Question

28. Use the standard reduction potenital to calculate equilibrium constants for the following reaction. H2O (1) H (aq) +OH (aq) at 298.15 K H2 (g)2H (aq) +2e 2H2O (aq) +2e - H2+ 20H (aq) A.K=1.0 x10 B. K=1.6 x10-1l C.K=1.0 x10-14 0 Define electrolvsis. E= 0V E -0.8277 V -10 spontaneous nonspontaneous nonspontaneous

Answer 1

Ans: Given half cell reactions:

H₂ (g) $\rightarrow$2H⁺ (aq) + 2e E^{$^{\circ}$}= 0 V

2H₂O (aq) + 2e $\rightarrow$ H₂ + 2OH^- (aq) E^{$^{\circ}$}= -0.8277 V

Overall reaction : H₂O (l) $\rightarrow$ H⁺ (aq) + OH^- (aq)

For the overall reaction, E^{$^{\circ}$}= 0 + (-0.8277)

E^{$^{\circ}$}= -0.8277 V

As the value of E^{$^{\circ}$}is negative, this means the reaction does not proceed in forward direction. Hence the reaction is non-spontaneous.

Calculation of value of K:

We know Nernst equation, E = E^{$^{\circ}$}- 2.303 * (RT/nF) * log K

At, T = 298.15 K, 2.303 * (RT/F) = 0.0591

Also, at equilibrium, E = 0 V, so we get,

0 = E^{$^{\circ}$}- (0.0591/n) * log K

In the overall reaction, number of electrons involved = 1. So, n = 1.

E^{$^{\circ}$}= 0.0591 * log K

-0.8277 = 0.0591* log K

-0.8277/0.0591 = log K

log K = -14.00

K = 10 ^ (-14.00) = 1.0 * 10^-14.

Hence, the answer is option C.