Question

For a cell involving one electron $E_{cell}^o=0.59Vat298K,$ the equilibrium constant for cell reaction is :

$[\text{Given that}\frac{2.303RT}{F}=0.059VatT=298K]$

(NEET-2019)

• A. $1.0\times {10}^2$
• B. $1.0\times {10}^5$
• C. $1.0\times {10}^{30}$
• D. $1.0\times {10}^{10}$

Answer 1

The correct option is D $1.0\times {10}^{10}$

$E_{cell}=E_{cell}^0-\frac{0.059}{n}\log Q.......\left(i\right)$

$(\text{At equilibrium,}Q=K_{eq}and{E}_{cell}=0)$

From equation (i)
$0=E_{cell}^0-\frac{0.059}{1}\log K_{eq}$

$\log K_{eq}=\frac{E_{cell}^0}{0.059}=\frac{0.59}{0.059}=10$

$K_{eq}={10}^{10}=1\times {10}^{10}$