Given: $M n^{2 +} + 2 e^{-} \to M n; E^{\circ} = - 1.20 V$
$Z n^{2 +} + 2 e^{-} \to Z n; E^{\circ} = - 0.80 V$
What is the expression for the equilibrium constant of the spontaneous oxidation-reduction reaction?

K_{e q} = 0.4

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K_{e q} = 0.8

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K_{e q} = 10^{0.4 / 059}

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K_{e q} = 10^{0.8 / 059}

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Solution

The correct option is D $K_{e q} = 10^{0.8 / 059}$
At equilibrium,

$E_{c e l l} = 0$

Now,

$E_{c e l l}^{o} = \frac{0.059}{2} l o g K_{e q}$

$E_{c e l l}^{o} = E_{C}^{o} - E_{A}^{o}$

$= - 0.80 - (- 1.20)$

$= 0.40 V$

$0.40 = \frac{0.059}{2} l o g K_{e q}$

$K_{e q} = 10^{0.8 / 0.059}$

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For a spontaneous reaction the $Δ G$ , equilibrium constant $(K_{e q})$ and $E_{c e l l}^{0}$ will be respectively

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K_{e q} = 10^{25}

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