Question

The standard emf of galvanic cell involving cell reaction with n=4 is found to be 0.295V at 25°C. The equilibrium constant of the reaction would be

Answer 1

1.0 x 10^20

Answer 2

The relationship between the standard emf of a galvanic cell ($E^\circ_{cell}$) and the equilibrium constant (K) of the cell reaction at 25°C is given by the Nernst equation at equilibrium:

$E^\circ_{cell} = \frac{2.303 RT}{nF} \log K$

At 25°C (298 K), the value of $\frac{2.303 RT}{F}$ is approximately 0.0591 V. So, the equation simplifies to:

$E^\circ_{cell} = \frac{0.0591}{n} \log K$

Given values: Standard emf ($E^\circ_{cell}$) = 0.295 V Number of electrons transferred (n) = 4 Temperature = 25°C

Substitute the given values into the equation: $0.295 = \frac{0.0591}{4} \log K$

Rearrange the equation to solve for $\log K$: $\log K = \frac{0.295 \times 4}{0.0591}$ $\log K = \frac{1.18}{0.0591}$

Perform the division: $\frac{1.18}{0.0591} \approx 19.966$

For practical purposes in such problems, this value is usually rounded to the nearest integer, which is 20. So, $\log K = 20$

To find K, take the antilog: $K = 10^{20}$

The equilibrium constant of the reaction is $10^{20}$.