A zinc rod dipped in n molar solution of `ZnSO_4` has an electrode potential of -0.56 V. The salt is 98 percent dissociated at room temperature. What is the molarity of the solution?` (E°(Zn^(+2)/Zn) = -0.5 V)`

Options

A. 8.44 × `10^(-3)` M
B. 9.44 × `10^-4` M
C. 8.44 × `10^-4` M
D. 9.44 × `10^-3` M

Answer

To solve the problem, we will use the Nernst equation and the information provided in the question. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have a zinc rod in an n molar solution of ZnSO4 with an electrode potential of -0.56 V. The salt is 98% dissociated, and we need to find the molarity (n) of the solution.

### Step 2: Write the Half-Cell Reaction
The half-cell reaction for zinc is:
\[ \text{Zn} \leftrightarrow \text{Zn}^{2+} + 2e^- \]

### Step 3: Identify Given Values
- Standard electrode potential, \( E^\circ (\text{Zn}^{2+}/\text{Zn}) = -0.5 \, \text{V} \)
- Measured electrode potential, \( E = -0.56 \, \text{V} \)
- Degree of dissociation = 98% or 0.98

### Step 4: Calculate the Concentration of Zn²⁺
Since the salt is 98% dissociated, the concentration of Zn²⁺ ions in the solution can be calculated as:
\[ [\text{Zn}^{2+}] = 0.98 \times n \]

### Step 5: Apply the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{1}{[\text{Zn}^{2+}]} \right) \]
Substituting the known values:
\[ -0.56 = -0.5 - \frac{0.059}{2} \log \left( \frac{1}{0.98n} \right) \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ -0.56 + 0.5 = - \frac{0.059}{2} \log \left( \frac{1}{0.98n} \right) \]
\[ -0.06 = - \frac{0.059}{2} \log \left( \frac{1}{0.98n} \right) \]

### Step 7: Solve for the Logarithm
Multiply both sides by -2:
\[ 0.12 = 0.059 \log \left( \frac{1}{0.98n} \right) \]
Now divide both sides by 0.059:
\[ \log \left( \frac{1}{0.98n} \right) = \frac{0.12}{0.059} \]
Calculating the right side:
\[ \log \left( \frac{1}{0.98n} \right) \approx 2.0339 \]

### Step 8: Exponentiate to Remove the Logarithm
Taking the antilogarithm:
\[ \frac{1}{0.98n} = 10^{2.0339} \]
Calculating \( 10^{2.0339} \):
\[ \frac{1}{0.98n} \approx 108.7 \]

### Step 9: Solve for n
Now, rearranging gives:
\[ 0.98n = \frac{1}{108.7} \]
\[ n \approx \frac{1}{108.7 \times 0.98} \]
Calculating \( n \):
\[ n \approx 9.44 \times 10^{-3} \, \text{M} \]

### Final Answer
The molarity of the solution is approximately:
\[ n \approx 9.44 \times 10^{-3} \, \text{M} \]

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