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 solve the problem regarding the electrochemical cell reaction, we will follow these steps:

### Step 1: Identify the Components of the Cell
The cell is represented as:
\[ \text{Pt}_{(s)}, \text{H}_2 (1 \text{ atm}) | \text{H}^+ (pH = 2) || \text{H}^+ (pH = 3) | \text{H}_2 (1 \text{ atm}), \text{Pt} \]

Here, we have:
- Anode (left side): \(\text{H}_2\) at 1 atm and \(\text{H}^+\) at pH = 2
- Cathode (right side): \(\text{H}^+\) at pH = 3 and \(\text{H}_2\) at 1 atm

### Step 2: Calculate the Concentration of \(\text{H}^+\)
Using the relationship between pH and \(\text{H}^+\) concentration:
\[ \text{pH} = -\log[\text{H}^+] \]
For pH = 2:
\[ [\text{H}^+] = 10^{-2} \text{ M} \]
For pH = 3:
\[ [\text{H}^+] = 10^{-3} \text{ M} \]

### Step 3: Write the Half-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: Write the Overall Cell Reaction
Combining the half-reactions, we have:
\[ \text{H}_2 (1 \text{ atm}) + 2 \text{H}^+ (10^{-2} \text{ M}) \rightarrow \text{H}_2 (1 \text{ atm}) + 2 \text{H}^+ (10^{-3} \text{ M}) \]

### Step 5: Determine the Reaction Quotient (Q)
The reaction quotient \(Q\) can be expressed as:
\[ Q = \frac{[\text{H}^+]^2_{\text{cathode}}}{[\text{H}^+]^2_{\text{anode}}} = \frac{(10^{-3})^2}{(10^{-2})^2} = \frac{10^{-6}}{10^{-4}} = 10^{-2} \]

### Step 6: Calculate the Cell Potential (E_cell)
Using the Nernst equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \]
Since both half-reactions involve hydrogen, \(E^\circ_{\text{cell}} = 0\).
Substituting values:
- \(R = 8.314 \, \text{J/(mol K)}\)
- \(T = 298 \, \text{K}\)
- \(n = 2\) (number of moles of electrons transferred)
- \(F = 96500 \, \text{C/mol}\)

Calculating:
\[ E_{\text{cell}} = 0 - \frac{(8.314)(298)}{(2)(96500)} \ln(10^{-2}) \]
\[ E_{\text{cell}} = -0.0257 \cdot (-4.605) \]
\[ E_{\text{cell}} \approx 0.059 \, \text{V} \]

### Step 7: Determine Spontaneity
Using the Gibbs free energy equation:
\[ \Delta G = -nFE_{\text{cell}} \]
Substituting values:
\[ \Delta G = -2 \cdot 96500 \cdot 0.059 \]
\[ \Delta G \approx -11300 \, \text{J} \]

Since \(\Delta G\) is negative, the reaction is spontaneous.

### Final Answer
The cell reaction is spontaneous.

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