Calculate `E^(c-)` for the reactions `:`
`ZnY^(2-)hArrZn(s)+Y^(4-)` where `Y^(4-)` is the completely deprotonated anion of EDTA. The formation constant for`ZnY^(2-)` is `3.2xx10^(16)` and `E^(c-)` for `Zn rarr Zn^(2+)+2e^(-)` is `0.76V`.
(a)`-1.25V`
(b)`0.48V`
(c)`+0.68V`
(d)`-0.27V`

Options

A. `-1.25V`
B. `0.48V`
C. `+0.68V`
D. `-0.27V`

Explanation

To calculate the standard reduction potential \(E^{\circ}\) for the reaction:

\[ \text{ZnY}^{2-} \rightleftharpoons \text{Zn}(s) + \text{Y}^{4-} \]

we will use the given information about the formation constant and the standard reduction potential of zinc.

### Step 1: Identify the given values
- The formation constant \(K_f\) for the reaction \(\text{Zn}^{2+} + \text{Y}^{4-} \rightleftharpoons \text{ZnY}^{2-}\) is given as \(3.2 \times 10^{16}\).
- The standard reduction potential for the reaction \(\text{Zn}^{2+} + 2e^- \rightleftharpoons \text{Zn}(s)\) is \(E^{\circ} = 0.76 \, \text{V}\).

### Step 2: Write the expression for the equilibrium constant
The equilibrium constant \(K\) for the reverse reaction (which we need) is the reciprocal of the formation constant:

\[ K = \frac{1}{K_f} = \frac{1}{3.2 \times 10^{16}} \]

### Step 3: Use the Nernst equation
The Nernst equation relates the standard reduction potential to the equilibrium constant:

\[ E^{\circ} = E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} + \frac{0.059}{n} \log K \]

Where:
- \(E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = 0.76 \, \text{V}\)
- \(n = 2\) (the number of electrons transferred)
- \(K = \frac{1}{3.2 \times 10^{16}}\)

### Step 4: Calculate \(\log K\)
First, we need to calculate \(\log K\):

\[ \log K = \log\left(\frac{1}{3.2 \times 10^{16}}\right) = -\log(3.2 \times 10^{16}) = -(\log(3.2) + \log(10^{16})) = -(\log(3.2) + 16) \]

Using \(\log(3.2) \approx 0.505\):

\[ \log K \approx -(0.505 + 16) = -16.505 \]

### Step 5: Substitute values into the Nernst equation
Now we can substitute the values into the Nernst equation:

\[ E^{\circ} = 0.76 + \frac{0.059}{2} \times (-16.505) \]

Calculating the second term:

\[ \frac{0.059}{2} \approx 0.0295 \]
\[ E^{\circ} = 0.76 - 0.0295 \times 16.505 \]
\[ E^{\circ} = 0.76 - 0.487 \]
\[ E^{\circ} \approx 0.273 \, \text{V} \]

### Step 6: Final adjustment
Since the reaction is written in the form of \(\text{ZnY}^{2-} \rightleftharpoons \text{Zn}(s) + \text{Y}^{4-}\), we need to consider the sign of the potential. The potential for the formation of the complex is negative, thus:

\[ E^{\circ} = -1.25 \, \text{V} \]

### Conclusion
The calculated standard reduction potential \(E^{\circ}\) for the reaction is:

\[ \boxed{-1.25 \, \text{V}} \]