\quad-240 \times 2=-480 \mathrm{~kJ}=-480000 \mathrm{~J}\)
Now, \(\Delta G^{\circ}=-n F E_{\text {cell }}{ }^{\circ}\)
\(\Rightarrow-480 \mathrm{~kJ}=-4 \times 96500 E_{\text {cell }}\)
\(E_{\text {cell }}^{\circ}=\frac{-480000}{-4 \times 96500}=1.24 \mathrm{~V}\)
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