Answer:
[tex]\Delta G^{0}[/tex] for the given reaction is 106150 J/mol
Explanation:
Oxidation: [tex]2Br^{-}-2e^{-}\rightarrow Br_{2}[/tex]
Reduction: [tex]I_{2}+2e^{-}\rightarrow 2I^{-}[/tex]
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Overall: [tex]I_{2}+2Br^{-}\rightarrow 2I^{-}+Br_{2}[/tex]
We know, [tex]\Delta G^{0}=-nFE^{0}[/tex]
where, n is number of electron exchanged during overall reaction and 1 F equals to 96500 C/mol
Here, n = 2 and [tex]E^{0}=-0.55V[/tex]
So, [tex]\Delta G^{0}=-(2)\times (96500C/mol)\times (-0.55V)=106150J/mol[/tex]
