Correct option is 4E1 = E2;ΔG1 = 2ΔG2

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Reaction Emf Gibbs free energy
$H_{2} + {Cl}_{2} \to 2 HCl$ $E_{1}$ ${ΔG}_{1}$
$\frac{1}{2} H_{2} + \frac{1}{2} {Cl}_{2} \to HCl$ $E_{2}$ ${ΔG}_{2}$
Correct relation is

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E1 = E2 ;ΔG1 = ΔG2

E1 = 2E2;ΔG1 = 2ΔG2

E1 = 2E2; ΔG1 = ΔG2

E1 = E2;ΔG1 = 2ΔG2

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detailed solution

Correct option is D

Cell potential and electro potentials are intensive properties …. uninfluenced by stoichiometry where as Gibbs free energy is an extensive property.∴ E1 = E2& ΔG1 = 2ΔG2

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	Column I		Column II
A.	${Fe}^{3 +} and I^{-}$	p.	$E_{cell}^{\circ} = + ve$
B.	${Fe}^{3 +} and {Br}^{-}$	q.	${ΔG}^{\circ} = + ve$
C.	$Zn and H^{+}$	r.	$E_{cell}^{\circ} = - E_{electrode}^{\circ}$
D.	$Br and {Fe}^{2 +}$	s.	${ΔG}^{\circ} = - ve$

Reaction	Emf	Gibbs free energy
$H_{2} + {Cl}_{2} \to 2 HCl$	$E_{1}$	${ΔG}_{1}$
$\frac{1}{2} H_{2} + \frac{1}{2} {Cl}_{2} \to HCl$	$E_{2}$	${ΔG}_{2}$

ReactionEmfGibbs free energyH2 + Cl2 → 2HClE1ΔG112H2+12Cl2 → HClE2ΔG2Correct relation is

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Reaction Emf Gibbs free energy
$H_{2} + {Cl}_{2} \to 2 HCl$ $E_{1}$ ${ΔG}_{1}$
$\frac{1}{2} H_{2} + \frac{1}{2} {Cl}_{2} \to HCl$ $E_{2}$ ${ΔG}_{2}$
Correct relation is