What would be the stands cell potntial, if equilibrium constant of cell reactiom is $1.9413 \times 10^{37}$ at $25^{o} C$ temperature? $(n = 2)$

2.2 V

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1.1 V

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0.0085 V

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0.013 V

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Solution

The correct option is C $1.1 V$
We know,
$Δ G = - R T l n K_{f} = - n F E^{0}$
$l o g K_{f} = \frac{n F}{R T \times 2.303} \times E^{0}$
$l o g (1.9413 \times 10^{37}) = \frac{n}{0.059} E^{0}$
$n = 2$
$\frac{37.28 \times 0.059}{2} = E^{0}$
$E^{0} = 1.1 V$

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What would be the stands cell potntial, if equilibrium constant of cell reactiom is 1.9413×1037 at 25oC temperature? (n=2)

The correct option is C 1.1 VWe know, ΔG=−RTlnKf=−nFE0logKf=nFRT×2.303×E0log(1.9413×1037)=n0.059E0n=237.28×0.0592=E0E0=1.1V

What would be the stands cell potntial, if equilibrium constant of cell reactiom is $1.9413 \times 10^{37}$ at $25^{o} C$ temperature? $(n = 2)$

The correct option is C $1.1 V$
We know,
$Δ G = - R T l n K_{f} = - n F E^{0}$
$l o g K_{f} = \frac{n F}{R T \times 2.303} \times E^{0}$
$l o g (1.9413 \times 10^{37}) = \frac{n}{0.059} E^{0}$
$n = 2$
$\frac{37.28 \times 0.059}{2} = E^{0}$
$E^{0} = 1.1 V$