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

Similar Questions

Match the following columns.

	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.	${ΔG}^{\circ} = - ve$

$[Given, E_{{Fe}^{3 +} / {Fe}^{2 +}}^{\circ} = 0 .77, E_{[\frac{1}{2} I_{2} / 1^{-}]}^{\circ} = + 0 .54$
$E_{1 / 2 {Br}_{2} / {Br}^{-}} = 1 .09, E_{{Zn}^{2 +} / Zn}^{\circ^{\circ}} = - 0 .76]$