Question involving voltage and galvanic batteries

[dungy]

I'm currently studying for the MCAT and most of the physics questions I get wrong, the explanations adequately help me. But, one question has me stumped.

"Question:"

What is the standard emf for the galvanic cell in which the following overall reaction occurs?

2Na(s) + Cl2(g) -> 2Na+(aq) + 2Cl–(aq)

Half-reaction (V) E° red
Na+(aq) + e– -> Na(s) –2.71
Cl2(g) + 2e– -> 2Cl–(aq) +1.36

A) –1.35 V
B) +1.35 V
C) +4.07 V
D) +6.78 V

"end of question"

The answer they provide is:
According to the table, the oxidation of Na has a potential of +2.71 V and the reduction of Cl2 has a potential of +1.36 V. The standard emf for the galvanic cell is therefore +2.71 V + 1.36 V = +4.07 V. Thus, C is the best answer.

My understanding was, to calculate the emf, I would have needed to double the charge during the reduction of sodium because 2 moles are used in the cell. But the Kaplan test clearly claims that the reduction of sodium is only counted once. Why is this? I chose D) +6.78 V

I'm tired of getting 12 every time I write the physics section practice, it's my weakest. I want better.

No, I'm not taking any silly expensive study courses to help me.

 

[bel]

I think you're right, it's like chemistry thermodynamics exercises, you reverse one equation, and add it to the other after multiplying (this time by two to the first half equation) to get the target equation and the voltage term.

 

[dungy]

That's what I thought...

I would have chalked it up to the test being wrong, but this is the second time it would have made exactly the same mistake. These are questions from the official Kaplan tests and I haven't seen any other answers that made no sense to me, so it seems like an awfully large coincidence.