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 at cathode?

Answer

To solve the problem, we need to determine the concentration of H⁺ ions at the cathode of a concentration cell consisting of two hydrogen electrodes. Here are the steps to find the solution:

### Step 1: Understand the Cell Configuration
The cell consists of two hydrogen electrodes. One electrode is in contact with a solution of pH = 6, and we need to find the concentration of H⁺ ions at the cathode.

### Step 2: Calculate the Concentration of H⁺ Ions at the Anode
Given that the pH of the solution at the anode is 6, we can calculate the concentration of H⁺ ions using the formula:
\[ \text{pH} = -\log[H^+] \]
Rearranging this gives:
\[ [H^+] = 10^{-\text{pH}} \]
Substituting the value of pH:
\[ [H^+] = 10^{-6} \, \text{M} \]

### Step 3: Write the Nernst Equation
For a concentration cell, the standard cell potential (E°cell) is 0. The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log \left(\frac{[H^+]_{anode}}{[H^+]_{cathode}}\right) \]
Where:
- \( E_{cell} \) is the cell potential (0.118 V in this case).
- \( n \) is the number of electrons transferred in the half-reaction (which is 1 for hydrogen).

### Step 4: Substitute Known Values into the Nernst Equation
Substituting the known values into the Nernst equation:
\[ 0.118 = 0 - \frac{0.0591}{1} \log \left(\frac{10^{-6}}{[H^+]_{cathode}}\right) \]

### Step 5: Simplify the Equation
Rearranging gives:
\[ 0.118 = -0.0591 \log \left(\frac{10^{-6}}{[H^+]_{cathode}}\right) \]
Dividing both sides by -0.0591:
\[ -\frac{0.118}{0.0591} = \log \left(\frac{10^{-6}}{[H^+]_{cathode}}\right) \]

### Step 6: Calculate the Logarithm
Calculating the left side:
\[ \log \left(\frac{10^{-6}}{[H^+]_{cathode}}\right) = -2.000 \] (approximately, from the calculation of -0.118/0.0591)

### Step 7: Solve for [H⁺] at the Cathode
Using the property of logarithms:
\[ \frac{10^{-6}}{[H^+]_{cathode}} = 10^{-2} \]
Rearranging gives:
\[ [H^+]_{cathode} = \frac{10^{-6}}{10^{-2}} = 10^{-4} \, \text{M} \]

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

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