For a cell involving one electron E_ cell ^ =0.59 V at 298 ;K, th… - Vidyadip

MCQ

For a cell involving one electron $E_{cell }^{\ominus} =0.59 \mathrm{V}$ at $298 \;\mathrm{K}$, the equilibrium constant for the cell reaction is

Given that $\left.\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{V} \text { at } \mathrm{T}=298 \mathrm{K}\right]$

A
$1.0 \times 10^{2}$
B
$1.0 \times 10^{5}$
✓
$1.0 \times 10^{10}$
D
$1.0 \times 10^{30}$

✓

Answer

Correct option: C.

$1.0 \times 10^{10}$

c
$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{\circ}-\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log _{10} \mathrm{Q}$

at equllbrium $\mathrm{E}_{\mathrm{cell}}=0, \mathrm{Q}=\mathrm{K}_{\mathrm{eq}}$

$0=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{0.0591}{1} \log _{10} \mathrm{K}_{\mathrm{eq}}$

$\mathrm{E}_{\text {cell }}^{\circ}=+0.0591 \log _{10} \mathrm{K}_{\mathrm{eq}}$

$0.59=+0.0591 \log _{10} \mathrm{K}_{\mathrm{eq} .}$

$+10=\log _{10} \mathrm{K}_{\mathrm{eq}}$

$\mathrm{K}_{\mathrm{eq}}=10^{+10}$

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