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Electrolytic Cell

A galvanic ce...

Question

A galvanic cell is formed by immersing $A g (s)$ in $A g N O_{3} (a q, 0.1 M)$ and $C u (s)$ in $C u S O_{4} (a q, 4 M)$ at $298 K$
Calculate the EMF of the cell.
Given:
$E_{A g^{+} / A g}^{0} = 0.80 V$
$E_{C u^{2 +} / C u}^{0} = 0.34 V$

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Solution

Standard reduction potential of $A g^{+} / A g$ couple has value higher value, hence, it will act as cathode and $C u^{2 +} / C u$ will act as anode.

Thus, the cell representation will be,
$C u (s) ∣ C u^{2 +} (a q, 4.0 M) ∥ A g^{+} (a q, 0.1 M) ∣ A g (s)$
Cell reaction is
$C u + 2 A g^{+} \to C u^{2 +} + 2 A g$
Here $n = 2$ ;
$E = E_{c e l l}^{0} - \frac{0.059}{2} l o g \frac{[C u^{2 +}]}{[A g^{+}]^{2}}$
$E_{c e l l}^{0} = E_{c a t h o d e}^{0} - E_{a n o d e}^{0}$
$E_{c e l l}^{0} = 0.80 - 0.34$
$E_{c e l l}^{0} = 0.46 V$

By Nernst equation,
$E_{c e l l} = 0.46 - \frac{0.059}{2} l o g \frac{[4.0]}{{[0.1]}^{2}}$
$E_{c e l l} = 0.46 - \frac{0.059}{2} l o g (4 \times 10^{2})$
$E_{c e l l} = 0.46 - \frac{0.059}{2} \times 0.602 - 0.059$
$E_{c e l l} = 0.46 - 0.018 - 0.059$
$E_{c e l l} = 0.46 - 0.077$
$E_{c e l l} = 0.383 V$

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