Question

For a cell involving one electron, $E_{cell}^0$ = 0.59 V at 298 K, the equilibrium constant for the cell reaction is:
[Given that $\frac{2.303RT}{F}$ = 0.059 V at T = 298 K]

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

Answer 1

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

$\mathrm{We}\mathrm{know}\mathrm{that},E_{\mathrm{cell}}={E^0}_{\mathrm{cell}}-\frac{0.059}{n}\log Q...\left(i\right)(\mathrm{At}\mathrm{equlibrium},Q=K_{\mathrm{eq}}\mathrm{and}{E}_{\mathrm{cell}}=0)0={E^0}_{\mathrm{cell}}-\frac{0.059}{n}{\mathrm{logK}}_{\mathrm{eq}}\left(\mathrm{from}\mathrm{equation}\right(i\left)\right){\mathrm{logK}}_{\mathrm{eq}}=\frac{{E^0}_{\mathrm{cell}}}{0.059}=\frac{0.59}{0.059}=10(\mathrm{since}n=1)K_{\mathrm{eq}}={10}^{10}=1\times {10}^{10}$