Q25 / 79 · NEST (NISER) 2026 · Chemistry
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 $Q_I = Q_{II} = Q_0$, the ratio of the EMF of the two cells ($E_I/E_{II}$) is exactly equal to the ratio of the change of Gibbs free energy for the two cells ($\Delta G_I/\Delta G_{II}$). Assuming standard cell reactions written in the spontaneous direction and reversible operation, the pair of cells that can satisfy these conditions is:
- ACell I: $\text{Zn(s)}|\text{Zn}^{2+}\text{(aq)}||\text{Cu}^{2+}\text{(aq)}|\text{Cu(s)}$
Cell II: $\text{Al(s)}|\text{Al}^{3+}\text{(aq)}||\text{Ag}^+\text{(aq)}|\text{Ag(s)}$ - Correct answerBCell I: $\text{Zn(s)}|\text{Zn}^{2+}\text{(aq)}||\text{Ag}^+\text{(aq)}|\text{Ag(s)}$
Cell II: $\text{Zn(s)}|\text{Zn}^{2+}\text{(aq)}||\text{Cu}^{2+}\text{(aq)}|\text{Cu(s)}$ - CCell I: $\text{Fe}^{2+}\text{(aq)}|\text{Fe}^{3+}\text{(aq)}||\text{Cu}^{2+}\text{(aq)}|\text{Cu}^+\text{(aq)}$
Cell II: $\text{Cu(s)}|\text{Cu}^{2+}\text{(aq)}||\text{Ag}^+\text{(aq)}|\text{Ag(s)}$ - DCell I: $\text{Ag}^+\text{(aq)}|\text{Ag(s)}||\text{Fe}^{2+}\text{(aq)}|\text{Fe}^{3+}\text{(aq)}$
Cell II: $\text{Al(s)}|\text{Al}^{3+}\text{(aq)}||\text{Ag}^+\text{(aq)}|\text{Ag(s)}$
Solution
The relationship between the Gibbs free energy change ($\Delta G$) and the electromotive force ($E$) of a cell is given by the equation:
$\Delta G = -nFE$
where $n$ is the number of moles of electrons transferred in the balanced cell reaction and $F$ is the Faraday constant.
For Cell I and Cell II, we can write:
$\Delta G_I = -n_I F E_I$
$\Delta G_{II} = -n_{II} F E_{II}$
Taking the ratio of the Gibbs free energy changes for the two cells:
$\dfrac{\Delta G_I}{\Delta G_{II}} = \dfrac{-n_I F E_I}{-n_{II} F E_{II}} = \left(\dfrac{n_I}{n_{II}}\right) \dfrac{E_I}{E_{II}}$
The problem states that $\dfrac{E_I}{E_{II}} = \dfrac{\Delta G_I}{\Delta G_{II}}$. Substituting this into our ratio equation gives:
$\dfrac{E_I}{E_{II}} = \left(\dfrac{n_I}{n_{II}}\right) \dfrac{E_I}{E_{II}}$
This implies that $\dfrac{n_I}{n_{II}} = 1$, which means $n_I = n_{II}$. Therefore, the two cells must have the same number of electrons involved in their balanced redox reactions.
Let us determine the number of electrons transferred ($n$) for the cells in each option:
In option (1):
Cell I: $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$ ($n_I = 2$)
Cell II: $Al + 3Ag^+ \rightarrow Al^{3+} + 3Ag$ ($n_{II} = 3$)
Here $n_I \neq n_{II}$.
In option (2):
Cell I: $Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$ ($n_I = 2$)
Cell II: $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$ ($n_{II} = 2$)
Here $n_I = n_{II} = 2$, satisfying the condition.
In option (3):
Cell I: $Fe^{2+} + Cu^{2+} \rightarrow Fe^{3+} + Cu^+$ ($n_I = 1$)
Cell II: $Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag$ ($n_{II} = 2$)
Here $n_I \neq n_{II}$.
In option (4):
Cell I: $Ag + Fe^{3+} \rightarrow Ag^+ + Fe^{2+}$ ($n_I = 1$)
Cell II: $Al + 3Ag^+ \rightarrow Al^{3+} + 3Ag$ ($n_{II} = 3$)
Here $n_I \neq n_{II}$.
Only the pair of cells in option (2) involves the same number of electrons transferred during the redox process.
Answer: Cell I: $\text{Zn(s)}|\text{Zn}^{2+}\text{(aq)}||\text{Ag}^+\text{(aq)}|\text{Ag(s)}$
Cell II: $\text{Zn(s)}|\text{Zn}^{2+}\text{(aq)}||\text{Cu}^{2+}\text{(aq)}|\text{Cu(s)}$
$\Delta G = -nFE$
where $n$ is the number of moles of electrons transferred in the balanced cell reaction and $F$ is the Faraday constant.
For Cell I and Cell II, we can write:
$\Delta G_I = -n_I F E_I$
$\Delta G_{II} = -n_{II} F E_{II}$
Taking the ratio of the Gibbs free energy changes for the two cells:
$\dfrac{\Delta G_I}{\Delta G_{II}} = \dfrac{-n_I F E_I}{-n_{II} F E_{II}} = \left(\dfrac{n_I}{n_{II}}\right) \dfrac{E_I}{E_{II}}$
The problem states that $\dfrac{E_I}{E_{II}} = \dfrac{\Delta G_I}{\Delta G_{II}}$. Substituting this into our ratio equation gives:
$\dfrac{E_I}{E_{II}} = \left(\dfrac{n_I}{n_{II}}\right) \dfrac{E_I}{E_{II}}$
This implies that $\dfrac{n_I}{n_{II}} = 1$, which means $n_I = n_{II}$. Therefore, the two cells must have the same number of electrons involved in their balanced redox reactions.
Let us determine the number of electrons transferred ($n$) for the cells in each option:
In option (1):
Cell I: $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$ ($n_I = 2$)
Cell II: $Al + 3Ag^+ \rightarrow Al^{3+} + 3Ag$ ($n_{II} = 3$)
Here $n_I \neq n_{II}$.
In option (2):
Cell I: $Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$ ($n_I = 2$)
Cell II: $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$ ($n_{II} = 2$)
Here $n_I = n_{II} = 2$, satisfying the condition.
In option (3):
Cell I: $Fe^{2+} + Cu^{2+} \rightarrow Fe^{3+} + Cu^+$ ($n_I = 1$)
Cell II: $Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag$ ($n_{II} = 2$)
Here $n_I \neq n_{II}$.
In option (4):
Cell I: $Ag + Fe^{3+} \rightarrow Ag^+ + Fe^{2+}$ ($n_I = 1$)
Cell II: $Al + 3Ag^+ \rightarrow Al^{3+} + 3Ag$ ($n_{II} = 3$)
Here $n_I \neq n_{II}$.
Only the pair of cells in option (2) involves the same number of electrons transferred during the redox process.
Answer: Cell I: $\text{Zn(s)}|\text{Zn}^{2+}\text{(aq)}||\text{Ag}^+\text{(aq)}|\text{Ag(s)}$
Cell II: $\text{Zn(s)}|\text{Zn}^{2+}\text{(aq)}||\text{Cu}^{2+}\text{(aq)}|\text{Cu(s)}$
Hence, the correct answer is option: 'B'.