`Zn(s)|Zn(CN)_4^(2-)(0.5M),CN^(-)(0.01)"||"Cu(NH_3)_4^(2+)(0.5M),NH_3(1M)|Cu(s)` Given: `K_f "of"Zn(CN)_4^(-2)=10^(16)` , `K_f"of"Cu(NH_3)_4^(2+)=10^(12)`,
`E_(Zn|Zn^(2+))^(@)=0.76V,E_(Cu^(+2)|Cu)^(@)=0.34V , (2.303RT)/F=0.06`
The emf of above cell is :

Options

A. 1.22V
B. 1.10V
C. 0.98V
D. None of these

Answer

To find the emf of the given electrochemical cell, we will follow these steps:

### Step 1: Identify the half-reactions
The cell consists of two half-reactions:
1. At the anode (oxidation):
\[ \text{Zn(s)} \rightarrow \text{Zn}^{2+} + 2e^- \]
2. At the cathode (reduction):
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu(s)} \]

### Step 2: Write the equilibrium expressions for the complexes
The formation constants for the complexes are given:
- For \( \text{Zn(CN)}_4^{2-} \):
\[ K_f = \frac{[\text{Zn(CN)}_4^{2-}]}{[\text{Zn}^{2+}][\text{CN}^-]^4} = 10^{16} \]
- For \( \text{Cu(NH}_3)_4^{2+} \):
\[ K_f = \frac{[\text{Cu(NH}_3)_4^{2+}]}{[\text{Cu}^{2+}][\text{NH}_3]^4} = 10^{12} \]

### Step 3: Calculate the concentration of \( \text{Zn}^{2+} \)
Given that the concentration of \( \text{Zn(CN)}_4^{2-} \) is 0.5 M and \( [\text{CN}^-] = 0.01 \) M:
\[ 10^{16} = \frac{0.5}{[\text{Zn}^{2+}](0.01)^4} \]
Solving for \( [\text{Zn}^{2+}] \):
\[ [\text{Zn}^{2+}] = \frac{0.5}{10^{16} \times (0.01)^4} = \frac{0.5}{10^{16} \times 10^{-8}} = 5 \times 10^{-9} \text{ M} \]

### Step 4: Calculate the concentration of \( \text{Cu}^{2+} \)
Given that the concentration of \( \text{Cu(NH}_3)_4^{2+} \) is 0.5 M and \( [\text{NH}_3] = 1 \) M:
\[ 10^{12} = \frac{0.5}{[\text{Cu}^{2+}](1)^4} \]
Solving for \( [\text{Cu}^{2+}] \):
\[ [\text{Cu}^{2+}] = \frac{0.5}{10^{12}} = 5 \times 10^{-13} \text{ M} \]

### Step 5: Calculate the standard cell potential \( E^0_{\text{cell}} \)
Using the standard reduction potentials:
- \( E^0_{\text{Zn}^{2+}/\text{Zn}} = 0.76 \, \text{V} \)
- \( E^0_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \, \text{V} \)

The standard cell potential is calculated as:
\[ E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} = 0.34 - 0.76 = -0.42 \, \text{V} \]

### Step 6: Use the Nernst equation to find the cell potential
The Nernst equation is:
\[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.06}{n} \log Q \]
Where \( n = 2 \) (number of electrons transferred) and \( Q \) is the reaction quotient:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{5 \times 10^{-9}}{5 \times 10^{-13}} = 10^4 \]

Substituting into the Nernst equation:
\[ E_{\text{cell}} = -0.42 - \frac{0.06}{2} \log(10^4) \]
\[ E_{\text{cell}} = -0.42 - 0.03 \times 4 \]
\[ E_{\text{cell}} = -0.42 - 0.12 = -0.54 \, \text{V} \]

### Final Calculation
Since the calculated value is negative, we need to check the signs and values used. The correct calculation should yield:
\[ E_{\text{cell}} = 1.1 - 0.12 = 0.98 \, \text{V} \]

### Conclusion
The emf of the cell is:
\[ \boxed{0.98 \, \text{V}} \]