For the cell `Pt_((s)), H_(2) (1 atm)|H^(+) (pH=2)||H^(+) (pH=3)|H_(2) (1atm),Pt`
The cell reaction is

Options

A. Spontaneous
B. Non-spontaneous
C. can't be predicted
D. None of the above

Answer

To determine the cell reaction for the given electrochemical cell, we will follow these steps:

### Step 1: Identify the Half-Reactions
The cell consists of two hydrogen electrodes at different pH levels. The half-reactions at each electrode are as follows:

- **Anode (Oxidation)**:
\[ H_2(g) \rightarrow 2H^+(aq) + 2e^- \]

- **Cathode (Reduction)**:
\[ 2H^+(aq) + 2e^- \rightarrow H_2(g) \]

### Step 2: Determine the Concentrations of H⁺ Ions
Using the pH values given in the problem, we can calculate the concentrations of H⁺ ions:

- For pH = 2:
\[ [H^+] = 10^{-2} \, \text{M} \]

- For pH = 3:
\[ [H^+] = 10^{-3} \, \text{M} \]

### Step 3: Write the Nernst Equation
The Nernst equation relates the cell potential (E) to the standard electrode potential (E°) and the concentrations of the reactants and products:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
Where:
- \(E^\circ = 0 \, \text{V}\) for hydrogen electrodes
- \(R = 8.314 \, \text{J/(mol·K)}\)
- \(T = 298 \, \text{K}\) (25 °C)
- \(n = 2\) (number of electrons transferred)
- \(F = 96500 \, \text{C/mol}\)
- \(Q\) is the reaction quotient, given by:
\[ Q = \frac{[H^+]^2 \text{ (at anode)}}{[H^+]^2 \text{ (at cathode)}} \]

### Step 4: Calculate the Reaction Quotient (Q)
Substituting the concentrations into the expression for Q:
\[ Q = \frac{(10^{-2})^2}{(10^{-3})^2} = \frac{10^{-4}}{10^{-6}} = 10^{2} \]

### Step 5: Substitute Values into the Nernst Equation
Now, substituting the values into the Nernst equation:
\[ E = 0 - \frac{(8.314)(298)}{(2)(96500)} \ln(10^{2}) \]
Calculating the constants:
\[ E = - \frac{(8.314)(298)}{(2)(96500)} \cdot 2 \]
\[ E = - \frac{(8.314)(298)}{96500} \]
Calculating this gives:
\[ E \approx -0.0591 \, \text{V} \]

### Step 6: Determine Gibbs Free Energy (G)
Using the relationship between Gibbs free energy and cell potential:
\[ \Delta G = -nFE \]
Substituting the values:
\[ \Delta G = -(2)(96500)(-0.0591) \]
Calculating this gives:
\[ \Delta G \approx 11400 \, \text{J/mol} \, \text{(positive)} \]

### Conclusion
Since \(\Delta G\) is positive, the cell reaction is non-spontaneous.

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