Question

For a cell involving one electron \[\ce{E^{\circ}_{cell}}\] = 0.59 V at 298 K, the equilibrium constant for the cell reaction is:

[Given that \[\ce{\frac{2.303 RT}{F}}\] = 0.059 V at T = 298 K]

Options

• 1.0 × 10²

• 1.0 × 10⁵

• 1.0 × 10¹⁰

• 1.0 × 10³⁰

Answer 1

1.0 × 10¹⁰

Explanation:

From the Nernst equation at equilibrium (E = 0):

\[\ce{E^{\circ}_{cell} = \frac{0.059}{n} log K}\]    ...(i)

Where:

n = number of electrons transferred

K = equilibrium constant

\[\ce{E^{\circ}_{cell}}\] = 0.59 V

n = 1

Then the equation (i) becomes:

\[\ce{E^{\circ}_{cell} = \frac{0.059}{1} log K}\]

0.59 = 0.059 log K

log K = \[\ce{\frac{0.59}{0.059}}\]

log k = 10

k = 10¹⁰

∴ The correct answer is 1.0 × 10¹⁰.