Given E deg(Cl2/Cl-) = 1.36 V and E deg(Br2/Br-) = 1.09 V, the standard Gibbs energy change for Cl2 + 2Br- -> 2Cl- + Br2 at 298 K (F = 96500 C/mol) is

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Given $E^{\circ} (Cl X_{2} / Cl X^{-}) = 1.36$ V and $E^{\circ} (Br X_{2} / Br X^{-}) = 1.09$ V, the standard Gibbs energy change for $Cl X_{2} + 2 Br X^{-} 2 Cl X^{-} + Br X_{2}$ at 298 K (F = 96500 C/mol) is

E_{ce ll}^{\circ} = 1.36 - 1.09 = 0.27

V and n = 2.

Δ G^{\circ} = - n F E^{\circ} = - 2 \times 96500 \times 0.27 = - 52110

J, which is

- 52.1

kJ/mol.