[Shafski]

This is for unit 5 chemistry.

I just dont understand them. Theres nothing about the equilibrium positions of redox reactions in the AQA nelson thornes book unless im missing it somewhere.

But for example, the question.


The cell represented below was set up.
Pt|Fe²⁺ (aq), Fe³⁺ (aq) || BrO₃- (aq), Br₂(aq)|Pt


(iii) Deduce what change in the concentration of Fe3+(aq) would cause an increase in the e.m.f. of the cell. Explain your answer.

Here is the EMF of the cells in question

Fe³⁺(aq) + e^- --> Fe²⁺(aq) ########################## +0.77
2BrO₃- (aq) + 12H⁺ (aq) + 10e^- --> Br₂(aq) + 6H₂O(l) ####### +1.52

[matthew769]

(Original post by Shafski)
This is for unit 5 chemistry.

I just dont understand them. Theres nothing about the equilibrium positions of redox reactions in the AQA nelson thornes book unless im missing it somewhere.

But for example, the question.


The cell represented below was set up.
Pt|Fe²⁺ (aq), Fe³⁺ (aq) || BrO₃- (aq), Br₂(aq)|Pt


(iii) Deduce what change in the concentration of Fe3+(aq) would cause an increase in the e.m.f. of the cell. Explain your answer.

Here is the EMF of the cells in question

Fe³⁺(aq) + e^- --> Fe²⁺(aq) ########################## +0.77
2BrO₃- (aq) + 12H⁺ (aq) + 10e^- --> Br₂(aq) + 6H₂O(l) ####### +1.52

If the concentration of Fe3+ increases, then the emf for that half cell will decrease as when written as a reduction potential Fe3+ + e- -> Fe2+ + 0.77V, an increase in Fe3+ concentration will cause an accumulation of negative charge as more electrons are lost. The emf of the cell is calculated by E = E(red) - E(ox) so as the emf of iron becomes more negative, the overall emf increases , sorry that's probably badly explained but it makes sense to me lol

[Shafski]

(Original post by matthew769)
If the concentration of Fe3+ increases, then the emf for that half cell will decrease as when written as a reduction potential Fe3+ + e- -> Fe2+ + 0.77V, an increase in Fe3+ concentration will cause an accumulation of negative charge as more electrons are lost. The emf of the cell is calculated by E = E(red) - E(ox) so as the emf of iron becomes more negative, the overall emf increases , sorry that's probably badly explained but it makes sense to me lol

Oo that makes sense, I can apply that to such questions. I understand it mainly, but just the last part I don't get. I've seen that formula somehwere, maybe in class when I wasn't paying much attention, the E = E(red) - E(ox) but how can you apply it to a question. What is the E(red) and E(ox) in a equation for example.

[matthew769]

(Original post by Shafski)
Oo that makes sense, I can apply that to such questions. I understand it mainly, but just the last part I don't get. I've seen that formula somehwere, maybe in class when I wasn't paying much attention, the E = E(red) - E(ox) but how can you apply it to a question. What is the E(red) and E(ox) in a equation for example.

I don't know if you need to know it for your exam board, I do CCEA. Emf of cell = (Emf of half cell being reduced) - (Emf of half cell being oxidised)