Estimate the cell potential of a Daniel cell having `1.0Zn^(++)` and originally having `1.0M Cu^(++)` after sufficient `NH_(3)` has been added to the cathode compartment to make `NH_(3)` concentration `2.0M`. Given `K_(f)` for `[Cu(NH_(4))_(4)]^(2+) = 1xx 10^(12), E^(@)` for the reaction, `Zn +Cu^(2+) rarr Zn^(2+) +Cu` is `1.1V`

Explanation

To estimate the cell potential of a Daniel cell with the given conditions, we will follow these steps:

### Step 1: Identify the given data
- Concentration of Zn²⁺ = 1.0 M
- Initial concentration of Cu²⁺ = 1.0 M
- Concentration of NH₃ = 2.0 M
- Formation constant (Kf) for [Cu(NH₃)₄]²⁺ = 1 × 10¹²
- Standard cell potential (E°) for the reaction Zn + Cu²⁺ → Zn²⁺ + Cu = 1.1 V

### Step 2: Write the equilibrium reaction for the complex formation
The complex formation reaction can be written as:
\[ \text{Cu}^{2+} + 4 \text{NH}_3 \rightleftharpoons [\text{Cu(NH}_3)_4]^{2+} \]

### Step 3: Write the expression for the formation constant (Kf)
The formation constant expression for the above reaction is:
\[ K_f = \frac{[\text{Cu(NH}_3)_4]^{2+}}{[\text{Cu}^{2+}][\text{NH}_3]^4} \]
Given that \( K_f = 1 \times 10^{12} \).

### Step 4: Set up the equation to find the concentration of Cu²⁺
Assuming that the concentration of the complex [Cu(NH₃)₄]²⁺ is approximately equal to the initial concentration of Cu²⁺ (1.0 M), we can rearrange the Kf expression:
\[ 1 \times 10^{12} = \frac{[\text{Cu(NH}_3)_4]^{2+}}{[\text{Cu}^{2+}](2.0)^4} \]
Let [Cu²⁺] = x, then:
\[ 1 \times 10^{12} = \frac{1.0}{x \cdot 16} \]

### Step 5: Solve for x (Cu²⁺ concentration)
Rearranging gives:
\[ x = \frac{1.0}{1 \times 10^{12} \cdot 16} \]
\[ x = \frac{1.0}{1.6 \times 10^{12}} \]
\[ x = 6.25 \times 10^{-14} \, \text{M} \]

### Step 6: Use the Nernst equation to find E_cell
The Nernst equation is given by:
\[ E = E^\circ + \frac{0.0591}{n} \log \frac{[\text{products}]}{[\text{reactants}]} \]
For the reaction, n = 2 (since 2 electrons are transferred):
\[ E_{cell} = E^\circ + \frac{0.0591}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]} \]
Substituting the values:
\[ E_{cell} = 1.1 + \frac{0.0591}{2} \log \frac{1.0}{6.25 \times 10^{-14}} \]

### Step 7: Calculate the logarithm
First, calculate the logarithm:
\[ \log \frac{1.0}{6.25 \times 10^{-14}} = \log(1.6 \times 10^{13}) \approx 13.2041 \]

### Step 8: Substitute back into the Nernst equation
Now substituting back:
\[ E_{cell} = 1.1 + \frac{0.0591}{2} \times 13.2041 \]
\[ E_{cell} = 1.1 + 0.02955 \times 13.2041 \]
\[ E_{cell} = 1.1 + 0.390 \]
\[ E_{cell} = 0.71 \, \text{V} \]

### Final Answer
The estimated cell potential of the Daniel cell is **0.71 V**.