Need Help on an Electrochem Q

Use half-equations from the table to deduce an equation for the reduction of VO2^(+) to VO^(2+) in aqueous solution by Iron.

[Fe(H2O)6]^2+ (aq) + 2e- --> Fe (s) + 6H2O (l) E' = -0.44V
H+ (aq) + e- --> 1/2H2 (g) E' = 0.00V
[Fe(H2O)6]^3+ (aq) + e- --> [Fe(H2O)6]^2+ (aq) E' = 0.77V
VO2^(+) (aq) + 2H+ (aq) + e- --> VO^(2+) (aq) + H2O (l) E' = 1.00V

The answer I got was:
2VO2^(+) + 4H+ + Fe + 4H2O ---> 2VO^(2+) + [Fe(H2O)6]^2+
But the correct answer on the mark scheme is:
3 VO2^(+) + 6 H+ + Fe + 3 H2O → 3 VO^(2+) + [Fe(H2O)6]^3+

Can someone please explain to me how to arrive at the correct answer? It seems I have combined the wrong half-equations together but I'm not sure why Fe3+ is used and not Fe2+ and why this would mean that the VO2+ equation is multiplied by 3 instead of 2.

Thankyou in advance

The electrode potential for [Fe(H2O)6]^3+ + e^- —> [Fe(H2O)6]^2+ is + 0.77 V and the electrode potential for VO2^+ + 2H^+ + e^- --> VO^2+ + H2O is +1.00 V.

Can you infer whether the [Fe(H2O)6]^2+ ions will be oxidised by VO2^+ from this information?

Yes? I mean its electrode potential is more negative (or less positive) so the equation would flip?

Ok, so if [Fe(H2O)6]^2+ is oxidisable by the VO2^+, then you will form [Fe(H2O)6]^3+.

How might you refine your answer to account for the fact more VO2^+ ions will react with the [Fe(H2O)6]^2+ formed initially?

VO2^(+) (aq) + 2H+ (aq) + [Fe(H2O)6]^2+ (aq) ---> VO^(2+) (aq) + H2O (l) + [Fe(H2O)6]^3+ (aq)

Now try adding together this equation and your original answer together. Cancel down anything that appears on each side of the equation

Now try adding together this equation and your original answer together. Cancel down anything that appears on each side of the equation

Please can you explain how to do this?
I understand how we got VO2^(+) (aq) + 2H+ (aq) + [Fe(H2O)6]^2+ (aq) ---> VO^(2+) (aq) + H2O (l) + [Fe(H2O)6]^3+ (aq)
But I don't get how it simplifies to 3VO2^(+) + 6H+ + Fe + 3H2O ---> 3VO^(2+) + [Fe(H2O)6]^3+
Why does Fe no longer have a 2+ charge and why has some of the balancing changed?

Look again at the electrode potentials.

Do they in any way suggest that [Fe(H2O)6]^2+ could react further with the VO2^+ ions?

You may wish to try doing some electrode potential calculations to see what happens.

Thank you for the quick response!!!

I got the right answer after I combined the equations:
Fe (s) + 6H2O (l) ---> [Fe(H2O)6]^2+ (aq) + 2e-
[Fe(H2O)6]^2+ (aq) ---> [Fe(H2O)6]^3+ (aq) + e-
VO2^(+) (aq) + 2H+ (aq) + e- --> VO^(2+) (aq) + H2O (l)

But I don't understand why I needed to use two oxidation equations and how would I know to combine 3 equations in future questions?

When I did the electrode potential calculation for when just this [Fe(H2O)6]^2+ (aq) ---> [Fe(H2O)6]^3+ (aq) + e- cell and this VO2^(+) (aq) + 2H+ (aq) + e- --> VO^(2+) (aq) + H2O (l) cell are connected I got 0.23 (1.00 - 0.77), so doesn't this mean its feasible without also having Fe (s) + 6H2O (l) ---> [Fe(H2O)6]^2+ (aq) + 2e- connected?

The big hint this time was that all three half-equations were given, indicating that all three may have been required.

You did the calculation correctly. If it makes more sense, you can think of the reaction as occurring in two step:

(1. The Fe is oxidised completely to [Fe(H2O)6]^2+ first.

(2. Since there is no Fe left, the [Fe(H2O)6]^2+ is oxidised all the way to [Fe(H2O)6]^3+.

This approach of one reaction occurring after the other is how I tend to think of stepwise reductions and it tends to help when describing what is observed during the demonstration when VO2^+ is reduced with zinc (an example that is used quite frequently in these types of exam questions).