Question

The cell in which the following reaction occurs:
$2F{e}_{\left(aq\right)}^{3+}+2I_{\left(aq\right)}^{-}\to 2F{e}_{\left(aq\right)}^{2+}+I_{2\left(s\right)}$ has $E_{cell}^o=0.236V$ at $298K$.
The equilibrium constant of the cell reaction is:

• A. $6.69\times {10}^{-7}$
• B. $7.69\times {10}^{-7}$
• C. $9.69\times {10}^7$
• D. $6.69\times {10}^7$

Answer 1

The correct option is C $9.69\times {10}^7$

We know

$\log K_c=\frac{nF{E}_{cell}^0}{2.303RT}$

where,

$n=2$

$F=96487$

$E_{cell}^0=0.236V$

$R=8.31$

$T=298$

Substituting the values, we get

$\log K_c=\frac{2\times 96487\times 0.236}{2.303\times 8.31\times 298}$

$\log K_c=7.9854$

$K_c=antilog\left(7.9854\right)$

$K_c=9.69\times {10}^7$