Question

The equilibrium constant of the reaction:
$A_{(s)}+2B_{(aq)}^{+}\leftrightharpoons A_{(aq)}^{2+}+2B_{(s)};E_{cell}^{\circ }=0.0295V$ is
$[\frac{2.303RT}{F}=0.059]$

• A. $10$
• B. $2\times 10^2$
• C. $3\times 10^2$
• D. $2\times 10^5$

Answer 1

Answer: (A)

$A(s)+2B^{+}(aq)\rightleftharpoons A^{2+}(aq)+2B(s)$

Here, $n=$ number of $e^{-}$ transfer= 2

$E^{\circ }$ cell $=0.295V$

$K_C=?$

$\because E_{\text{cell }}^{\circ }=\frac{0.059}{n}\log K_c$

$\therefore 0.0295=\frac{0.059}{2}\log K_c$

$\therefore \log K_c=1=\log 10$

$\Rightarrow K_c=10$