Consider the cell `:`
`Pt|H_(2)(p_(1)atm)|H^(o+)(x_(1)M) || H^(o+)(x_(2)M)|H_(2)(p_(2)atm)Pt`.
The cell reaction be spontaneous if

Options

A. `p_(1)=p_(2)` and `x_(1) lt x_(2)`
B. `p_(1)=p_(2)` and `x_(1) lt x_(2)`
C. `x_(1)=x_(2)` and `p_(1) gt p_(2)`
D. `x_(1)=x_(2)` and `p_(1)ltp_(2)`

Explanation

To determine the conditions under which the given electrochemical cell reaction is spontaneous, we can follow these steps:

### Step 1: Identify the Cell Reaction
The cell is represented as:
\[ \text{Pt} | \text{H}_2 (p_1 \text{ atm}) | \text{H}^+ (x_1 \text{ M}) || \text{H}^+ (x_2 \text{ M}) | \text{H}_2 (p_2 \text{ atm}) | \text{Pt} \]

In this cell, the left side is the anode where oxidation occurs and the right side is the cathode where reduction occurs.

### Step 2: Write the Half-Reactions
At the anode (oxidation):
\[ \text{H}_2 (g) \rightarrow 2 \text{H}^+ + 2 e^- \]
At the cathode (reduction):
\[ 2 \text{H}^+ + 2 e^- \rightarrow \text{H}_2 (g) \]

### Step 3: Write the Overall Cell Reaction
Adding the two half-reactions gives:
\[ \text{H}_2 (p_1 \text{ atm}) + 2 \text{H}^+ (x_1 \text{ M}) \rightarrow \text{H}_2 (p_2 \text{ atm}) + 2 \text{H}^+ (x_2 \text{ M}) \]

### Step 4: Determine the Reaction Quotient (Q)
The reaction quotient \( Q \) is defined as the ratio of the concentrations of products to reactants:
\[ Q = \frac{[H^+]^2_{(x_2)}}{[H^+]^2_{(x_1)} \cdot P_{H_2 (p_1)}} \]

This can be simplified to:
\[ Q = \frac{x_2^2 \cdot p_2}{x_1^2 \cdot p_1} \]

### Step 5: Determine Conditions for Spontaneity
For the cell reaction to be spontaneous, the cell potential \( E \) must be positive. The relationship between cell potential and the reaction quotient is given by:
\[ E = E^0 - \frac{0.059}{n} \log Q \]

Since \( E^0 = 0 \) for a concentration cell, we have:
\[ E = -\frac{0.059}{n} \log Q \]

For \( E \) to be positive, \( \log Q \) must be negative, which implies:
\[ Q < 1 \]

### Step 6: Analyze the Conditions for Q < 1
From the expression for \( Q \):
\[ \frac{x_2^2 \cdot p_2}{x_1^2 \cdot p_1} < 1 \]

This can be rearranged to:
\[ x_2^2 \cdot p_2 < x_1^2 \cdot p_1 \]

### Step 7: Analyze Given Options
1. If \( p_1 = p_2 \), then \( x_2^2 < x_1^2 \) implies \( x_2 < x_1 \).
2. If \( x_1 = x_2 \), then \( p_2 < p_1 \).

Thus, the conditions for spontaneity are:
- \( x_2 < x_1 \) (when pressures are equal)
- \( p_2 < p_1 \) (when concentrations are equal)

### Final Answer
The cell reaction will be spontaneous if:
- \( x_2 < x_1 \) or \( p_2 < p_1 \)