[John623]

I got these questions from the book "Calculations for A-level Chemistry". I think there is a mistake in the book, am I right?

Calculate the standard e.m.f.s of the following cells at 298 K:
1) Ni(s)|Ni2+(aq) :: Sn2+(aq), Sn4+(aq)|Pt

Both are sides are oxidized. I thought with these equations, on the right side of the cell equation is supposed to be a reduction and the left side there is supposed to be an oxidation. The book claims the answer to be +0.40V

I was taught that for these problems you take the voltage of the reduced compound and minus the voltage of the oxidised compound from it, giving the final voltage of the cell equation


2) Pt|I2(s),I-(aq) :: Ag+(aq)|Ag(s)

On the left we have a reduction, on the right we have another reduction. The book claims the answer to be +0.26V
Also there is a comma between I2 solid and I- aquarius, can you even do that since they are not the same state?


3) Pt|Cl2(g),Cl-(aq)::Br2(l),Br-(aq)|Pt

On the left we have a reduction, and on the right we have another reduction. The book claims the answer to be -0.27V
Once again commas between species of different states

[Borek]

Both are sides are oxidized. I thought with these equations, on the right side of the cell equation is supposed to be a reduction and the left side there is supposed to be an oxidation.

Whether reaction classifies as oxidation or reduction depends on which way it goes - but it is still the same reaction. If it is reversible, it can go both ways depending on the conditions.

Standard EMF of a cell is just the distance between standard reaction potentials for both half cells (distance - that is, higher one minus the lower one).

I was taught that for these problems you take the voltage of the reduced compound and minus the voltage of the oxidized compound from it, giving the final voltage of the cell equation

And that's almost exactly what I told you above - you just should not treat a reaction as "oxidation" or "reduction" without looking at the other half cell. Cu/Cu²⁺ will work as a reductor in a solution containing Ag⁺, but as an oxidizer in a solution containing iron nail.

The same applies to other questions.

Also there is a comma between I2 solid and I- aquarius, can you even do that since they are not the same state?

Which means I₂(s) is present in the solution, and the solution is saturated with the dissolved I₂.

Once again commas between species of different states

And the same meaning - solution contains Cl^- and there is a gaseous Cl₂ present, this time bubbled through the solution. In effect solution is saturated with the dissolved Cl₂.

[John623]

I thought the cell equation tells you which way the reaction is going. We are doing galvanic cells in class.

Let's take Ni(s)|Ni2+(aq) :: Sn2+(aq), Sn4+(aq)|Pt

I was taught that what is before the single line is what you start with, and what is after the single line is what you finish with. So I would assume that the above cell equation came from an ionic equation such as Ni(s) + Sn2+ -> Ni2+ + Sn4+ which doesn't look possible.

Such as:
Ni(s) -> Ni2+ +2e- oxidised
Sn2+ -> Sn4+ +2e- oxidised

Whereas if the equation was Ni2+(aq)|Ni(s) :: Sn2+(aq), Sn4+(aq)|Pt

Ni2+ +2e- -> Ni reduced
Sn2+ -> Sn4+ +2e- oxidised

Our teacher told us one of the reasons to do the reduced voltage - the oxidised voltage is to predict if certain reactions are possible or not

[Borek]

I always treat the half cell notation as meaning "these things are present", and not saying anything about how the reaction goes. The latter can be deduced from the standard redox potential tables and the Nernst equation.

[John623]

Ni(s)|Ni2+(aq) :: Sn2+(aq), Sn4+(aq)|Pt
Ni2+(aq)|Ni(s) :: Sn2+(aq), Sn4+(aq)|Pt
Pt|Sn2+(aq), Sn4+(aq) :: Ni2+(aq)|Ni(s)
Pt|Sn2+(aq), Sn4+(aq) :: Ni(s)|Ni2+(aq)

I see, so you would say that the above equations are the same and that the order is not relevant at all? Where you would work out which one is oxidized and which one is reduced by looking at the voltages on the table?

[Borek]

I would always put Ni(s) on the outside, to make it clear it is an electrode. Other than that, for me all these are equivalent and describe the same cell. Could be I am missing some nitty-gritty details of the convention used to describe the cell, but I don't care too much about it. Definitely writing reagents in some particular order reflecting reactions/anode/cathode set up can be helpful, but it is not always possible. Sometimes just changing concentrations will reverse the cell (which is why I have mentioned Nernst equation).