NEST2026ChemistryElectrochemistry
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 = Q₀ , 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 ( G_I/ G_ II ). Assuming standard cell reactions written in the spontaneous direction
Options
- ACell I: Zn(s) | Zn ²⁺ (aq) || Cu ²⁺ (aq) | Cu(s) Cell II: Al(s) | Al ³⁺ (aq) || Ag ^+ (aq) | Ag(s)
- BCell I: Zn(s) | Zn ²⁺ (aq) || Ag ^+ (aq) | Ag(s) Cell II: Zn(s) | Zn ²⁺ (aq) || Cu ²⁺ (aq) | Cu(s)
- CCell I: Fe ²⁺ (aq) | Fe ³⁺ (aq) || Cu ²⁺ (aq) | Cu ^+ (aq) Cell II: Cu(s) | Cu ²⁺ (aq) || Ag ^+ (aq) | Ag(s)
- DCell I: Ag ^+ (aq) | Ag(s) || Fe ²⁺ (aq) | Fe ³⁺ (aq) Cell II: Al(s) | Al ³⁺ (aq) || Ag ^+ (aq) | Ag(s)
Correct answer
B. Cell I: Zn(s) | Zn ²⁺ (aq) || Ag ^+ (aq) | Ag(s) Cell II: Zn(s) | Zn ²⁺ (aq) || Cu ²⁺ (aq) | Cu(s)
Step-by-step solution
The relationship between the Gibbs free energy change ( G ) and the electromotive force ( E ) of a cell is given by the equation: 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: G_I = -n_I F E_I G_ II = -n_ II F E_ II Taking the ratio of the Gibbs free energy changes for the two cells: G_I G_ II = -n_I F E_I -n_ II F E_ II = ( n_I n_ II ) E_I E_ II The problem states that E_I E_ II = G_I G_ II . Subst