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Gibbs Free Energy & Spontaneity

In an electro...

Question

In an electrochemical cell, if $E^{o}$ is the emf of the cell involving $n$ mole of electrons, then $Δ G^{o}$ is:

Δ G^{o} = n F E^{o}

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Δ G^{o} = - n F E^{o}

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E^{o} = n F Δ G^{o}

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Δ G^{o} = n F / E^{o}

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Solution

The correct option is B $Δ G^{o} = - n F E^{o}$
Relation between ${Δ G}^{0}$ and $E M F$ of the cell is
${Δ G}^{0} = - n F E^{0}$
where, $n =$ number of electrons lost or gained
$F =$ Faraday's constant
$E^{0} =$ standard electrode potential
for spontaneous reaction, $Δ G < 0 \Rightarrow E^{0} > 0$
for non-spontaneous reaction, $Δ G > 0 \Rightarrow E^{0} > 0$

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