# Identifying the anode/cathode

1. Aug 23, 2007

### big man

1. The problem statement, all variables and given/known data
For the following metal couples, indicate which will be the anode and cathode
respectively in an electrochemical cell. Write balanced equations for the
reactions, and calculate the electrode potential for the cell in each case, assuming
standard conditions. Note which metal will corrode if they are in contact in a
moist environment.

(a) Sn/Sn2+ and Cu/Cu2+
(b) Cd/Cd2+ and Ag/Ag+
(c) Sn/Sn2+ and Al/Al3+
(d) Ag/Ag+ and Cu/Cu2+

3. The attempt at a solution
Now if I look at my reduction potentials for (c) lets say, Sn/Sn2+ will have a potential of -0.14 V and Al/Al3+ has a potential of -1.66 V. From this I would say that Al would be the anode since it has the lower reducion potential. However, the half equation then becomes:

$$Al--> Al^3^+ + 3e^-$$

This is the reverse reaction that appears in the standard reduction table so wouldn't the potential now be 1.66 V???

This would then lead to an Emf of the cell of

$$E_c_e_l_l=E_c- E_a=-0.14-1.66=-1.80V$$

Where the c means cathode and the a means anode.

I would really appreciate if someone could just look over this one problem and tell me if I'm doing it correctly.

2. Aug 23, 2007

### rocomath

Ecell must always be positive, E > 0

i don't have my chemistry book at my side, but if you give me the standard potential for Sn/Sn2+ i can write you a balanced reaction.

Ecell = Ec - Ea; Ec = Sn/Sn2+, Ea = Al/Al3+

you want to manipulate it so that you get a positive Ecell. but in general, the greatest positive E or least negative should be the cathode.

Ecell = -.14 - (-1.66) = +1.52v

*This is the reverse reaction that appears in the standard reduction table so wouldn't the potential now be 1.66 V

you can do it 2 different ways. if you're going to plug into the equation, then keep the charges given (and remember, EMF is an intensive property just like density, if you balance your equation so you can cancel out the electrons for your overall reaction, you do not multiply your EMF by that value). say you have 2 positive Ep, then make the smallest Ep negative; if you have 2 negative Ep, make the most negative Ep positive ... then just simply add the two for your EMF.

when i say positive/negative, it's in respects to making them your cathode/anode.

Last edited: Aug 23, 2007
3. Aug 23, 2007

### big man

I know how to balance the equations and write them out, but what I'm struggling to come to terms with is why you keep the sign the same for the Aluminium reaction. The reaction is going in the opposite direction to how it is shown in the reduction tables (which gives a potential of -1.66 V). So wouldn't the potential be 1.66 V. I've found a site that says that you would reverse the sign, but then the equation it gives for the cell Emf is:

$$E_c_e_l_l = E_c + E_a$$

Am I mixing up two different conventions here?

4. Aug 23, 2007

### rocomath

no, bc Ea becomes positive

a negative times a negative = positive

plugging into the equation with given data, -.14 and -1.66

Ec = -.14 - (-1.66) = +EMF

no equation but keeping in mind that you flipped the most negative Ep = flip it's sign

-.14 + 1.66 = +EMF

i know what you mean, it is weird. just like Kc, i thought if Kc for the fwd rxn is 5, then Kc(rev) is -5, but no ... it's the inverse and the reason is bc Kc can't be negative (i think), and also when you i guess in a sense "Proof" it, it is the inverse. and there are a few others at first you think it should be this, but it's not.

Last edited: Aug 23, 2007
5. Aug 23, 2007

### big man

Yeah I understand that a negative times a negative is a positive, but what I'm saying is that there seems to be two methods. One (where E=Ec-Ea) keeps the electric potentials the same sign irregardless of the direction of the reaction. The other method where (E=Ea +Ec) says to reverse the sign of the electric potential

Method one:
http://www.ualberta.ca/~jplambec/che/p102/p02064.htm
http://en.wikipedia.org/wiki/Electrochemistry#Electrochemical_cells

Method two:
http://www.ausetute.com.au/calcelemf.html
http://hyperphysics.phy-astr.gsu.edu/HBASE/Chemical/electrode.html#c1

If you have a look at these sites you might undersand what I'm trying to say. Essentially I vaguely remember having to reverse the signs of the electric potential if the directin of the reaction was reversed, but when I applied that process using the equation in method one I kept getting the wrong answer.

Sorry for making such a confusing post!

6. Aug 23, 2007

### rocomath

sorry but i don't have time to read those, if you can paste what is confusing you i'll read it

idk how else to say it, you flip the sign on the most negative Ep when you have 2 negative Eps bc you want a positive EMF, E > 0