# A2 redox exam question help!!

Can someone help me with question 3c please, unit 5.pdf
thanks xx
VO2+ is the final state you can get to from these equations, with total E still > 0(infact i got +0.44V) hence is thermodynamically allowed.

combine the equations
1, 3, 6 and 8
where your overall change takes place stepwise, and you deduct E from value of eqn 8 based on
Ered - Eox.

so (8) - (1) then -(2) then (-6); check at each stage, E is sufficiently positive to allow the redox.
The manganate ions will be reduced, so their standard electrode potential values must be more positive than those of vanadium half-equations if they are to act as oxidising agents. As you can see, all the vanadium half-equations shown have more negative potentials than the manganates, so the vanadium will oxidise through to the highest oxidation state shown. Which is in VO2+.

You're starting with V2+ and ending with VO2+.

Half equation: V2+ + zH2O----> VO2+ + ye + xH+

Balance z,y, and x and perhaps rewrite as a reduction half-equation depending on your boards conventions.
klgyal
Can someone help me with question 3c please, unit 5.pdf
thanks xx

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An excess of acidified potassium manganate(VII) was added to a solution containing V2+(aq) ions. Use the data given in the table to determine the vanadium species present in the solution at the end of this reaction. State the oxidation state of vanadium in this species and write a half-equation for its formation from V2+(aq).
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set out the half equation for MnO4-. It reacts as an oxidising agent with the V2+ and is reduced (therefore the V2+ is a reducing agent and is oxidised).

E = E(red) - E(ox) = +1.52 - - 0.26 = +1.78V

This is positive and greater than 0.3V and so is spontaneous.

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Now look at the species formed by oxidation of V2+. This is V3+.

Now perform the same analysis only using V3+ as the reducing agent with MnO4-

E = E(red) - E(ox) = +1.52 - +0.34 = +1.18V

This is positive and greater than 0.3V and so is spontaneous.

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Now look at the species formed by oxidation of V3+. This is VO2+.

Now perform the same analysis only using VO
Original post by charco

Original post by rimz3y
why do u keep comparing to 0.3v

It's a rather arbitary value used to assess the probability of the reaction going to completion.

A positive value for means that the forward reaction is spontaneous.

When the value is between 0 and 0.3V is is likely that an equilibrium is established.

Above 0.3V the reaction is likely to go 'all the way'
Original post by shengoc
VO2+ is the final state you can get to from these equations, with total E still > 0(infact i got +0.44V) hence is thermodynamically allowed.

combine the equations
1, 3, 6 and 8
where your overall change takes place stepwise, and you deduct E from value of eqn 8 based on
Ered - Eox.

so (8) - (1) then -(2) then (-6); check at each stage, E is sufficiently positive to allow the redox.

I think the potassium manganate is in excess so it doesn't matter if E is -ve...
I don't get how they did the half equationswhich 2 equations did they combine?
Original post by LearningMath
The manganate ions will be reduced, so their standard electrode potential values must be more positive than those of vanadium half-equations if they are to act as oxidising agents. As you can see, all the vanadium half-equations shown have more negative potentials than the manganates, so the vanadium will oxidise through to the highest oxidation state shown. Which is in VO2+.

You're starting with V2+ and ending with VO2+.

Half equation: V2+ + zH2O----> VO2+ + ye + xH+

Balance z,y, and x and perhaps rewrite as a reduction half-equation depending on your boards conventions.

how did they get the half equation
?