How to determine which chemicals produce a high voltage

[Viridiana G [3M]]

My question refers to the Winter 2014 Final, #3A.

How do we determine which chemicals will produce a higher voltage? Do we look at the given half reactions and voltage values that correspond with the given chemicals in the problem?

Here's all of the chemicals:
Zn(s), Fe(s), (aq), (aq), Cu(s), Pt(s), (aq), (aq)


Thank you!

[Neil DSilva 1L]

You are essentially correct. You basically just look at all the half reactions you can form with the given chemicals and compare them to find the highest combination. You would get the highest combination by taking the half reaction with the largest negative value and the half reaction with the largest positive value. The possible reduction half reactions that can be made from the chemicals given are as follows:

- reduction of Zn²⁺ to Zn (-0.76 V)
- reduction of Fe³⁺ to Fe²⁺ (0.77 V)
- reduction of Fe²⁺ to Fe (-0.44 V)
- reduction of Cu²⁺ to Cu (0.34 V)

You should now be able to see that the combination that would yield the largest voltage involves the Zn²⁺, Zn, Fe²⁺, and Fe³⁺. In addition to those chemicals, you would also need the Pt to use as an electrode on the side with the Fe ions.

[Viridiana G [3M]]

Thanks, Neil!

[Chem_Mod]

Thanks Neil for your nice explanation!

[ysteve8]

Hi,

I am not sure as to why we must also use the Fe(NO3)2 (Fe2+) for this design since we need a large positive value half reaction and a large negative value half reaction (so i thought we would use Zn2+ and Fe3+).

Also, why can't a Fe(s) be used to conduct electricity instead of Pt(s)? I thought Fe could conduct electricity.

Any clarifications would be much appreciated :)

[Neil DSilva 1L]

There was a typo in my original post.
It read:
- reduction of Fe2+ to Fe3+ (0.77 V)
It should have said (and it's updated):
- reduction of Fe3+ to Fe2+ (0.77 V)

To answer your other questions:
The zinc will be oxidized in the anode half-cell and the iron (iii) will be reduced in the cathode half-cell, but there are still zinc ions and iron (ii) ions in solution, so that may be why they are included in the solutions of their half-cells.

You can't have iron as an electrode because that may cause the wrong reactions to occur; it may be active in the cell and screw up the voltage.

[Ashley Van Belle 2B]

So, in order to solve this problem, we would need to be given the half reactions, right? We wouldn't be expected to know which elements is the higher/lower oxidizing or reducing agent.

[Rodri_Karla_1J]

Yes. I'm pretty sure we'll be given the half reactions on the constant sheets.

So, in order to solve this problem, we would need to be given the half reactions, right? We wouldn't be expected to know which elements is the higher/lower oxidizing or reducing agent.