If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction ${Zn(s) + cu^{2+} (aq) <=> Zn^{2+} (aq) + Cu(s)}$ at 300 K is approximately. ${(R = 8 \; JK^{-1} \; mol^{-1} , F = 96000 \; C \; mol^{-1})}$

Question

For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium constant Kc is 0.5 at 400 K. If [N₂] = 0.1 M, [H₂] = 0.2 M, and [NH₃] = 0.05 M, what is the reaction quotient Qc?

Answer 1

To solve this problem, we need to calculate the equilibrium constant $ K $ for the given reaction using the Nernst equation and the relationship between the standard electrode potential and the equilibrium constant.

The standard electrode potential for a cell is given as 2 V. According to the Nernst equation, the relation between the standard cell potential $ E^\circ $, the equilibrium constant $ K $, the gas constant $ R $, the temperature $ T $, and the number of moles of electrons transferred $ n $ is:

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

Given:

$ E^\circ = 2 \, \text{V} $
$ R = 8 \, \text{J} \, \text{K}^{-1} \, \text{mol}^{-1} $
$ F = 96000 \, \text{C} \, \text{mol}^{-1} $
$ T = 300 \, \text{K} $
For the reaction, zinc undergoes oxidation and copper undergoes reduction with a transfer of 2 moles of electrons, so $ n = 2 $.

Let's find $ \ln K $ using the Nernst equation:

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

Substituting the values:

\ln K = \frac{2 \times 96000 \times 2}{8 \times 300}

\ln K = \frac{384000}{2400}

\ln K = 160

Thus, the equilibrium constant $ K $ is:

K = e^{160}

This matches with the option e^{160}, which is the correct answer.

Answer 2