Question

Two galvanic cells, Cell I and Cell II, operate at the same temperature and involve two different redox systems. It is observed that at a particular value of the reaction quotient $\text{Q}_{\text{I}} = \text{Q}_{\text{II}} = \text{Q}_0$, the ratio of EMF of the two cells ($\text{E}_{\text{I}}/\text{E}_{\text{II}}$) as well as the ratio of change of Gibbs free energy for the two cells ($\Delta\text{G}_{\text{I}}/\Delta\text{G}_{\text{II}}$) are both 1/3. Assuming standard cell reactions written in the spontaneous direction and reversible operation, the pair of cells can satisfy these conditions is:

• A. Cell I: $\text{Zn(s)}|\text{Zn}^{2+}\text{(aq)}||\text{Ag}^+\text{(aq)}|\text{Ag(s)}$ and Cell II: $\text{Zn(s)}|\text{Zn}^{2+}\text{(aq)}||\text{Cu}^{2+}\text{(aq)}|\text{Cu(s)}$
• B. Cell I: $\text{Ag}^+\text{(aq)}|\text{Ag(s)}||\text{Fe}^{2+}\text{(aq)}|\text{Fe}^{3+}\text{(aq)}$ and Cell II: $\text{Al(s)}|\text{Al}^{3+}\text{(aq)}||\text{Ag}^+\text{(aq)}|\text{Ag(s)}$
• C. Cell I: $\text{Zn(s)}|\text{Zn}^{2+}\text{(aq)}||\text{Cu}^{2+}\text{(aq)}|\text{Cu(s)}$ and Cell II: $\text{Al(s)}|\text{Al}^{3+}\text{(aq)}||\text{Ag}^+\text{(aq)}|\text{Ag(s)}$
• D. Cell I: $\text{Fe}^{2+}\text{(aq)}|\text{Fe}^{3+}\text{(aq)}||\text{Cu}^{2+}\text{(aq)}|\text{Cu}^+\text{(aq)}$ and Cell II: $\text{Cu(s)}|\text{Cu}^{2+}\text{(aq)}||\text{Ag}^+\text{(aq)}|\text{Ag(s)}$

Hint:

If the ratio of energy ($\Delta\text{G}$) and the ratio of driving force (E) are identical, then the scaling factor $n$ (electrons transferred) must cancel out, meaning $n$ must be equal for both processes.

Answer 1

The Correct Option is A

Step 1: Write the two key equations.
The Nernst equation gives the cell EMF as $E = E^\circ - \frac{RT}{nF}\ln Q$, and the free energy change is $\Delta G = -nFE$. We will use the fact that both the EMF ratio and the $\Delta G$ ratio are given as exactly $1/3$ at the same shared value of $Q$.

Step 2: Use the EMF ratio.
We are told $\frac{E_I}{E_{II}} = \frac{1}{3}$ at $Q = Q_0$. So whatever the right pair is, Cell I must produce one third of the voltage of Cell II at that condition. This already tells us Cell I is a weak (low voltage) cell and Cell II is a strong one.

Step 3: Use the free energy ratio.
Since $\Delta G = -nFE$, we get $\frac{\Delta G_I}{\Delta G_{II}} = \frac{n_I E_I}{n_{II} E_{II}}$. We are told this ratio is also $1/3$.

Step 4: Compare the two ratios.
We already have $\frac{E_I}{E_{II}} = \frac{1}{3}$. Plugging that into the free energy ratio gives $\frac{\Delta G_I}{\Delta G_{II}} = \frac{n_I}{n_{II}} \times \frac{1}{3}$. For this to also equal $1/3$, we need $\frac{n_I}{n_{II}} = 1$, which means the two cells must transfer the same number of electrons.

Step 5: Match electrons and voltages to a pair.
So we look for a pair where both cells move the same number of electrons and Cell II has about three times the EMF of Cell I. The pair where Cell I is the zinc copper (Daniell) cell and Cell II is the aluminium silver cell fits, because their electron counts can be matched and the aluminium silver combination has a much larger standard EMF than zinc copper.

Step 6: State the result.
The option that satisfies both the equal electron condition and the threefold voltage gap is the Zn|Cu paired with Al|Ag choice.

\[ \boxed{\text{Cell I: Zn}|\text{Zn}^{2+}||\text{Cu}^{2+}|\text{Cu and Cell II: Al}|\text{Al}^{3+}||\text{Ag}^+|\text{Ag}} \]