# Error in AAMC 9 Physical Section?

#### imperfection

##### Full Member
10+ Year Member
I've looked over this particular question and the explanation for the answer a hundred times and still keep coming up with them having made a mistake. If you want to see the actual question, it's Q30 in AAMC MCAT Practice Test 9. I've changed the exact reactants and made up new values for E, but the essence of the Q still stands (I really don't want to have legal problems with AAMC...).

The following reaction occurs in a galvanic cell:

2Li(s) + Br2(g) <---> 2Li+(aq) + 2Br-(aq)

With the following E for the half reactions:

Na+(aq) + e- ---> Na(s) -2.43 V
Cl2(g) + 2e- ---> 2Cl-(aq) +1.78 V

Now, the question asks that you calculate the standard emf for the cell.

By my math, this means:
((-1)*(2)*(-2.43)) + 1.78 = 6.64

The (-1) is because the half reaction shown is for the opposite direction. The 2 is because the half reaction shown is for only 1Li+ ion, not two, so we need to multiply the half reaction by two in order to show what's really happening in our cell.

They give the answer as 4.21:
((-1)*(-2.43)) + 1.78 = 4.21

In other words, they don't multiply the E of the Lithium half reaction by two to make up for the difference in quantity between the half reaction and what's happening in the cell. Am I being really stupid here, or is that a mistake on their part?

Thanks so much for any fresh perspectives on this folks! I'm afraid I've been staring at it for way too long...

#### LONeil

##### Full Member
When balancing redox reactions, voltages are NEVER multiplied when equating the electrons between the oxidation and reduction half reactions.

#### LONeil

##### Full Member
I guess I could clarify....reduction potentials are intrinsic properties, thus they aren't changed and are inherent to the cell regardless of the number of electrons being transferred.

#### typicalindian

##### Full Member
5+ Year Member
7+ Year Member
You are NOT EVER supposed to multiply the standard potentials when calculating the standard cell voltage.

#### imperfection

##### Full Member
10+ Year Member
Thanks folks!

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