Pt/I2,I-1 //Au+3/Au
according to the given galvanic cell representation
left side electrode acts as anode and right side electrode acts
as cathode.
oxidation half reaction at
anode           Â
[ 2 I1- -------------------- I2 + 2e-]x3
reduction half reaction at
cathode      [ Au+3 +3 e-
----------------- Au]x2
                                               Â
-----------------------------------------------------------------
                                                     Â
6 I1-Â Â + 2 Au+3Â Â
------------------- 3 I2 + 2 Au
                                        Â
----------------------------------------------------------------------------
E0 of I2/I1- = 0.535V
                                   Â
E0 of Au+3/Au = 1.50V
E0 cell = E0 cathode - E0 anode
E0cell = 1.50 - 0.535
E0cell = 0.965V
Totally 6 electrons are transferred .