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

Question

Alkyl halide and aryl halide react in the presence of sodium metal and dry ether. The name of the reaction is?

Answer 1

To solve this problem, we need to calculate the equilibrium constant (K_{eq}) for the given cell reaction at 298 K. The relationship between the standard cell potential (E^{\circ}_{cell}) and the equilibrium constant is given by the Nernst equation for standard conditions:

E^{\circ}_{cell} = \frac{2.303 \, RT}{nF} \log K_{eq}

In this problem:

E^{\circ}_{cell} = 0.59 \, \text{V}
\frac{2.303 \, RT}{F} = 0.059 \, \text{V} (given)
n = 1 (since it involves one electron)

Substitute the given values into the Nernst equation:

0.59 = 0.059 \log K_{eq}

To find \log K_{eq}, divide both sides by 0.059:

\log K_{eq} = \frac{0.59}{0.059} = 10

Now, solve for K_{eq} by taking the antilogarithm (base 10):

K_{eq} = 10^{10}

Therefore, the equilibrium constant for the cell reaction is 1.0 \times 10^{10}, which corresponds to the correct answer from the options given.

Correct Answer: 1.0 \times 10^{10}

Conclusion: The relationship between the Nernst equation and equilibrium constant underpins electrochemical reactions, helping us to deduce the extent to which a reaction proceeds to completion. A high equilibrium constant like 10^{10} indicates a strong tendency towards completion or product formation in the cell reaction.

Answer 2