An electrochemical cell is shown below `Pt, H_(2)(1 "atm")|HCl(0.1 M)|CH_(3)COOH(0.1 M)|H_(2)(1 "atm")`, The emf of the cell will not be zero, because

Options

A. EMF depends on molarities of acids used
B. pH of 0.1 M HCl and 0.1 M `CH_(3)COOH` is same
C. the temperature is constant
D. acids used in two compartments are different

Explanation

To determine why the emf of the electrochemical cell `Pt, H2(1 atm)|HCl(0.1 M)|CH3COOH(0.1 M)|H2(1 atm)` is not zero, we can analyze the components and conditions of the cell step by step.

### Step 1: Identify the Components of the Cell
The cell consists of:
- A platinum electrode in contact with hydrogen gas (H2) at 1 atm pressure.
- An acidic solution of hydrochloric acid (HCl) at a concentration of 0.1 M.
- A solution of acetic acid (CH3COOH) at a concentration of 0.1 M.
- Another platinum electrode in contact with hydrogen gas (H2) at 1 atm pressure.

### Step 2: Understand the Reactions Taking Place
In this electrochemical cell, the following half-reactions can occur:
1. At the anode (oxidation):
\[ H_2 \rightarrow 2H^+ + 2e^- \]
2. At the cathode (reduction):
\[ 2H^+ + 2e^- \rightarrow H_2 \]

### Step 3: Determine the Standard Cell Potential (E°cell)
Since both half-reactions involve hydrogen, the standard cell potential (E°cell) can be calculated. The standard electrode potential for the hydrogen electrode is defined as 0 V. Therefore, the overall standard cell potential is:
\[ E°_{cell} = E°_{cathode} - E°_{anode} = 0 - 0 = 0 \, V \]

### Step 4: Apply the Nernst Equation
The Nernst equation relates the emf of the cell to the concentrations of the reactants and products:
\[ E_{cell} = E°_{cell} - \frac{0.059}{n} \log(Q) \]
Where:
- \( n \) = number of moles of electrons transferred (which is 2 in this case).
- \( Q \) = reaction quotient.

### Step 5: Calculate the Reaction Quotient (Q)
The reaction quotient \( Q \) for this cell can be calculated based on the concentrations of H+ ions from both HCl and acetic acid. Since both solutions are 0.1 M:
\[ Q = \frac{[H^+]^2}{[H_2]} = \frac{(0.1)^2}{1} = 0.01 \]

### Step 6: Substitute into the Nernst Equation
Substituting the values into the Nernst equation:
\[ E_{cell} = 0 - \frac{0.059}{2} \log(0.01) \]
\[ = -0.0295 \times (-2) \]
\[ = 0.059 \, V \]

### Conclusion
The emf of the cell is not zero because the concentrations of H+ ions from HCl and acetic acid, although equal, result in a non-zero reaction quotient, leading to a positive emf.