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Physics

EMF

The efficienc...

Question

The efficiency of a fuel cell is $80 %$ and the standard heat of reaction is $- 300 k J$ . The reaction involves two electrons in redox change. $E^{⊖}$ for the cell is:

1.24 V

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2.48 V

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0 V

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0.62 V

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Solution

The correct option is A $1.24 V$
$E f f i c i e n c y = \frac{Δ G^{⊖}}{Δ H^{⊖}} = - \frac{n E^{⊖} F}{Δ H} = 80$
$E^{⊖} = \frac{80 \times (- 300) \times 10^{3}}{2 \times 96500 \times 100} = 1.24 V$

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The efficiency of a fuel cell is 80% and the standard heat of reaction is −300kJ. The reaction involves two electrons in redox change. E⊖ for the cell is:

The correct option is A 1.24VEfficiency=ΔG⊖ΔH⊖=−nE⊖FΔH=80E⊖=80×(−300)×1032×96500×100=1.24V

The efficiency of a fuel cell is $80 %$ and the standard heat of reaction is $- 300 k J$ . The reaction involves two electrons in redox change. $E^{⊖}$ for the cell is:

The correct option is A $1.24 V$
$E f f i c i e n c y = \frac{Δ G^{⊖}}{Δ H^{⊖}} = - \frac{n E^{⊖} F}{Δ H} = 80$
$E^{⊖} = \frac{80 \times (- 300) \times 10^{3}}{2 \times 96500 \times 100} = 1.24 V$