E CELL question

12

Q8.
Which ion cannot catalyse the reaction between iodide (I–
) and peroxodisulfate
(S2O8
2–
)?
Use the data below to help you answer this question.
Half-equation EO / V
S2O8
2– + 2e– → 2SO4
2– +2.01
Co3+ + e– → Co2+ +1.82
Fe3+ + e– → Fe2+ +0.77
I2 + 2e– → 2I– +0.54
Cr3+ + e– → Cr2+ –0.41
A Co2+
B Cr2+
C Fe2+
D Fe3+
the answer is B, I understand this because the ion can reduce them both. but what about the ones in between, can they not oxidise the iodine? why can we consider reduction only here

Reply 1

UtterlyUseless69

12

Original post

by mitostudent

Q8.
Which ion cannot catalyse the reaction between iodide (I–
) and peroxodisulfate
(S2O8
2–
)?
Use the data below to help you answer this question.
Half-equation EO / V
S2O8
2– + 2e– → 2SO4
2– +2.01
Co3+ + e– → Co2+ +1.82
Fe3+ + e– → Fe2+ +0.77
I2 + 2e– → 2I– +0.54
Cr3+ + e– → Cr2+ –0.41
A Co2+
B Cr2+
C Fe2+
D Fe3+
the answer is B, I understand this because the ion can reduce them both. but what about the ones in between, can they not oxidise the iodine? why can we consider reduction only here

It would be good practice to consider both steps in the mechanism and whether both can take place.

We can see that the oxidation step will always be feasible, as the standard reduction potential for S2O8^2- + 2e^- —> 2SO4^2- is more positive than the standard reduction potentials for all the other half-equations. As such, it isn’t really a concern. As such, it is quicker to just consider the reduction step.

To convince yourself of the above recall the following

E(cell) = E(reduction) - E(oxidation)

Where E(reduction) is the standard reduction potential for the half equation involving the oxidising agent and E(oxidation) is the standard reduction potential for the half equation involving the reducing agent.

In this case, the oxidising agent will be the S2O8^2- and the reducing agent will be the metal ion used. Feel free to plug in the numbers and you’ll observe that you get a +ve E(cell), which implies the reaction has a -ve ΔG and so the reaction will be (thermodynamically) favourable.

Now let’s re-try using the same formula for the reduction step. Only this time, the metal cations act as the oxidising agent and the iodide ions act as the reducing agent. Which one gives a -ve E(cell) and so has a +ve ΔG?

Reply 2

Nitrotoluene

12

Original post

by mitostudent

Q8.
Which ion cannot catalyse the reaction between iodide (I–
) and peroxodisulfate
(S2O8
2–
)?
Use the data below to help you answer this question.
Half-equation EO / V
S2O8
2– + 2e– → 2SO4
2– +2.01
Co3+ + e– → Co2+ +1.82
Fe3+ + e– → Fe2+ +0.77
I2 + 2e– → 2I– +0.54
Cr3+ + e– → Cr2+ –0.41
A Co2+
B Cr2+
C Fe2+
D Fe3+
the answer is B, I understand this because the ion can reduce them both. but what about the ones in between, can they not oxidise the iodine? why can we consider reduction only here

Hello mitostudent!

Each ion pair must be analysed as follows:

Co^3+/Co^2+: Co^3+ can be reduced to Co^2+ w/ a potential of +1.82 V.
This high potential shows that Co^3+ is a strong oxidising reagent, which accelerates the reaction.

Cr^3+/Cr^2+: The half equation for the reduction of Cr^3+ gives a potential of -0.41 V.
This low potential means that Cr^3+ is a weak oxidant and is therefore unable to accelerate the reaction.
Fe^3+/Fe^2+, Fe^3+ can be reduced to Fe^2+ w/ a potential of +0.77 V.
This potential is favourable for acceleration of the redox as Fe^3+ can efficiently oxidise iodide.

- Correct answer: B, Cr^2+.

- Explanation: The potential of Cr^3+ to Cr^2+ (-0.41 V) shows that it is not a strong enough oxidising reactant to efficiently facilitate electron transfer from iodide to peroxodisulphate.

In contrast, the other ions (Co^3+, Fe^3+) have higher reducing potentials, which means that they can better accept electrons and help to drive the reaction forward.

- Conclusion: The ion in the list that cannot accelerate the reaction between iodide (I^-) and peroxodisulphate (S2O8^2-) is B, Cr^2+. Its low oxidising potential is indicated by the negative potential.

ITALY FLAG.png

Sandro

(edited 1 year ago)

Reply 3

mitostudent

OP

12

Very late reply but thank you for your explanations! I understood that the value for Cr was inbetween the two involved in the catalysis - so it couldn't catalyse. I suppose if they gave Cr3+ too, how would you decide?

Reply 4

mitostudent

OP

12

Very late reply but thank you for your explanations! I understood that the value for Cr was inbetween the two involved in the catalysis - so it couldn't catalyse. I suppose if they gave Cr3+ too, how would you decide?

Reply 5

Nitrotoluene

12

Original post

by mitostudent

Very late reply but thank you for your explanations! I understood that the value for Cr was inbetween the two involved in the catalysis - so it couldn't catalyse. I suppose if they gave Cr3+ too, how would you decide?

We don't necessarily have to talk only about reduction.. Oxidation is important too. The things that speed up the reaction must be able to do both. Then S2O8^2- has to oxidize Cr^3+. This can react also with I^⁻ (which implies the oxidized form can reduce I^⁻).2What if the chemical can switch between oxidation states a couple of times (for example Cr^3+)?
So if you had Cr^3+, you wanted to see if that could be reduced and then if that reduced form could be oxidized.
Cr^3+ + e^⁻ → Cr^2- (-0.41V) Therefore, Cr^3+ is reducible.
In order to be a reaction helper, that Cr^2- would then have to be oxidized by the S2O8^2-. But, as we said earlier, it isn't going to happen. Thus, Cr^3+ is not useful for the reaction.
Why Focus on Reduction?
We're not exclusively focusing on reduction. We're considering both oxidation and reduction potentials. The catalyst must be able to undergo both processes in a cycle.
-The catalyst must be oxidized by S2O8^2- (meaning S2O8^2- must be able to reduce the oxidized form of the catalyst).
-The catalyst must be reduced by I^- (meaning the oxidized form of the catalyst must be able to oxidize I^-).
Okay, so basically:
-A catalyst has to help both I^- lose electrons and S2O8^2- gain them.
-The E° values tell us if these reactions will actually happen.
-If S2O8^2- can't oxidize an ion, that ion can't be a catalyst here.

Bye,
Sandro

Reply 6

TypicalNerd

18

Original post

by mitostudent

Very late reply but thank you for your explanations! I understood that the value for Cr was inbetween the two involved in the catalysis - so it couldn't catalyse. I suppose if they gave Cr3+ too, how would you decide?

If they gave you Cr^3+ as well, you’d run into a similar problem as with Cr^2+.

Recall that the mechanism of catalysis, in no particular order is as follows:

2M^2+ + S2O8^2- —> 2M^3+ + 2SO4^2-
2M^3+ + 2I^- —> 2M^2+ + I2

(Where M is a generic transition metal)

That is to say that if M^2+ can catalyse the reaction, so can M^3+ as the steps in the mechanism would just occur in the reverse order. Similarly, if M^2+ cannot catalyse the reaction, nor can M^3+ by the same principle.

Supposing you had both Cr^2+ and Cr^3+ as options, you could reasonably eliminate them both based on this logic as they would both have the same effect on the rate of reaction.

E CELL question

Quick Reply

Related discussions

Articles for you

How The Student Room is moderated