Calculate the voltage `E` of the cell at `25^(@)C`
`Mn(s) |Mn(OH_(2))(s)|Mn^(2+)(x,M)OH^(-) (1.00 xx 10^(-4)M)||Cu^(2+) (0.675M)|Cu(s)`
given that `K_(sp) = 1.9 xx 1-^(-13)` for `Mn(OH)_(2)(s) E^(@) (Mn^(2+)//Mn) =- 1.18 V, E^(@) (Cu^(+2)//C) = +0.34V`

Explanation

To calculate the voltage \( E \) of the cell at \( 25^\circ C \), we will follow these steps:

### Step 1: Identify the half-cell reactions
The half-cell reactions for the given cell are:
1. For copper:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (E^\circ = +0.34 \, \text{V}) \]
2. For manganese:
\[ \text{Mn} \rightarrow \text{Mn}^{2+} + 2e^- \quad (E^\circ = -1.18 \, \text{V}) \]

### Step 2: Calculate the standard cell potential \( E^\circ_{\text{cell}} \)
The standard cell potential can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, copper is the cathode and manganese is the anode:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Cu}} - E^\circ_{\text{Mn}} = 0.34 \, \text{V} - (-1.18 \, \text{V}) = 0.34 \, \text{V} + 1.18 \, \text{V} = 1.52 \, \text{V} \]

### Step 3: Determine the concentration of \( \text{Mn}^{2+} \)
Given the solubility product \( K_{sp} \) for \( \text{Mn(OH)}_2 \):
\[ K_{sp} = [\text{Mn}^{2+}][\text{OH}^-]^2 \]
Given \( K_{sp} = 1.9 \times 10^{-13} \) and \( [\text{OH}^-] = 1.00 \times 10^{-4} \, \text{M} \):
\[ 1.9 \times 10^{-13} = [\text{Mn}^{2+}] \times (1.00 \times 10^{-4})^2 \]
\[ 1.9 \times 10^{-13} = [\text{Mn}^{2+}] \times 1.00 \times 10^{-8} \]
\[ [\text{Mn}^{2+}] = \frac{1.9 \times 10^{-13}}{1.00 \times 10^{-8}} = 1.9 \times 10^{-5} \, \text{M} \]

### Step 4: Calculate the reaction quotient \( Q \)
The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{Mn}^{2+}]}{[\text{Cu}^{2+}]} \]
Substituting the values:
\[ Q = \frac{1.9 \times 10^{-5}}{0.675} \approx 2.81 \times 10^{-5} \]

### Step 5: Calculate the Nernst equation
The Nernst equation is given by:
\[ E = E^\circ_{\text{cell}} - \frac{0.059}{n} \log Q \]
Where \( n = 2 \) (number of electrons transferred):
\[ E = 1.52 \, \text{V} - \frac{0.059}{2} \log(2.81 \times 10^{-5}) \]
Calculating \( \log(2.81 \times 10^{-5}) \):
\[ \log(2.81 \times 10^{-5}) \approx -4.55 \]
Substituting this value back into the Nernst equation:
\[ E = 1.52 \, \text{V} - \frac{0.059}{2} \times (-4.55) \]
\[ E = 1.52 \, \text{V} + 0.059 \times 2.275 \approx 1.52 \, \text{V} + 0.134 \, \text{V} \approx 1.66 \, \text{V} \]

### Final Answer
Thus, the voltage \( E \) of the cell at \( 25^\circ C \) is approximately:
\[ \boxed{1.66 \, \text{V}} \]