A cell is containing two H electrode The negative electrode is in contact with a solution of pH =6 eMF of the cell is 0.118 V at `25^(@)C` calculate conc of H ions ?

Answer

To solve the problem of calculating the concentration of H⁺ ions in a cell with two hydrogen electrodes, we can follow these steps:

### Step 1: Understand the given information
We have:
- Two hydrogen electrodes (anode and cathode)
- The negative electrode is in contact with a solution of pH = 6
- The EMF of the cell (E_cell) is 0.118 V at 25°C

### Step 2: Calculate the concentration of H⁺ ions from pH
The pH of the solution is given as 6. The concentration of H⁺ ions can be calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Rearranging this gives:
\[ [\text{H}^+] = 10^{-\text{pH}} \]
Substituting the value of pH:
\[ [\text{H}^+] = 10^{-6} \, \text{M} \]

### Step 3: Write the half-reactions for the electrodes
For the hydrogen electrode reactions:
- At the anode (oxidation):
\[ \text{H}_2 \rightarrow 2\text{H}^+ + 2e^- \]
- At the cathode (reduction):
\[ 2\text{H}^+ + 2e^- \rightarrow \text{H}_2 \]

### Step 4: Use the Nernst equation
The Nernst equation for a hydrogen electrode can be expressed as:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
Where:
- \( E \) is the cell potential (0.118 V)
- \( E^\circ \) for hydrogen electrode is 0 V
- \( R \) is the universal gas constant (8.314 J/(mol·K))
- \( T \) is the temperature in Kelvin (25°C = 298 K)
- \( n \) is the number of moles of electrons transferred (2)
- \( F \) is Faraday's constant (96485 C/mol)
- \( Q \) is the reaction quotient, which in this case is \(\frac{[\text{H}^+]^2}{[\text{H}^+]^2}\) (since both sides are the same, it simplifies)

### Step 5: Substitute values into the Nernst equation
Since the concentration of H⁺ at the anode is \(10^{-6}\) M, we can express the Nernst equation as:
\[ 0.118 = 0 - \frac{0.0591}{2} \log \left( \frac{[\text{H}^+]_{\text{cathode}}^2}{10^{-6}^2} \right) \]

### Step 6: Simplify the equation
This simplifies to:
\[ 0.118 = -0.02955 \log \left( \frac{[\text{H}^+]_{\text{cathode}}^2}{10^{-12}} \right) \]

### Step 7: Solve for \([\text{H}^+]_{\text{cathode}}\)
Rearranging gives:
\[ 0.118 = -0.02955 \left( 2 \log [\text{H}^+]_{\text{cathode}} - 2 \log 10^{-6} \right) \]
\[ 0.118 = -0.02955 \left( 2 \log [\text{H}^+]_{\text{cathode}} + 12 \right) \]

### Step 8: Isolate the log term
Dividing both sides by -0.02955:
\[ \frac{0.118}{-0.02955} = 2 \log [\text{H}^+]_{\text{cathode}} + 12 \]

### Step 9: Calculate the logarithm
Calculating the left side:
\[ -4.00 = 2 \log [\text{H}^+]_{\text{cathode}} + 12 \]

### Step 10: Solve for \([\text{H}^+]_{\text{cathode}}\)
Isolating the logarithm:
\[ 2 \log [\text{H}^+]_{\text{cathode}} = -4.00 - 12 \]
\[ 2 \log [\text{H}^+]_{\text{cathode}} = -16.00 \]
\[ \log [\text{H}^+]_{\text{cathode}} = -8.00 \]
Thus:
\[ [\text{H}^+]_{\text{cathode}} = 10^{-8} \, \text{M} \]

### Final Result
The concentration of H⁺ ions at the cathode is:
\[ [\text{H}^+]_{\text{cathode}} = 10^{-4} \, \text{M} \]