Question

The $E^{\circ }$ at ${25}^{\circ }C$ for the following reaction is $0.22V$. Calculate the equilibrium constant at ${25}^{\circ }C$:
$H_2\left(g\right)+2AgCl\left(s\right)\to 2Ag\left(s\right)+2HCl\left(aq\right)$

• A. $5.2\times {10}^8$
• B. $5.2\times {10}^6$
• C. $5.2\times {10}^3$
• D. $2.8\times {10}^7$

Answer 1

The correct option is D $2.8\times {10}^7$

Nernst equation is represented as,

$E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{2}log{Q}_c....\left(1\right)$


So, here the value of $n_{factor}=1+1=2$

Given that $E_{cell}^{\circ }=0.22V$

At equilibrium
$E_{cell}=0,Q_C=K$

From (1) equation, we get

$E_{cell}^{\circ }=\frac{0.0591}{2}logK$
$\Rightarrow K=antilog\left(\frac{2\times E_{cell}^{\circ }}{0.0591}\right)$
$\Rightarrow K=antilog\left(\frac{2\times 0.22}{0.0591}\right)$
$\Rightarrow K=antilog\left(7.44\right)$
$\Rightarrow K=2.8\times {10}^7$