Question

For the cell reaction$:4{\mathrm{Br}}^{-}+{\mathrm{O}}_2+4{\mathrm{H}}^{+}\rightleftharpoons 2{\mathrm{Br}}_2+2{\mathrm{H}}_2\mathrm{O};\mathrm{E}°=0.18\mathrm{V}$. The value of log $\left(\log \mathrm{KC}\right)$ at $298\mathrm{K}$ is $\left[\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.06\right]$

• A. $12$
• B. $6$
• C. $18$
• D. $13$

Answer 1

The correct option is A

$12$

The explanation for the correct answer:-

Option $\left(\mathrm{A}\right)$ $12$:

Step 1: Given data:-

Cell reaction- $4{\mathrm{Br}}^{-}+{\mathrm{O}}_2+4{\mathrm{H}}^{+}\rightleftharpoons 2{\mathrm{Br}}_2+2{\mathrm{H}}_2\mathrm{O}$

Value of $\mathrm{E}°=0.18\mathrm{V}$

Temperature $=298\mathrm{K}$

Value of $\left[\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.06\right]$

Step 2: Creating a balanced half-cell reaction:

Anode: $4{\mathrm{Br}}^{-}\to 2{\mathrm{Br}}_2+4{\mathrm{e}}^{-}$

Cathode$:{\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}\to 2{\mathrm{H}}_2\mathrm{O}$

Step 3: Applying cell equation:-

We know that,

$\mathrm{E}°=\frac{2.303\mathrm{RT}}{\mathrm{nF}\log {\mathrm{K}}_{\mathrm{c}}}$

$0.18\mathrm{V}=\frac{0.06}{4\log \mathrm{Kc}}$

$\log K_c=12$

Therefore, the option $\left(\mathrm{A}\right)$ is correct.