Electrochemistry Question

What is the concentration of Al3+ in the following cell at 25°C if the cell voltage is 1.486 V?
Al | Al3+(?) || Sn2+(3.21 × 10-4 M) | Sn
Al3+ + 3 e- --> Al(s) E° = -1.676 V
Sn2+ + 2 e- --> Sn(s) E° = -0.137 V

A) 4.5 × 10-5 M
B) 2.4 × 10-2 M
C) 2.8 × 10-3 M
D) 7.8 × 10-6 M
E) 1.6 × 10-5 M

Please Explain

Ok... that's really odd... I don't see the answer anywhere in those multiple choice.
The answer should be 75.53 M

And it comes from the nerst equation which is:

E = E(naught) - .0592/n (log q)

E would be 1.486 V

To find E(naught) you would add the E(naughts) of both your half reactions....

First of all, you have to establish that

Al l Al3+ is ANODE since it is at the beginning of the notation, and since it is the ANODE that means it's oxidizing... that means the half reaction should be

Al (s) ---> Al3+ + 3e-

so the E(naught) is reversed and becomes E(naught) = 1.676 V

The Sn l Sn2+ is the CATHODE being at the end of the notation, and the CATHODE is reduced... so its

Sn2+ + 2e- ---> Sn(s)

the e(naught) stays the same, so it is -0.137


When you add those two E(naught) together,

1.676 + -0.137 = 1.539 V

The E(naught) for the entire equation is 1.539 V...

Your half reactions have different amount of e- transfer, so to balance them, you need to multiply both by a number to have the same amt. of electron transfer

Al(s) ----> Al3+ + 3e-
Sn2+ + 2e- ---> Sn(s)

Since 3 and 2, the least common multiple is 6, you multiply each rxn by those numbers to get 6e- transfer...

2(Al(s) ---> Al3+ + 3e- ) = 2Al(s) ---> 2Al3+ + 6e-
3(Sn2+ + 2e- ---> Sn(s) ) = 3Sn2+ + 6e- ----> 3Sn(s)

The overall equation is


2Al(s) + 3Sn2+ ----> 2Al3+ + 3Sn(s)

Remember, when you multiply the half reactions, it does not affect the E(naught) values AT ALL.

After you've done this, you find out the e- transfer is 6 e-...

So... n = 6e-

Q = ANODE CONCENTRATION/CATHODE CONCENTRATION


So, you plug in all your values in the NERST equation, which is:

E = E(naught) - .0592/n (log q)

1.486 V = 1.539 V - .0592/6 (log ANODE CONCENTRATION/3.21E-4)

-.053 = -.0592/6 (log ANODE CONCENTRATION/3.21E-4)

5.37162160 = log ANODE/3.21E-4

10^5.37162160 = ANODE/3.21E-4

ANODE CONCENTRATION (AL3+) = 75.53 M

The answer should be 75.53 M.... or else it makes no sense...