Question

The standard e.m.f. of a galvanic cell involving cell reaction with n = 2 is found to be 0.295 V at 25ºC. The equilibrium constant of the reaction would be

• A. 4.0 × 10¹²
• B. 1.0 × 10²
• C. 1.0 × 10¹⁰
• D. 2.0 × 10¹¹

Answer 1

The Correct Option is C

To find the equilibrium constant of the reaction, we use the Nernst equation which relates the standard e.m.f. (electromotive force) of a galvanic cell to the equilibrium constant \( K \) of the reaction it undergoes. The equation is given by:

E^\circ = \frac{RT}{nF} \ln K

Where:

• E^\circ is the standard e.m.f. (0.295 V)
• R is the universal gas constant (8.314 J/mol·K)
• T is the temperature in Kelvin (298 K, at 25°C)
• n is the number of moles of electrons transferred in the reaction (2)
• F is Faraday's constant (96485 C/mol)
• K is the equilibrium constant we need to find.

We first convert the natural logarithm to base 10 logarithm using the relation:

\ln K = 2.303 \log K

Substituting the known values into the Nernst equation gives:

0.295 = \frac{8.314 \times 298}{2 \times 96485} \times 2.303 \log K

Simplifying the equation:

0.295 = \frac{688.372}{192970} \times 2.303 \log K

0.295 = 0.00711 \times 2.303 \log K

0.295 = 0.016378 \log K

Solving for \log K gives:

\log K = \frac{0.295}{0.016378} \approx 18.006

Taking the antilogarithm to find K:

K \approx 10^{18.006} \approx 1.0 \times 10^{10}

Thus, the equilibrium constant of the reaction is 1.0 \times 10^{10}. Therefore, the correct answer is 1.0 \times 10^{10}.