Galvanic Cells

[darkheavens]

Hey guys, could any of you please help me with this question?

A simple galvanic cell based on the reactants MnO4-/H+ and Cu is shown in the diagram below...

So basically the question was asking to write the overall balanced equation that would represent the reaction occuring in this cell. How do you work out which species is being oxidised/reduced? MnO4- isn't on the electrochemical series but Mn2+ is.. And would the H+ take any part in the reaction? Ie how would you know whether or not the H+ reduces into hydrogen gas?

Any ideas?

[scocliffe09]

To me it seems like the question needs more information.
This (wikipedia) is what happens:
In an acidic solution, manganate(VII) is reduced to the colourless +2 oxidation state of the manganese(II) (Mn2+) ion.
8 H+ + MnO4− + 5 e− → Mn2+ + 4 H2O

Then...
Possible oxidants: H+, H2O, Mn2+, Cu2+
Possible reductants: H2O ?Cu (you didn't say what the electrode is made of)

In any case, you have no "top left bottom right" relationship so there should be no reaction. As far as I can tell. It would be helpful to know what the electrodes are made of and the exact composition of each cell. Does it state anywhere that it's galvanic and not electrolytic?

[darkheavens]

Thanks for the reply!
Yup, it says in the question that it's a galvanic cell, not electrolytic.
And the answer states that the MnO4- has been reduced to Mn2+...so I guess that's a fact that we have to remember then?
A later question asked us to work out the electrodes...electrode 1 being Pt and electrode 2 being Cu
However, since we don't know the relative positions of MnO4- and Cu2+, we can't be sure of the answers then?

[scocliffe09]

Dodgy little question.
From what wikipedia says, I think the reduction of MnO4- is spontaneous in acidic solution, which means that the MnO4- is going to (almost) all turn into Mn2+ without sending any electrons through the wire, because it already has access to a reductant (H+). Which would lead me to say that species left for reaction are not going to react - because Cu is both the strongest oxidant and strongest reductant.
But yeah. I think it's not a great question and I wouldn't worry about it too much.

[Mao]

Dodgy little question.
From what wikipedia says, I think the reduction of MnO4- is spontaneous in acidic solution, which means that the MnO4- is going to (almost) all turn into Mn2+ without sending any electrons through the wire, because it already has access to a reductant (H+). Which would lead me to say that species left for reaction are not going to react - because Cu is both the strongest oxidant and strongest reductant.
But yeah. I think it's not a great question and I wouldn't worry about it too much.

I disagree. Reduction of MnO4- can only occur when there is a reductant (such as Cu(s)), the half reaction requires electrons. A solution of acidified permanganate solution is stable.

You are expected to know the reaction of permanganate and dichromate. You are expected to know that both of these are strong oxidants, permanganate reduces to Mn(II), and dichromate reduces to Cr(III).

[scocliffe09]

ok - but then how are we supposed to know the E0 value for the MnO4- --> Mn2+ reaction and predict what will happen?

[Mao]

ok - but then how are we supposed to know the E0 value for the MnO4- --> Mn2+ reaction and predict what will happen?

It is a strong oxidant ==> It will react with most reductants, such as Cu(s)