In the Daniel cell, if the concentrations of `Zn^(2+) ` and `Cu^(2+)` ions are doubled at 298 K , the e.m.f. Of the cell __________.

Options

A. is doubled
B. is reduced to half
C. remains same
D. becomes four times

Answer

To solve the problem regarding the effect of doubling the concentrations of `Zn^(2+)` and `Cu^(2+)` ions in a Daniel cell on its electromotive force (e.m.f.), we can use the Nernst equation. Here’s a step-by-step solution:

### Step 1: Write the Nernst Equation
The Nernst equation for a cell can be expressed as:
\[ E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right) \]
Where:
- \( E_{cell} \) = e.m.f. of the cell
- \( E^0_{cell} \) = standard e.m.f. of the cell
- \( R \) = universal gas constant (8.314 J/mol·K)
- \( T \) = temperature in Kelvin (298 K in this case)
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (96485 C/mol)
- \([Zn^{2+}]\) and \([Cu^{2+}]\) = concentrations of the ions

### Step 2: Determine Initial Conditions
Let’s denote the initial concentrations of \( Zn^{2+} \) and \( Cu^{2+} \) as:
- \([Zn^{2+}]_1 = [Zn^{2+}]\)
- \([Cu^{2+}]_1 = [Cu^{2+}]\)

Using these concentrations, the initial e.m.f. \( E_1 \) can be expressed as:
\[ E_1 = E^0_{cell} - \frac{RT}{nF} \ln \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right) \]

### Step 3: Change the Concentrations
When the concentrations are doubled, we have:
- \([Zn^{2+}]_2 = 2[Zn^{2+}]\)
- \([Cu^{2+}]_2 = 2[Cu^{2+}]\)

Now, we can express the new e.m.f. \( E_2 \) as:
\[ E_2 = E^0_{cell} - \frac{RT}{nF} \ln \left( \frac{[Zn^{2+}]_2}{[Cu^{2+}]_2} \right) \]

### Step 4: Substitute the New Concentrations
Substituting the new concentrations into the equation:
\[ E_2 = E^0_{cell} - \frac{RT}{nF} \ln \left( \frac{2[Zn^{2+}]}{2[Cu^{2+}]} \right) \]

### Step 5: Simplify the Equation
The factors of 2 cancel out:
\[ E_2 = E^0_{cell} - \frac{RT}{nF} \ln \left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right) \]
This shows that:
\[ E_2 = E_1 \]

### Conclusion
From the above steps, we conclude that the e.m.f. of the cell remains the same when the concentrations of \( Zn^{2+} \) and \( Cu^{2+} \) ions are doubled. Therefore, the answer is that the e.m.f. of the cell remains constant.

### Final Answer
The e.m.f. of the cell remains the same.

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