First slide

Nernst Equation

Question

Calculate the equilibrium constant for the reaction,
$Zn (s) + {Cu}^{2 +} (aq) ⟶ {Zn}^{2 +} (aq) + Cu (s)$
$[Given, E_{cell}^{\circ} = 1 .1 V]$

Moderate

A

$2 \times 10^{32}$
B

$2 \times 10^{34}$
C

$2 \times 10^{37}$
D

$2 \times 10^{39}$

Solution

$Zn (s) + {Cu}^{2 +} (aq) ⟶ {Zn}^{2 +} (aq) + Cu (s)$
$0 = E_{cell} = E_{cell}^{\circ} - \frac{2 .303 RT}{2 F} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]}$
or $E_{cell}^{\circ} = \frac{2 .303 RT}{2 F} \log \frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]}$
But at equilibrium $\frac{[{Zn}^{2 +}]}{[{Cu}^{2 +}]} = K_{C}$
$\begin{array}{l} E_{cell}^{\circ} = \frac{0 .059 V}{2} \log K_{C} = 1 .1 V \\ \log K_{C} = \frac{1 .1 V \times 2}{0 .059 V} = 37 .288 \\ K_{C} = 2 \times 10^{37} at 298 K \end{array}$

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