[Catherina]

Hi all!

I'm very confused on two questions, and I'm hoping someone can help explain things a bit more clearly.

1). Predict the cathode, anode and net cell reactions and minimum potential difernce for each of the following electrolytic cells.

a) C | Ni(2+) , I(-) | C

b) Pt | Na(+), OH(-) | Pt


And second

2).  What is the minimum electric potential difference of an external power supply that produces chemical changes in the following electrolytic cells?

a) C | Cr(3+), Br(-) | C

b) Cu | Cu(2+) , SO4(2-)  | Cu


My first question is, what exactly is the difference between these two questions? (besidse not having to write the anode, cathode equations.?)

What I have so far for 1a) is this:

Ni(2+) + 2e- --> Ni     -0.26 V

2[I(-) --> I + 1e-]    +0.54 V

net equation: Ni(2+) + 2I(-)  --> Ni + I

I have no idea if that's correct, or if I'm even on the right track.   When it comes to the b question, the OH- is what is throwing me off.  Do you start adding H2O and OH-? 

*delete me*    Please!! I have been working on this particular lesson for a week and am starting to go batty!!

[DevaDevil]

1) I think you should assume the cell is charging. In other words that the nickel in a) is being reduced and the Iodine is being oxidised. Reduction takes place at the cathode, oxidation at the anode.

The minimum potential required can be found by finding the standard electrode potentials for the components, and setting up the equation for the whole cell

for a) half cell reactions are correct. However the potential of the I₂/I^- couple is reversed as the reaction goes the other way.
Ni²⁺ + 2 e^- --> Ni    E = -0.25 V
2 I^- --> I₂ + 2 e^-    E = -0.54 V

This means for the cell E_cell = -0.25 + -0.54 = -0.79 V, in other words the minimum potential difference to drive the reaction is 0.79 V

I will leave b) for you to try. the cell is assumed to contain water as well of course. (NaOH solution in water)


2)
This goes the same way as in 1) as far as I can tell.

[Catherina]

Ok, for question b) , this is what I came up with:

4[ Na(+) _ 1e- --> Na]     -2.71V

4OH(-) --> O2 + 2H2O + 4e-  +0.40

So the min. potention would be -2.71 + .40 = -2.31 ?

Why exactly do you end up reversing the charge at the end?

[DevaDevil]

the reason I "reverse the charge" is the following:

for the following reaction:
O₂(g) + 2H₂O(l) + 4e^– → 4OH^–(aq)    E = +0.40 V

which means that for the reverse reaction:
4OH^–(aq) → O₂(g) + 2H₂O(l) + 4e^–   E = - 0.40 V
and this is the one you have.

The sodium side looks fine

So the minumum potential would be: -2.71 + -0.40 = -3.11 V

[Catherina]

Ok.. I think I get it.  Thank you!!