H2 Chemistry: Electrolysis

Audi SG

In this topic, the formula for Ecell is Ecell= E(reduction half cell) - E (oxidation half cell). So usually we look at the standard electrode potential, the one with the more +ve value undergoes reduction, fit into formula to get ans. Besides questions asking us to determine whether a rxn is feasible (that one dont have to see which one is more +ve or -ve, we look at the reaction equation to see which rxtant undergoes oxid. or reduction), I chanced upon this question which wasnt properly explained to us in school and which I dont understand why we DONT have to see which eqn is MORE +ve or -ve to determine whether they undergo oxid or reduc. Here goes:

When aq solution of Copper II sulfate and potassium iodide are mixed, a brown ppt is formed. On addition of aq sodium thiosulfate to this mixture, the brown colour disappears and a white ppt REMAINS (means it is formed previously and stays there or the white ppt has just formed and stayed there)?

What is the std cell potential of the rxn of ppotassium iodide with copper II sulfate?

During lecture we werent given much explanation regarding this. The teacher just said well, I2 was formed and Cu+ was formed...

then write this equation

2I- ---> I2 + 2e- +0.54V

Cu2+ +e --->Cu+ +0.15V

Ecell = (+0.15) - (+0.54) = -0.39V

From the 2 equations of cuz i can see that Cu2+ undergoes reduc and I- undergoes oxidation. But why in this case we look at the equation to determine which one undergoes oxid or reduc? Why not the numbers? (quite silly question but still want to ask)

If the brown colour I2 disappears eventually, why do we still use the equation 2I- ---> I2 + 2e- ?? shouldnt it be the opposite way round?

If possible, can someone write for me the entire equations of this whole reaction? (for better understanding) thanks

UltimaOnline SG

In the 1st reaction :

Cu2+ reduced to Cu+, and I- oxidized to I2

Ppt formed is copper(I) iodide, which is white; but because iodine is brown, the result appears to be a brown ppt (which is really a mixture of two separate species, one white one brown).

In the 2nd reaction :

S2O3 2- oxidized to S4O6 2-, and I2 reduced to I-.

Hence the brown colour dissapears (I2 reduced to colourless I-), while the white ppt (copper(I) iodide) remains.

The following calculations assume standard molarities for all species (per molar quantities specified by equation), which of course isn't actually the case (based on the stoichiometry of the balanced redox eqns; furthermore the following reaction would not be feasible or spontaneous in the forward direction with a negative cell potential). So the following is "just for argument's sake".

For 1st reaction :

2I- ---> I2 + 2e- -0.54V oxidation potential at anode

Cu2+ + e- ---> Cu+ +0.15V reduction potential at cathode

E cell = Reduction potential at Cathode + Oxidation potential at Anode

= (+0.15) + (-0.54) = -0.39V

For 2nd reaction :

2S2O3 2- ---> S4O6 2- + 2e- -0.09V oxidation potential at anode

I2 + 2e- ---> 2I- +0.54V reduction potential at cathode

E cell = Reduction potential at Cathode + Oxidation potential at Anode

= (+0.54) + (-0.09) = +0.45V

T3eee SG

i suck.