Chem question help please

Hi, please could I have help on part f, I’m not sure how to do it?
Question: https://ibb.co/XfrJ3Xjv
Thanks!

Reply 1

Hi, please could I have help on part f, I’m not sure how to do it?
Question: https://ibb.co/XfrJ3Xjv
Thanks!

A lot of iron(III) ions can be mixed with vanadium metal in a redox reaction. Here the more reactive vanadium reduces the less reactive iron(III) ions. The balanced reaction is represented as follows:

Fe^3+(aq) + V(s) ==> Fe(s) + V^3+(aq)

Here’s what the symbols mean:
This is the additional iron(III) ions present in the solution, Fe^3+(aq).
V(s) is the vanadium metal.
Fe(s) is the iron metal that is deposited out of solution.
The extracted vanadium(III) ions are V^3+(aq).
Essentially the Iron(III) ions become iron metal, and the vanadium metal becomes vanadium(III) ions.
Bye,
Sandro

(edited 1 month ago)

Reply 2

TypicalNerd

Original post by anonymous56754

Hi, please could I have help on part f, I’m not sure how to do it?
Question: https://ibb.co/XfrJ3Xjv
Thanks!

There are actually several potentially useful equations in the electrode potentials section:

Possible reduction half-reactions:

Fe^3+ (aq) + e^- <=> Fe^2+ (aq) (E° = +0.77 V)
Fe^3+ (aq) + 3e^- <=> Fe (s) (E° = -0.04 V)
Fe^2+ (aq) + 2e^- <=> Fe (s) (E° = -0.44 V)

Possible reversed oxidation half-reactions:

V^2+ (aq) + 2e^- <=> V (s) (E° = -1.20 V)
V^3+ (aq) + e^- <=> V^2+ (aq) (E° = -0.26 V)
VO^2+ (aq) + 2H^+ (aq) + e^- <=> V^3+ (aq) + H2O (l) (E° = +0.34 V)
VO2^+ (aq) + 2H^+ (aq) + e^- <=> VO^2+ (aq) + H2O (l) (E° = +1.00 V)

You want to find reactions where E°(cell) is greater than, or equal to zero.

Recall that E°(cell) = E°(reduction) - E°(reverse oxidation)

So for example, let’s consider whether it is possible for Fe^3+ to be reduced straight to Fe and for V to be oxidised to V^2+.

E°(cell) = (-0.04 V) - (-1.20 V) = +1.16 V > 0.

This reaction is possible, but what’s to say that the vanadium cannot be oxidised further? Let’s put it to the test and see if it can be oxidised to vanadium(III) or vanadium(IV) if the iron(III) reduces fully:

Vanadium(III):

E°(cell) = (-0.04 V) - (-0.26 V) = +0.22 V > 0.

Vanadium(IV):

E°(cell) = (-0.04 V) - (+0.34 V) = -0.38 V < 0.

So if the iron(III) ions reduce as far as iron metal, the vanadium is only oxidised as far as the +3 oxidation state.

You can repeat this process in the event that the iron(III) is only reduced as far as iron(II):

E°(V^2+ made) = (+0.77 V) - (-1.20 V) = +1.97 V > 0
E°(V^3+ made) = (+0.77 V) - (-0.26 V) = +1.03 V > 0
E° (VO^2+ made) = (+0.77 V) - (+0.34 V) = +0.43 V > 0
E° (VO2^+ made) = (+0.77 V) - (+1.00 V) = -0.23 V < 0

So this reduction would only go as far as producing vanadium(IV). However, what if the iron(II) product can oxidise the vanadium(IV)?

E° (cell) = (-0.44 V) - (+1.00 V) = -1.44 V < 0

So this process is not possible.

So we have two possible equations:

(1) Fe^3+ (aq) + V (s) —> Fe (s) + V^3+ (aq)

(2) 4Fe^3+ (aq) + V (s) + H2O (l) —> 4Fe^2+ (aq) + VO^2+ (aq) + 2H^+ (aq)

In reality, because there is excess Fe^3+ present and Fe^3+ reacts with Fe to give Fe^2+, I’d think the second reaction is more likely.

(edited 1 month ago)

Reply 3

anonymous56754

Original post by TypicalNerd

There are actually several potentially useful equations in the electrode potentials section:
Possible reduction half-reactions:
Fe^3+ (aq) + e^- <=> Fe^2+ (aq) (E° = +0.77 V)
Fe^3+ (aq) + 3e^- <=> Fe (s) (E° = -0.04 V)
Fe^2+ (aq) + 2e^- <=> Fe (s) (E° = -0.44 V)
Possible reversed oxidation half-reactions:
V^2+ (aq) + 2e^- <=> V (s) (E° = -1.20 V)
V^3+ (aq) + e^- <=> V^2+ (aq) (E° = -0.26 V)
VO^2+ (aq) + 2H^+ (aq) + e^- <=> V^3+ (aq) + H2O (l) (E° = +0.34 V)
VO2^+ (aq) + 2H^+ (aq) + e^- <=> VO^2+ (aq) + H2O (l) (E° = +1.00 V)
You want to find reactions where E°(cell) is greater than, or equal to zero.
Recall that E°(cell) = E°(reduction) - E°(reverse oxidation)
So for example, let’s consider whether it is possible for Fe^3+ to be reduced straight to Fe and for V to be oxidised to V^2+.
E°(cell) = (-0.04 V) - (-1.20 V) = +1.16 V > 0.
This reaction is possible, but what’s to say that the vanadium cannot be oxidised further? Let’s put it to the test and see if it can be oxidised to vanadium(III) or vanadium(IV) if the iron(III) reduces fully:
Vanadium(III):
E°(cell) = (-0.04 V) - (-0.26 V) = +0.22 V > 0.
Vanadium(IV):
E°(cell) = (-0.04 V) - (+0.34 V) = -0.38 V < 0.
So if the iron(III) ions reduce as far as iron metal, the vanadium is only oxidised as far as the +3 oxidation state.
You can repeat this process in the event that the iron(III) is only reduced as far as iron(II):
E°(V^2+ made) = (+0.77 V) - (-1.20 V) = +1.97 V > 0
E°(V^3+ made) = (+0.77 V) - (-0.26 V) = +1.03 V > 0
E° (VO^2+ made) = (+0.77 V) - (+0.34 V) = +0.43 V > 0
E° (VO2^+ made) = (+0.77 V) - (+1.00 V) = -0.23 V < 0
So this reduction would only go as far as producing vanadium(IV). However, what if the iron(II) product can oxidise the vanadium(IV)?
E° (cell) = (-0.44 V) - (+1.00 V) = -1.44 V < 0
So this process is not possible.
So we have two possible equations:
(1) Fe^3+ (aq) + V (s) —> Fe (s) + V^3+ (aq)
(2) 4Fe^3+ (aq) + V (s) + H2O (l) —> 4Fe^2+ (aq) + VO^2+ (aq) + 2H^+ (aq)
In reality, because there is excess Fe^3+ present and Fe^3+ reacts with Fe to give Fe^2+, I’d think the second reaction is more likely.

I got the first equation but you are right, the answer is the second equation because V2+ is further oxidised to VO2+? But how did you write the equation? I tried to combined the Fe3+ half equation and VO2+ half equation but i'm not getting the correct answer? Thanks!

Reply 4

TypicalNerd

Original post by anonymous56754

Well, I worked out that these equations were needed in my above post from the electrode potentials in the CIE data booklet:

Fe^3+ (aq) + e^- —> Fe^2+ (aq) (as is)

V^2+ (aq) + 2e^- —> V (s) (to reverse)

V^3+ (aq) + e^- —> V^2+ (to reverse)

VO^2+ (aq) + 2H^+ (aq) + e^- —> V^3+ (aq) + H2O (l) (to reverse)

Combining all the equations to reverse and then reversing the result:

V (s) + H2O (l) —> VO^2+ (aq) + 2H^+ (aq) + 4e^-

Now notice how each Fe^3+ accepts 1 electron and that this half-reaction produces 4? That is to say we need to add 4Fe^3+ to the LHS and 4Fe^2+ to the RHS:

4Fe^3+ (aq) + V (s) + H2O (l) —> 4Fe^2+ (aq) + VO^2+ (aq) + 2H^+ (aq)

Reply 5

anonymous56754

Original post by TypicalNerd

Well, I worked out that these equations were needed in my above post from the electrode potentials in the CIE data booklet:
Fe^3+ (aq) + e^- —> Fe^2+ (aq) (as is)
V^2+ (aq) + 2e^- —> V (s) (to reverse)
V^3+ (aq) + e^- —> V^2+ (to reverse)
VO^2+ (aq) + 2H^+ (aq) + e^- —> V^3+ (aq) + H2O (l) (to reverse)
Combining all the equations to reverse and then reversing the result:
V (s) + H2O (l) —> VO^2+ (aq) + 2H^+ (aq) + 4e^-
Now notice how each Fe^3+ accepts 1 electron and that this half-reaction produces 4? That is to say we need to add 4Fe^3+ to the LHS and 4Fe^2+ to the RHS:
4Fe^3+ (aq) + V (s) + H2O (l) —> 4Fe^2+ (aq) + VO^2+ (aq) + 2H^+ (aq)

sorry, please could you clarify why we are combining the equations of V? Can't we just use the last one because that is what it is finally oxidised to?

Reply 6

TypicalNerd

Original post by anonymous56754

sorry, please could you clarify why we are combining the equations of V? Can't we just use the last one because that is what it is finally oxidised to?

You are starting from metallic vanadium and then sequentially oxidising it to V^2+, then V^3+ and finally VO^2+. As such, you need to combine all the oxidation equations determined to be useful to get the overall.