Question

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

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

Answer 1

Answer: (A)

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

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

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

$K _{ C }=?$

$\because E_{\text {cell }}^{\circ}=\frac{0.059}{n} \log\, K _{ c }$

$\therefore 0.0295=\frac{0.059}{2} \log\, K _{ c }$

$\therefore \log K _{ c }=1=\log \,10$

$\Rightarrow K _{ c }=10$