The standard Gibbs free energy change (in joule) forthe given cell reaction isMg(s) Ni2 (aq) - Mg2 (aq)Ni(s)Given:E Mg /Mg2.36 V; E 2/N = 0.25 V)2+li(1) 5.22 F(2)-6.11 F(3) -2.11 F(4) 4.22 F

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Q. The magnitude of standard Gibbs free energy for the given cell reaction in

K J m o l^{- 1} a t 298 K

Z n (s) + C u^{2 +} (a q) \to Z n^{2 +} (a q) + C u (s), E^{\circ} = 2 V a t 298 K

(F a r a d a y^{'} s c o n s t a n t, F = 96000 C m o l^{- 1})

The cell in which the following reaction occurs:

2 F e^{3 +} (a q) + 2 I^{-} (a q) \to 2 F e^{2 +} (a q) + I_{2} (s)

has

E_{c e l l}^{\circ} = 0.236 V

298 K

Calculate the standard Gibbs energy of the cell reaction. (Given :

1 F = 96, 500 C m o l^{- 1})

Q. For the cell reaction

2 {Fe}^{3 +} (aq) + 2 I^{-} (aq) \to 2 {Fe}^{2 +} (aq) + I_{2} (aq) E °_{cell} = 0.24 V at 298 K

The standard Gibbs energy

△ G^{o}

of the cell reaction is:
[Given that Faraday constant F = 96500 C mol–1]

Q. Determine the standard emf of the cell and standard free energy change of the cell reaction.

Z n {/ Z n}^{2 +} ∥ {N i}^{2 +} / N

E^{o}

{Z n}^{2 +} / Z n = - 0.76 V

E^{o}

{N i}^{2 +} / N i = - 0.25 V

The standard electrode potential for an electrochemical cell is $2.2 V$ . Calculate the standard Gibbs energy change for reaction:
$X (s) + Y^{2 +} (a q) \to X^{2 +} (a q) + Y (s)$
(Given, $1 F = 96500 C m o l^{- 1}$ )
(correct answer +1, wrong answer -0.25)

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