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Calculate the...

Question

Calculate the value of $E_{c e l l}$ at $298 K$ for the following cell:
$A l / A l^{3 +} (0.01 M) ∥ S n^{2 +} (0.015 M) / S n$
$E_{A l^{3 +} / A l}^{0} = - 1.66 v o l t$ and $E_{S n^{2 +} / S n}^{0} = - 0.14 v o l t$

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Solution

Given cell is:
$A l / A l^{3 +} (0.01 M) ∥ S n^{2 +} (0.015 M) / S n$
given, $E_{A l^{3 +} / A l}^{0} = - 1.66 V$ ,
$E_{S n^{2 +} / S n}^{0} = - 0.14 V$
$∴ E_{c e l l}^{0} = E_{R}^{0} - E_{L}^{0}$
$= - 0.14 - (- 1.66)$
$= - 0.14 + 1.66$
$= 1.52 V$
and net cell reaction is
$2 A . + 3 S n^{2 +} \to 2 A l^{3 +} + 3 S n$
so net cell reaction involves transfer of $6 m o l e$ of electrons
$∴ n = 6$
According to Nernst Equation, at $298 K$
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.059}{n} log \frac{[A l^{3 +}]^{2}}{[S n^{2 +}]^{3}}$
$= 1.52 - \frac{0.059}{6} log \frac{[0.01]^{2}}{[0.015]^{3}}$
$= 1.52 - 0.01447$
$E_{c e l l} = 1.50553 V$ .

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