For the following two half-cell reaction Fe3+ + e- --> Fe2+ and I3- + 2e- --> 3 I- determine the spontaneous reaction (balanced). Must give a short reason for choice. Calculate the standard cell potential. Calculate the free energy of the reaction at standard conditions
first find the reduction potentials
Fe+3 + e- ---> Fe+2 Eo = 0.77
I3- + 2e- --> 3I- Eo = 0.53
we know that
the electrode with more reduction potential is cathode
so
iron is cathode
now
cathode : reduction
2Fe+3 + 2e ---> 2Fe+2
anode : oxidation
3I- + 2e- --> I3-
so
the balanced reaction is
3I- + 2Fe+3 ---> I3- + 2Fe+2
now
Eo cell = Eo cathode - Eo anode
Eo cell = 0.77 - 0.53
Eo cell = 0.24 V
now
dGo = - n x F x Eo
here
n =2 as two electrons are transferred
so
dGo = - 2 x 96485 x 0.24
dGo = -46313
dGo = - 46.313 kJ
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