Question

For a cell involving one electron \(E^\ominus cell=0.59V\) at 298 K, the equilibrium constant for the cell reaction is: [Given that \(\frac{ 2.303 RT }{F}\) \(=0.059 V\) at\( T 298 K \)] 

• A. \(1.0\times 10^{2}\)
• B. \(1.0\times 10^{5}\)
• C. \(1.0\times 10^{10}\)
• D. \(1.0\times 10^{30}\)

Answer 1

The Correct Option is C

We need to calculate the equilibrium constant (\(K\)) for the given cell reaction using the Nernst equation at standard conditions. The Nernst equation relates the cell potential at any point in time to the equilibrium constant and is given by:

E = E^\ominus - \frac{RT}{nF} \ln K

At equilibrium, the cell potential \(E\) is zero, so we can rearrange the equation as:

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

Given:

• \(E^\ominus_{\text{cell}} = 0.59 \, \text{V}\)
• Number of electrons involved, \(n = 1\)
• \(\frac{2.303RT}{F} = 0.059 \, \text{V}\) at \(T = 298 \, \text{K}\)

We substitute the given values into the rearranged equation:

0.59 = \frac{0.059}{1} \log K

Solving for \(\log K\):

\log K = \frac{0.59}{0.059} = 10

Therefore, the equilibrium constant \(K\) is calculated as follows:

K = 10^{10}

The correct answer is:

1.0\times 10^{10}