Question

The emf values of three galvanic cells I, II and III are \(E_1\), \(E_2\) and \(E_3\) respectively. Determine the correct order among them. \[ \text{(I)}\quad Zn|Zn^{2+}(1M)||Cu^{2+}(0.1M)|Cu \] \[ \text{(II)}\quad Zn|Zn^{2+}(1M)||Cu^{2+}(1M)|Cu \] \[ \text{(III)}\quad Zn|Zn^{2+}(0.1M)||Cu^{2+}(1M)|Cu \]

• A. \(E_3 > E_2 > E_1\)
• B. \(E_1 > E_2 > E_3\)
• C. \(E_2 > E_3 > E_1\)
• D. \(E_1 > E_3 > E_2\)

Hint:

For the Daniell cell, \[ Zn + Cu^{2+} \rightarrow Zn^{2+}+Cu \] emf increases when:

• \(Cu^{2+}\) concentration increases.

• \(Zn^{2+}\) concentration decreases.
A quick way is to compare the ratio \[ \frac{[Zn^{2+}]}{[Cu^{2+}]} \] Smaller ratio means larger emf.

Answer 1

The Correct Option is A

Step 1: Understand the cell reaction.
In each cell zinc gives electrons and copper ions take them: $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$. The standard emf $E^\circ$ is the same for all three. Only the ion concentrations differ.

Step 2: Write the Nernst equation.
\[ E = E^\circ - \frac{0.0591}{2}\log\frac{[Zn^{2+}]}{[Cu^{2+}]} \] So a bigger $\dfrac{[Cu^{2+}]}{[Zn^{2+}]}$ ratio gives a larger emf.

Step 3: Work out cell I.
$[Zn^{2+}]=1$, $[Cu^{2+}]=0.1$. Ratio $\dfrac{[Zn^{2+}]}{[Cu^{2+}]}=\dfrac{1}{0.1}=10$. This is the largest log term, so it lowers emf the most. So $E_1$ is the smallest.

Step 4: Work out cell II.
$[Zn^{2+}]=1$, $[Cu^{2+}]=1$. Ratio $=1$, so $\log 1 = 0$ and $E_2 = E^\circ$.

Step 5: Work out cell III.
$[Zn^{2+}]=0.1$, $[Cu^{2+}]=1$. Ratio $=\dfrac{0.1}{1}=0.1$, so $\log(0.1)=-1$, which adds to emf. So $E_3$ is the largest.

Step 6: Order the three emf values.
Largest is $E_3$, then $E_2$, then $E_1$.

Step 7: State the final answer.
The correct order is:
\[ \boxed{E_3 > E_2 > E_1} \]