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Chemistry Question on Electrochemistry

For a cell reaction involving two electron change, the standard e.m.f. of the cell is found to be 0.295 Vat 298 K. The equilibrium constant of the reaction at the same temperature will be

A

2.95×1022.95 \times 10^2

B

10

C

1×10101 \times 10^{10}

D

1×10−101 \times 10^{-10}

Answer

1×10101 \times 10^{10}

Explanation

Solution

Ecell∘=0.295VE^{^\circ}_{cell} = 0.295 V
T=298 KT =298 \,K
Ecell∘=2.303RTnFlog⁡KcE^{^{\circ} }_{cell} = \frac{2.303 RT}{nF} \log K_{c}
log⁡Kc=Ecell∘×nF2.303RT=0.295×2×965002.303×8.314×298\log K_{c} = \frac{E^{^{\circ} }_{cell} \times nF}{2.303 RT} = \frac{0.295 \times2\times96500}{2.303 \times 8.314 \times 298}
log⁡Kc=9.978\log K_{c} = 9.978
Kc=9.51×109K_{c} = 9.51 \times 10^{9}
=0.95×1010≈1×1010= 0.95 \times 10^{10} \approx 1\times 10^{10}