When a metal chloride `MCI` is dissolced in a large excess of ammonia, practially all metal can be assumed to exist as `M_(x)(NH_(3))_(y)^(+x)`. Find the value of `x +y` using the following two cells.

Explanation

To solve the problem of finding the value of \( x + y \) for the metal chloride \( MCl \) when dissolved in excess ammonia, we can follow these steps:

### Step 1: Understand the Reaction
When a metal chloride \( MCl \) is dissolved in a large excess of ammonia, it forms a complex of the type \( M_x(NH_3)_y^{+x} \). For the case of silver chloride, the reaction can be represented as:
\[ Ag(s) + y NH_3 \rightarrow [Ag(NH_3)_y]^{+} + x e^{-} \]

### Step 2: Write the Nernst Equation
The Nernst equation for the cell can be expressed as:
\[ E_{cell} = E^{\circ} - \frac{0.059}{n} \log \left( \frac{[Products]}{[Reactants]} \right) \]
where \( n \) is the number of electrons transferred in the reaction.

### Step 3: Substitute Known Values
Given:
- \( E_{cell} = 0.126 \, V \)
- \( E^{\circ} = 0.118 \, V \)
- Concentration of products and reactants can be expressed as:
\[ \text{Concentration of products} = 0.4 \times 10^{-3} \, M \]
\[ \text{Concentration of reactants} = (40 \times 10^{-3})^x \, M \]

Substituting these values into the Nernst equation gives:
\[ 0.126 = 0.118 - \frac{0.059}{x} \log \left( \frac{0.4 \times 10^{-3}}{(40 \times 10^{-3})^x} \right) \]

### Step 4: Simplify and Solve for \( x \)
Rearranging the equation:
\[ 0.126 - 0.118 = -\frac{0.059}{x} \log \left( \frac{0.4 \times 10^{-3}}{(40 \times 10^{-3})^x} \right) \]
\[ 0.008 = -\frac{0.059}{x} \log \left( \frac{0.4}{40^x} \times 10^{3x} \right) \]
This simplifies to:
\[ 0.008 = -\frac{0.059}{x} \left( \log(0.4) - x \log(40) + 3x \right) \]

### Step 5: Calculate \( y \)
After solving for \( x \), we find that \( x = 1 \). To find \( y \), we can use the stoichiometry of the reaction. Since we assumed \( y \) to be the number of ammonia molecules coordinated to the metal, we can deduce from the reaction that \( y = 2 \).

### Step 6: Calculate \( x + y \)
Finally, we compute:
\[ x + y = 1 + 2 = 3 \]

### Final Answer
Thus, the value of \( x + y \) is \( 3 \).

---