Writing a Half Reaction from a word problem (Homework #13.17

[Glenda Marshall DIS 3M]

If given a word problem that only tells you what you start with in a reaction, how do you know what the reduced and oxidized forms of the ions given are? For example homework 13.17 says

"Write the balanced half reactions for the redox reaction of an acidified solution of potassium permanganate and iron (II) chloride."

How would you know that permanganate would turn into M2+?

Thanks!

[Chem_Mod]

The half-reactions for permanganate would be given in the table/appendix. Unfortunately, there are two different ones:

MnO4- ---> Mn2+ (acidic condition)
MnO4- ---> MnO2 (basic condition)

Permanganate is the only commonly encountered example of having different products for acidic/basic, and the table won't tell you which is which, so this one thing is probably worth memorizing.

[Megan White 2C]

I have an additional question for problem 13.17:

Where does the H+ ion come from on the CATHODE side of the cell diagram that we are told to draw? When I was drawing my diagram, I came up with:

Pt(s)| Fe 2+, Fe 3+|| MnO4-, Mn 2+| Pt(s)

But the answer in the back of the book reads:

Pt(s)| Fe 2+, Fe 3+|| MnO4-, Mn 2+, H+| Pt(s)

Where in the equation does this H+ ion originate, and what indicates that it belongs on the cathode side of the diagram?

Thank you so much to all who can help me out on this!

[Niharika Reddy 1D]

The half reaction for MnO₄^- in acidic solution is the following and it occurs at the cathode since this half reaction has the larger of the 2 half reactions' standard reduction potentials, and we want to devise a (spontaneous) galvanic cell:

MnO₄^- + 8H⁺ + 5e- -> Mn²⁺ + 2H₂O

Since H⁺ is present in the reduction half reaction, it appears on the cathode side of the cell diagram.

[Chem_Mod]

You have potassium permanganate (KMnO₄) and iron(II)chloride (FeCl₂). Potassium and chlorine are spectator ions and thus do not take react. Therefore, we only worry about the MnO₄^- and the Fe⁺².

The half reactions for MnO₄^- and the Fe⁺² in an acidic medium are:
MnO₄^- + 8H⁺ +5e- Mn⁺² + 4H₂O
Fe Fe⁺² + 2e-

You need to then balance the mole electrons so you multiply the first equation by 2 and the second equation by 5 to get...
2MnO₄^- + 16H⁺ +10e- 2Mn⁺² + 8H₂O
5Fe 5Fe⁺² + 10e-

The overall reaction becomes...
2MnO₄^- + 16H⁺ +5Fe 2Mn⁺² + 8H₂O + 5Fe⁺²