AQA A Level Chemistry Electrochemistry

https://irp-cdn.multiscreensite.com/22e27f06/files/uploaded/Electrochemical%20cell%202.pdf

For Q3.b)How did they work out the vanadium species and how did they form the half equations please?

And how would you know that the cell emf increases?

Reply 1

Snails:)3721

I hope this is vaguely helpful. I put charges in brackets- sorry that it’s hard to read
For i) initially you have V(2+) —->V(3+)
This is feasible as E is >0 (1.52-(-0.26)=+1.78V))
Then, because you have excess MnO4- and the half cell potential of the MnO4- is still greater than for the vanadium (which is now all V(3+),
V(3+) —> VO(2+)
And in a similar way,
VO(2+) —> VO2(+)
Now all the vanadium is VO2+ and there is no reaction involving vanadium with a Eo greater than 0, so it will stay like that.
The half equation is just the overall change in the vanadium- you start with V(2+) and finish with VO2(+) and balance it like any other half equation (lmk if you want me to explain)

For ii) basically I think the way my teacher said (can’t actually remember but it seems to work, even if the chemistry is wrong). If you add more Fe3+, more Fe3+ will be converted to Fe2+, so more electrons will be used up in the positive electrode, so there will be a greater p.d because the negative electrode still has the same number of electrons as before.
I know that that is not very well explained. If there’s anything you think I can explain better please say :smile: