A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of `10^(-6)M` hydrogen ions. The `EMF` of the cell is `0.118V` at `25^(@)C`. Calculate the concentration of hydrogen ions at the positive electrode.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the setup of the electrochemical cell
We have a cell with two hydrogen electrodes. The negative electrode (anode) is in contact with a solution of \(10^{-6} M\) hydrogen ions, while the positive electrode (cathode) has an unknown concentration of hydrogen ions.

### Step 2: Write the Nernst equation for the cell
The Nernst equation for a hydrogen electrode can be expressed as:
\[ E_{\text{cell}} = E^{\circ} - \frac{0.0591}{n} \log \frac{[H^+]_{\text{cathode}}}{[H^+]_{\text{anode}}} \]
For hydrogen electrodes, \(E^{\circ} = 0\) V, and \(n = 2\) (since 2 electrons are involved in the half-reaction). Thus, the equation simplifies to:
\[ E_{\text{cell}} = -0.0591 \log \frac{[H^+]_{\text{cathode}}}{[H^+]_{\text{anode}}} \]

### Step 3: Substitute known values into the Nernst equation
We know:
- \(E_{\text{cell}} = 0.118 \, V\)
- \([H^+]_{\text{anode}} = 10^{-6} M\)

Substituting these values into the Nernst equation gives:
\[ 0.118 = -0.0591 \log \frac{[H^+]_{\text{cathode}}}{10^{-6}} \]

### Step 4: Rearranging the equation
First, divide both sides by \(-0.0591\):
\[ \frac{0.118}{-0.0591} = \log \frac{[H^+]_{\text{cathode}}}{10^{-6}} \]
Calculating the left side:
\[ -1.996 = \log \frac{[H^+]_{\text{cathode}}}{10^{-6}} \]

### Step 5: Convert from logarithmic to exponential form
Using the property of logarithms, we can convert this to:
\[ \frac{[H^+]_{\text{cathode}}}{10^{-6}} = 10^{-1.996} \]
Calculating \(10^{-1.996}\):
\[ 10^{-1.996} \approx 0.0100 \]

### Step 6: Solve for \([H^+]_{\text{cathode}}\)
Now, multiply both sides by \(10^{-6}\):
\[ [H^+]_{\text{cathode}} = 0.0100 \times 10^{-6} = 10^{-4} M \]

### Final Answer
The concentration of hydrogen ions at the positive electrode (cathode) is:
\[ \boxed{10^{-4} M} \]