Equilibrium Constant Calculation

The question asks us to find the equilibrium constant (\(K_{eq}\)) for the formation of the complex ion \(\rm Zn(NH_3)_4^{2+}\) at 25°C, given the standard electrode potentials (\(E^0\)) for two half-reactions involving zinc.

The formation reaction for the complex \(\rm Zn(NH_3)_4^{2+}\) from the simple ion \(\rm Zn^{2+}\) and ammonia (\(\rm NH_3\)) is:

\(\rm Zn^{2+}\text{(aq)} + 4NH_3\text{(aq)} \rightleftharpoons Zn(NH_3)_4^{2+}\text{(aq)}\)

We are given the following standard reduction potentials:

• \(\rm Zn^{2+}\text{(aq)} + 2e^- \rightleftharpoons Zn\text{(s)}\); \(E^0_1 = -0.763 \text{ V}\)
• \(\rm Zn(NH_3)_4^{2+}\text{(aq)} + 2e^- \rightleftharpoons Zn\text{(s)} + 4NH_3\text{(aq)}\); \(E^0_2 = -1.03 \text{ V}\)

To find the equilibrium constant for the formation reaction, we need to relate it to a standard cell potential (\(E^0_{cell}\)). The formation reaction can be obtained by combining the two given half-reactions. Specifically, we can reverse the second reaction and add the first reaction:

Reverse the second reaction (Oxidation half-reaction):

\(\rm Zn\text{(s)} + 4NH_3\text{(aq)} \rightleftharpoons Zn(NH_3)_4^{2+}\text{(aq)} + 2e^-\); \(E^0_{oxidation} = -E^0_2 = -(-1.03 \text{ V}) = +1.03 \text{ V}\)

Keep the first reaction (Reduction half-reaction):

\(\rm Zn^{2+}\text{(aq)} + 2e^- \rightleftharpoons Zn\text{(s)}\); \(E^0_{reduction} = E^0_1 = -0.763 \text{ V}\)

Adding these two half-reactions gives the overall reaction for complex formation:

\(\rm Zn^{2+}\text{(aq)} + \cancel{Zn\text{(s)}} + 4NH_3\text{(aq)} + \cancel{2e^-} \rightleftharpoons \cancel{Zn\text{(s)}} + Zn(NH_3)_4^{2+}\text{(aq)} + \cancel{2e^-}\)

Simplified overall reaction:

\(\rm Zn^{2+}\text{(aq)} + 4NH_3\text{(aq)} \rightleftharpoons Zn(NH_3)_4^{2+}\text{(aq)}\)

The standard cell potential for this overall reaction is calculated as:

\(E^0_{cell} = E^0_{reduction} - E^0_{oxidation}\)

\(E^0_{cell} = E^0_1 - E^0_2\)

\(E^0_{cell} = (-0.763 \text{ V}) - (-1.03 \text{ V})\)

\(E^0_{cell} = -0.763 \text{ V} + 1.03 \text{ V}\)

\(E^0_{cell} = 0.267 \text{ V}\)

The number of electrons transferred in this reaction is \(n=2\).

At 25°C (298 K), the standard cell potential is related to the equilibrium constant (\(K_{eq}\)) by the equation:

\(E^0_{cell} = \frac{0.0591}{n} \log K_{eq}\)

We can rearrange this equation to solve for \(\log K_{eq}\):

\(\log K_{eq} = \frac{n \times E^0_{cell}}{0.0591}\)

Now, substitute the values:

\(\log K_{eq} = \frac{2 \times 0.267 \text{ V}}{0.0591}\)

\(\log K_{eq} = \frac{0.534}{0.0591}\)

\(\log K_{eq} \approx 9.0355\)

To find \(K_{eq}\), we take the antilogarithm:

\(K_{eq} = 10^{9.0355}\)

\(K_{eq} = 10^{0.0355} \times 10^9\)

\(K_{eq} \approx 1.085 \times 10^9\)

Comparing this calculated value with the given options:

• 2 \(\times 10^8\)

• 5.5 \(\times 10^{10}\)

• 1 \(\times 10^9\)

• 6.6 \(\times 10^7\)

The calculated equilibrium constant, \(1.085 \times 10^9\), is closest to the option \(1 \times 10^9\).

The large value of \(K_{eq}\) indicates that the formation of the \(\rm Zn(NH_3)_4^{2+}\) complex is highly favorable at equilibrium.

Summary of Steps

1. Identify the chemical reaction for complex formation.
2. Determine the standard cell potential (\(E^0_{cell}\)) for this reaction using the given standard reduction potentials.
3. Use the Nernst equation relationship between \(E^0_{cell}\) and \(K_{eq}\) at 25°C.
4. Solve for the equilibrium constant \(K_{eq}\).
5. Compare the result with the given options.