I2 (s) + 5Cu^2+(aq) + 6H2O(l) ----> 2IO3^-(aq) + 5Cu(s) + 12H^+ (aq) reduction h
ID: 800235 • Letter: I
Question
I2 (s) + 5Cu^2+(aq) + 6H2O(l) ----> 2IO3^-(aq) + 5Cu(s) + 12H^+ (aq)
reduction half reactions
I2(s) + 2e^- --> 2I^-(aq) =+.54
Cu^2+(aq) + 2e^- --> Cu(s) =+.34
i know the answer that it is nonspontaneous which means that:
Ecell= (reduction process(cathode))-(oxidation process(anode)) is negative (-.2) where the copper cell is the reduction process but how do i know to choose the copper cell as the cathode? Why isn't the Iodine cell the cathode? I think the iodine cell should be the cathode/reduction because it is more positive.
MAIN QUESTION: Both Half-Reactions are being reduced in the larger reaction above so how do i know when to pick which is cathode and which is anode?
Explanation / Answer
First reduction half cells are given by
Cu2+ + 2e- ---> Cu , Eo = 0.34
IO3- + 6H+ + 5e- -----------> I2 + 3H2O , Eo = 1.2 V ( this is not simple reduction its a equation hence different from I2---> 2I- , here I from + 5 changed to 0)
Eo cell = 0.34 -1.2 = -0.86 , hence non spontaneous .
In the process Iodine is oxidised from 0 oxidation state to + 5 , Yes reverse reaction is more spontaneous than forward. If spontaneous cell is required Iodine should be cathode where reduction happens and Copper should be anode where oxidation ocuurs .
In reverse process Eo = 1.2-0.34 = 0.86 ,
For spontaneous cell reactions to occur we need cathode of element which has more reduction than others. Like in this case Iodine.
For anode we choose element. which has lesser redcution potential .
for example in copper - zinc cell
where Cu2+ + 2e- ------> Cu , Eo = 0.34
Zn2+ + 2e- --------> Zn , Eo = -0.76
we here choose Copper as cathode where Cu2+ + 2e- -------> Cu reaction occurs
and Zn as anode where oxidation occurs Zn -> Zn2+ + 2e- ,
cell reaction is Cu2+ + Zn ----> Zn2+ + Cu
Eo cell = ( 0.34) -(-0.76) = 1.1 ( spontaneous)