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

The problem requires finding the equilibrium constant of a cell reaction, given the standard electromotive force (\(E^\ominus_{cell}\)) of the cell. To solve this, we will use the Nernst equation, which relates the equilibrium constant to the electromotive force.

The Nernst equation is given by:

\[ E_{cell} = E^\ominus_{cell} - \frac{RT}{nF} \ln K \]

At equilibrium, \(E_{cell} = 0\), so the equation simplifies to:

\[ 0 = E^\ominus_{cell} - \frac{RT}{nF} \ln K \]

Rearrange to solve for \(K\):

\[ \ln K = \frac{nF}{RT} E^\ominus_{cell} \]

Convert the logarithm from natural log to base 10:

\[ \log K = \frac{nF}{2.303RT} E^\ominus_{cell} \]

Given that \(\frac{2.303RT}{F} = 0.059 \, \text{V}\) at \(T = 298 \, \text{K}\), and \(n = 1\) (one electron involved), substitute these into the equation:

\[ \log K = \frac{E^\ominus_{cell}}{0.059} \]

Plug in the given \(E^\ominus_{cell} = 0.59 \, \text{V}\):

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

Now, convert the logarithm to an exponential form to find \(K\):

\[ K = 10^{10} \]

Thus, the equilibrium constant for the cell reaction is \(1.0 \times 10^{10}\).