A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of `10^(-6)M` hydrogen ion. The EMF of the cell is 0.118 V at 298 K. The concentration of `H^(+)` ion at the positive electrode is `10^(-x)`, The value of 'x' is

Answer

To solve the problem, we need to calculate the value of \( x \) given the conditions of the electrochemical cell. Let's break down the steps:

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

### Step 2: Write the Half-Reactions
At the anode (negative electrode), oxidation occurs:
\[ \text{H}_2 \rightarrow 2 \text{H}^+ + 2 e^- \]

At the cathode (positive electrode), reduction occurs:
\[ 2 \text{H}^+ + 2 e^- \rightarrow \text{H}_2 \]

### Step 3: Determine the Cell Reaction
The overall cell reaction can be represented as:
\[ \text{H}_2 (g) \rightarrow 2 \text{H}^+ (aq) + 2 e^- \quad \text{(anode)} \]
\[ 2 \text{H}^+ (aq) + 2 e^- \rightarrow \text{H}_2 (g) \quad \text{(cathode)} \]

### Step 4: Calculate the Reaction Quotient \( Q \)
The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{H}^+]^2_{\text{anode}}}{[\text{H}^+]^2_{\text{cathode}}} \]
Substituting the concentrations:
\[ Q = \frac{(10^{-6})^2}{(10^{-x})^2} = \frac{10^{-12}}{10^{-2x}} = 10^{-12 + 2x} \]

### Step 5: Use the Nernst Equation
The Nernst equation relates the cell potential \( E \) to the standard potential \( E^\circ \) and the reaction quotient \( Q \):
\[ E = E^\circ - \frac{0.059}{n} \log Q \]
For hydrogen electrodes, \( E^\circ = 0 \) and \( n = 2 \):
\[ E = 0 - \frac{0.059}{2} \log(10^{-12 + 2x}) \]
Given \( E = 0.118 \, V \):
\[ 0.118 = -\frac{0.059}{2} \log(10^{-12 + 2x}) \]

### Step 6: Simplify the Equation
Multiply both sides by -2:
\[ -2 \times 0.118 = 0.059 \log(10^{-12 + 2x}) \]
\[ -0.236 = 0.059 \log(10^{-12 + 2x}) \]

### Step 7: Divide by 0.059
\[ \frac{-0.236}{0.059} = \log(10^{-12 + 2x}) \]
Calculating the left side:
\[ -4 = \log(10^{-12 + 2x}) \]

### Step 8: Convert from Logarithmic to Exponential Form
Using the property of logarithms:
\[ 10^{-4} = 10^{-12 + 2x} \]

### Step 9: Set the Exponents Equal
Since the bases are the same:
\[ -4 = -12 + 2x \]

### Step 10: Solve for \( x \)
Rearranging gives:
\[ 2x = -4 + 12 \]
\[ 2x = 8 \]
\[ x = 4 \]

### Final Answer
The value of \( x \) is \( 4 \).

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