If the EMF of a cell :

$M | M^{2 +} (0.02 M) | | H^{+} (0.1 M) | H_{2}$ (1 atm)

at $25^{\circ} C$ is 0.81V. Find the standard reduction potential of $M$ .
(Write a value to the nearest integer. If an answer is negative then write its magnitude)

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Solution

From Nearst's equation-
$E_{c e l l} = E_{c e l l}^{o} - \frac{0.0593}{n} l o g (\frac{p_{H_{2}} \times [M^{2 +}]}{[H^{+}]^{2}})$

$0.81 = E_{c}^{⊖} - E_{a}^{⊖} - \frac{0.059}{2} l o g \frac{(0.02)}{(0.1)^{2}}$ .
$\Rightarrow E_{a}^{⊖} = 0 - 0.81 - 0.0089 = - 0.8189 V$
$E_{M^{2 +} | M}^{⊖} = - 0.8189 V \approx - 1 V$

So, the correct answer is $| - 1 | = 1$ .
Ans $1.$

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