[sinthreck]

Which reaction takes place at the cathode and the anode when molten MnI₂ is electrolyed?

The solutions then states that, "The cathode reaction will be the reaction with the most positive reduction potential".

Cathode: Mn²⁺ + 2e^- Mn        E⁰ = -1.18 V
Anode: 2I^- I₂ + 2e^-            -E⁰ = -0.54


The statement provided contradicts the solution. Clearly, The reduction of potential of I₂ (+0.54) is greater than Mn²⁺ (-1.18).

Yet the solution has Mn2+ at the cathode.

Are they wrong?

[Chymysta]

Remember there are 2 types of electrochemical cells: electrolytic cells and galvanic cells. What is the difference between them and which type are you dealing with here?

[sinthreck]

With a galvanic cell a redox reaction occurs spontaneously and electric current is produced.

An electrolytic cell requires electric current to drive a non-spontaneous redox reaction.

With my question, if you had I₂ being reduced at the cathode and Mn being oxidised at the anode, you would have a galvanic cell (E_cell = 1.74V).

I just don't like how the solution (and the explanation in both my textbooks) always says, "The cathode reaction will be the reaction with the most positive reduction potential.

I think it would be more correct to say:

"The cathode reaction will be the reaction with a negative potential that is closest to zero".  What do you guys think?

[Chymysta]

The cathode is always where the electrons go to, where reduction occurs.

Because the MgI2 is being electrolyzed, you are putting work in to make this reaction happen. So which type of cell is it?

[sinthreck]

The question states which kind of cell it is!

[Chymysta]

Then something is definitely missing or wrong. The cathode in a galvanic cell is the half reaction more likely to reduce (most positive reduction potential) but if it's an electrolytic cell, its the opposite.  "Which reaction takes place at the cathode and the anode when molten MnI2 is electrolyed?" implies an electrolytic cell, unless there is more to the question.

[Borek]

The statement provided contradicts the solution. Clearly, The reduction of potential of I₂ (+0.54) is greater than Mn²⁺ (-1.18).

There is no I₂ that can be reduced in the solution, there is I^-, and it can't be reduced any further.