calculate the value of equilibrium constant `(K_f)` for the reaction:
`Zn^(2+)(aq)+4OH^(-)(aq)iffZn(OH)_4^(2-)(aq)`
Given: `Zn^(2+)(aq)+2e^(-)toZn(s0,E^(@)=-0.76 V`
`Zn(OH)_4^(2-)(aq)+2e^(-)toZn(s)+4OH^(-)(aq),E^(@)=-1.36 V`
`2.303(RT)/F=0.06`

Options

A. `10^10`
B. `2xx10^10`
C. `10^20`
D. None of these

Answer

To calculate the value of the equilibrium constant \( K_f \) for the reaction:

\[ \text{Zn}^{2+}(aq) + 4\text{OH}^-(aq) \iff \text{Zn(OH)}_4^{2-}(aq) \]

we will follow these steps:

### Step 1: Write the half-reactions and their standard reduction potentials.

We have the following half-reactions:

1. \( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}(s) \) with \( E^\circ = -0.76 \, \text{V} \)
2. \( \text{Zn(OH)}_4^{2-} + 2e^- \rightarrow \text{Zn}(s) + 4\text{OH}^- \) with \( E^\circ = -1.36 \, \text{V} \)

### Step 2: Reverse the first half-reaction.

To align with the overall reaction, we need to reverse the first half-reaction:

\[ \text{Zn}(s) + 4\text{OH}^- \rightarrow \text{Zn(OH)}_4^{2-} + 2e^- \]

When we reverse the reaction, the sign of the standard potential changes:

\[ E^\circ = +0.76 \, \text{V} \]

### Step 3: Combine the half-reactions.

Now we can add the two half-reactions together:

1. \( \text{Zn}(s) + 4\text{OH}^- \rightarrow \text{Zn(OH)}_4^{2-} + 2e^- \) (oxidation, \( E^\circ = +0.76 \, \text{V} \))
2. \( \text{Zn(OH)}_4^{2-} + 2e^- \rightarrow \text{Zn}(s) + 4\text{OH}^- \) (reduction, \( E^\circ = -1.36 \, \text{V} \))

Adding these gives:

\[ \text{Zn}^{2+} + 4\text{OH}^- \rightarrow \text{Zn(OH)}_4^{2-} \]

### Step 4: Calculate the cell potential \( E^\circ_{cell} \).

The overall standard cell potential \( E^\circ_{cell} \) is calculated as follows:

\[ E^\circ_{cell} = E^\circ_{reduction} + E^\circ_{oxidation} \]

Substituting the values:

\[ E^\circ_{cell} = (-1.36) + (+0.76) = -0.60 \, \text{V} \]

### Step 5: Relate \( E^\circ_{cell} \) to the equilibrium constant \( K_f \).

Using the Nernst equation at equilibrium, we have:

\[ E^\circ_{cell} = \frac{RT}{nF} \ln K_f \]

Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T \) is the temperature in Kelvin (assume 298 K for standard conditions)
- \( n = 2 \) (number of electrons transferred)
- \( F = 96485 \, \text{C/mol} \)

Given that \( \frac{2.303RT}{F} = 0.06 \), we can rewrite the equation as:

\[ E^\circ_{cell} = -0.60 = \frac{0.06}{2} \ln K_f \]

### Step 6: Solve for \( K_f \).

Rearranging gives:

\[ \ln K_f = \frac{-0.60}{0.06} \cdot 2 \]

Calculating this:

\[ \ln K_f = -10 \]

Exponentiating both sides:

\[ K_f = e^{-10} \]

Using the approximation \( e^{-10} \approx 10^{-20} \):

\[ K_f \approx 10^{20} \]

### Final Answer:

Thus, the equilibrium constant \( K_f \) for the reaction is:

\[ K_f \approx 10^{20} \]