For the galvanic cell `Zn(s)abs(Zn^(2+) ) abs(Cd^(2+)) Cd(s),E_(cell ) = 0.30 V and E_(cell)^(o) = 0.36 V,` then value of `[Cd^(2+)]//[Zn^(2+)]` will be:

Options

A. 10
B. 1
C. `0.1`
D. 100

Answer

To solve the problem, we will use the Nernst equation, which relates the cell potential (E) to the standard cell potential (E°) and the concentrations of the reactants and products involved in the electrochemical reaction.

### Step-by-Step Solution:

1. **Write the Nernst Equation**:
The Nernst equation is given by:
\[ E = E^{\circ} - \frac{2.303RT}{nF} \log \left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]
where:
- \(E\) = cell potential
- \(E^{\circ}\) = standard cell potential
- \(R\) = universal gas constant (8.314 J/(mol·K))
- \(T\) = temperature in Kelvin (298 K at 25°C)
- \(n\) = number of moles of electrons transferred in the reaction
- \(F\) = Faraday's constant (96500 C/mol)

2. **Identify the Given Values**:
From the problem, we have:
- \(E = 0.30 \, V\)
- \(E^{\circ} = 0.36 \, V\)
- \(n = 2\) (since 2 electrons are involved in the reaction)

3. **Calculate the Value of \(\frac{2.303RT}{nF}\)**:
Plugging in the values:
\[ \frac{2.303 \times 8.314 \times 298}{2 \times 96500} \]
This simplifies to:
\[ \frac{2.303 \times 8.314 \times 298}{193000} \approx 0.0591 \, V \]

4. **Substitute into the Nernst Equation**:
Now we can substitute the known values into the Nernst equation:
\[ 0.30 = 0.36 - 0.0591 \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cd}^{2+}]}\right) \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ 0.0591 \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cd}^{2+}]}\right) = 0.36 - 0.30 \]
\[ 0.0591 \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cd}^{2+}]}\right) = 0.06 \]

6. **Solving for the Logarithm**:
Dividing both sides by 0.0591:
\[ \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cd}^{2+}]}\right) = \frac{0.06}{0.0591} \approx 1.014 \]

7. **Finding the Ratio**:
Converting from logarithmic form:
\[ \frac{[\text{Zn}^{2+}]}{[\text{Cd}^{2+}]} = 10^{1.014} \approx 10.26 \]

8. **Finding the Inverse Ratio**:
Since we need \(\frac{[\text{Cd}^{2+}]}{[\text{Zn}^{2+}]}\):
\[ \frac{[\text{Cd}^{2+}]}{[\text{Zn}^{2+}]} = \frac{1}{10.26} \approx 0.0975 \approx 0.01 \]

### Final Answer:
The value of \(\frac{[\text{Cd}^{2+}]}{[\text{Zn}^{2+}]}\) is approximately **0.01**.