Question

The $E^0$ at ${25}^{0}C$ for the following reaction is 0.22V. Calculate the equilibrium constant at ${25}^0$C:$H_2\left(g\right)+2AgCl\left(s\right)⟶2Ag\left(s\right)+2HCl\left(aq\right)$

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

Answer 1

Answer: (A)

$2.8\times {10}^7$

We have,

$logK={\left(\frac{0.0591}{n}\right)}^{-1}{E}^{0}cell$

Now, for $H_2\left(g\right)+2AgCl\left(s\right)\to 2Ag\left(s\right)+2HCl\left(g\right)$

Here, $n=2$ (as, ${Ag}^{2+}+2e^{-}\to 2Ag$)

Now, $logK=\frac{2}{0.0591}\times 0.22$

$\Rightarrow K={10}^7\times 2.71$