Given the following half-reactions and associated standard reduction potentials:

Question

AuBr−4(aq)+3e−→Au(s)+4Br−(aq)
E∘red=−0.858V
Eu3+(aq)+e−→Eu2+(aq)
E∘red=−0.43V
IO−(aq)+H2O(l)+2e−→I−(aq)+2OH−(aq)
E∘red=+0.49V
Sn2+(aq)+2e−→Sn(s)
E∘red=−0.14V

Write the cell reaction for the combination of half-cell reactions that leads to the smallest positive cell emf.

Express your answer as a chemical equation. Identify all of the phases in your answer.

I figured out that
Sn2+(aq)+2e−→Sn(s)
E∘red=−0.14V
-AND-
Eu3+(aq)+e−→Eu2+(aq)
E∘red=−0.43V
add up to be the smallest, but I am unsure how to go from there...

Answer 1

If that is the smallest Ecell, then turn those half cells into a reaction. Arrange the half cells so they produce a positive voltage.
2Eu^2+ + Sn^2+ ==> 2Eu^3+ + Sn

Answer 2

So I multiply Eu3+(aq)+e−→Eu2+(aq) by two so there is two e- in both half reactions. I also remember that I need to add H+ and OH- onto one or both sides depending if the reaction is acid or basic. How do I determine that?

Answer 3

The Eo voltage in tables may changes as a result of acid or basic conditions; however, there are no H or OH ions in the half reactions you've written so you need not worry about that.