Homework Answers
Solution:-
I2 + 2Br- ----------? 2I- + Br2
I2 + 2e- ----------> 2I- ....Eo = 0.53 V .........(1)
Br2 + 2e- ----------> 2Br- ........Eo = 1.07 V ........(2)
subtracting (2) from (1).....
I2 + 2e- - ( Br2 +2e- ) ---------> 2I- - 2Br- .......Eo = 0.53 - 1.07 = -0.54 V
I2 + 2e- - Br2 - 2e- ----------> 2I- - 2Br- .....Eo = -0.54 V
I2 - Br2 --------> 2I- - 2Br-.......Eo = -0.54 V
I2 + 2Br- -------> 2I- + Br2.........Eo = -0.54 V
and delta Go and Eo are connected through the following relation .......
ΔGo= −nFEo
here n = no.of electrons involved in electrode reaction = 2
F = 9.6484 x 10^4 C/mol
putting the values...
ΔGo = -(2 X 9.6484 X 10^4) X (-0.54) =
= (-192968) x (-0.54)= 104202.72 J/mole
ΔGo =1.0x105J/mole
Hence second is correct option.
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