### 1. Break Down the Problem
We need to calculate the electromotive force (emf) of the electrochemical cell based on the provided reaction. The reaction involves copper ions, hydrogen gas, and products that include solid copper and hydrogen ions.

### 2. Relevant Concepts
To calculate the emf, we can use the Nernst equation:
$$
E = E^0 - \frac{RT}{nF} \ln Q
$$

Where:
- $E$ is the cell potential (emf),
- $E^0$ is the standard reduction potential,
- $R$ is the universal gas constant (8.314 J/(mol·K)),
- $T$ is the temperature in Kelvin (assumed to be 298 K),
- $n$ is the number of moles of electrons transferred,
- $F$ is the Faraday constant (96485 C/mol),
- $Q$ is the reaction quotient.

### 3. Analysis and Detail
**Step 3.1: Determine Standard Cell Potential $E^0$**
The half-reactions are:
- Cu²⁺ + 2e⁻ → Cu (s) $\therefore E^0 = +0.34\,V$
- 2H⁺ + 2e⁻ → H₂ (g) $\therefore E^0 = 0\,V$

We can find the average standard potential:
$$
E^0 = E^0_{\text{cathode}} - E^0_{\text{anode}} = 0.34\,V - 0\,V = 0.34\,V
$$

**Step 3.2: Determine $Q$**
The reaction quotient $Q$ is given by:
$$
Q = \frac{[\text{H}^+]^2}{[\text{Cu}^{2+}][\text{H}_2]} = \frac{(10^{-3})^2}{(0.1)(1)} = \frac{10^{-6}}{0.1} = 10^{-5}
$$

**Step 3.3: Plug values into the Nernst Equation**
Assuming $T = 298 \, K$, $n = 2$:
$$
E = 0.34 - \frac{(8.314)(298)}{(2)(96485)} \ln(10^{-5})
$$

Calculating the logarithm:
$$
\ln(10^{-5}) = -5 \ln(10) \approx -5 \times 2.303 = -11.515
$$

**Step 3.4: Calculate the Nernst Equation**
$$
E = 0.34 - \frac{(8.314)(298)}{(2)(96485)} \times (-11.515)
$$
Calculating the fraction:
$$
\approx \frac{2478.772}{192970} \approx 0.01284
$$
Now calculate $E$:
$$
E = 0.34 + 0.01284 \times 11.515 \approx 0.34 + 0.147 \approx 0.487\,V
$$

### 4. Verify and Summarize
1. Verified the reaction quotient $Q$.
2. Verified standard reduction potentials.
3. Confirmed calculations of values used in the Nernst equation.

### Final Answer
The electromotive force (emf) for the cell is approximately $E \approx 0.487\,V$.

Question

### 1. Break Down the Problem
We need to calculate the electromotive force (emf) of the electrochemical cell based on the provided reaction. The reaction involves copper ions, hydrogen gas, and products that include solid copper and hydrogen ions.

### 2. Relevant Concepts
To calculate the emf, we can use the Nernst equation:
$$
E = E^0 - \frac{RT}{nF} \ln Q
$$

Where:
- $E$ is the cell potential (emf),
- $E^0$ is the standard reduction potential,
- $R$ is the universal gas constant (8.314 J/(mol·K)),
- $T$ is the temperature in Kelvin (assumed to be 298 K),
- $n$ is the number of moles of electrons transferred,
- $F$ is the Faraday constant (96485 C/mol),
- $Q$ is the reaction quotient.

### 3. Analysis and Detail
**Step 3.1: Determine Standard Cell Potential $E^0$**  
The half-reactions are:
- Cu²⁺ + 2e⁻ → Cu (s)  $\therefore E^0 = +0.34\,V$
- 2H⁺ + 2e⁻ → H₂ (g)  $\therefore E^0 = 0\,V$

We can find the average standard potential:
$$
E^0 = E^0_{\text{cathode}} - E^0_{\text{anode}} = 0.34\,V - 0\,V = 0.34\,V
$$

**Step 3.2: Determine $Q$**  
The reaction quotient $Q$ is given by:
$$
Q = \frac{[\text{H}^+]^2}{[\text{Cu}^{2+}][\text{H}_2]} = \frac{(10^{-3})^2}{(0.1)(1)} = \frac{10^{-6}}{0.1} = 10^{-5}
$$

**Step 3.3: Plug values into the Nernst Equation**  
Assuming $T = 298 \, K$, $n = 2$:
$$
E = 0.34 - \frac{(8.314)(298)}{(2)(96485)} \ln(10^{-5})
$$

Calculating the logarithm:
$$
\ln(10^{-5}) = -5 \ln(10) \approx -5 \times 2.303 = -11.515
$$

**Step 3.4: Calculate the Nernst Equation**  
$$
E = 0.34 - \frac{(8.314)(298)}{(2)(96485)} \times (-11.515)
$$
Calculating the fraction:
$$
\approx \frac{2478.772}{192970} \approx 0.01284
$$
Now calculate $E$:
$$
E = 0.34 + 0.01284 \times 11.515 \approx 0.34 + 0.147 \approx 0.487\,V
$$

### 4. Verify and Summarize
1. Verified the reaction quotient $Q$.
2. Verified standard reduction potentials.
3. Confirmed calculations of values used in the Nernst equation.

### Final Answer
The electromotive force (emf) for the cell is approximately $E \approx 0.487\,V$.

Knowee AI · Accepted Answer