# Chem 2, Cell diagrams for oxidation/reduction reactions

## Homework Statement

All I'm trying to do is translate the cell diagram below into a into a an equation so I can determine which is the anode and which is the cathode. Here is the cell diagram. I know how to balance it, I just want to make sure I'm putting reactants and products on their correct sides:

Al/ Al+3 // CU+2 / CU

## The Attempt at a Solution

The unbalanced equation from the diagram above is Either Al+Cu+2--->Al+3 +Cu or

Cu+Al+3---->Cu+2+Al

In the first case Aluminum is the cathode because it reduces the copper ion to copper metal, and Copper is the anode because it oxides aluminum metal to the aluminum ion. In the second case it's the opposite. Assuming the last two statements are correct, my only problem here is figuring out which equation is represented by the cell diagram.

symbolipoint
Homework Helper
Gold Member
So you need to know which metal would rather be reduced compared to the other. Look in the table of reduction potentials. Best guess without looking is that the copper would rather more be reduced than the aluminum. Check how you find those in the table and see which has the larger reduction potential.

Oh yeah I forgot about that. So using the table and the diagnoal rule the correct equation is:
Al+Cu+2--->Al+3 +Cu

But this leads me to another question. When I find the standard EMF of the cell using E cathode minus E anode (E Aluminum minus E Copper) I get a negative number, meaning the reaction will not occur. The rest of the problem I'm trying to do seems to assume that the reaction takes place.

Borek
Mentor
I get a negative number, meaning the reaction will not occur.

No, reaction takes place as long as this difference is not zero. When the reaction takes place potential of one half-cell goes up (Nernst equation), of the second goes down - till they are both the same. Then - and only then - system is at theromodynamic equilibrium.