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Nernst Equation

At 25oC, Ne...

Question

At $25^{o}$ C, Nernst equation is:

E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{n} l o g \frac{[i o n]_{R H S}}{[i o n]_{L H S}}

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E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{n} l o g \frac{[M]_{R H S}}{[M]_{L H S}}

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E_{c e l l} = E_{c e l l}^{o} + \frac{0.0591}{n} l o g \frac{[i o n]_{R H S}}{[i o n]_{L H S}}

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E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{n} l o g \frac{[i o n]_{L H S}}{[i o n]_{R H S}}

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Solution

The correct option is A $E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{n} l o g \frac{[i o n]_{R H S}}{[i o n]_{L H S}}$
Nernst eauation is given by,
$E_{c e l l} = E_{c e l l}^{0} - \frac{R T}{n F} l n Q$
At $T = 25^{0} C$ , on substituting values of R is 8.314 / mol. K and F =96485 C mol^-¹
^{We get,}
^{$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{n} l o g Q$}
^{where, $Q = \frac{i o n_{R H S}}{i o n_{L H S}}$}
$E_{c e l l}$ = $E_{c e l l}^{0}$ - $\frac{0.0591}{n}$ $l o g \frac{i o n_{R H S}}{i o n_{L H S}}$

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