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Chemistry Question on Electrochemistry

For a cell reaction involving two electron changes, Ecell∘=0.3 V{E^{\circ}_{cell} = 0.3 \, V} at 25∘C 25^{\circ} C. The equilibrium constant of the reaction is

A

10−1010^{-10}

B

3×10−23 \times 10^{-2}

C

10

D

101010^{10}

Answer

101010^{10}

Explanation

Solution

log⁡ Kc=nE∘0.0591\log\, K_c = \frac{nE^{\circ}}{0.0591}
log⁡ Kc=2×0.30.0591\log \,K_c = \frac{2 \times 0.3}{0.0591}
log⁡ Kc=10.15\log \, K_c = 10.15
Kc=antilog 10.15K_c = \text{antilog} \, 10.15
Kc=1.419×1010K_c = 1.419 \times 10^{10}