So the hint with which one to choose lies with the other reactant. Key word reactant meaning that it needs to be on left hand side of any half reaction or redox reaction you write. So if you go through the reduction potential page you will end up finding the equation with so2 in it which is: So4^2- + 4H^+ + 2e^- --> so2 + 2h2o
But notice it is an equation with so2 on the right as a product but we want it as a reactant so you take the reversed reaction. so2 + 2h2o -->So4^2- + 4H^+ + 2e^-
which means you also reverse potential giving -0.16 volts. Also, since we reversed the reaction it is now an oxidation reaction so Eoxidation = -0.16 volts.
This is very important to realise this because it then tells us that the other equation must be a reduction equation (because these redox reactions are referred to as half reactions and can't actually happen independently of one another but rather have to happen simultaneously with one another).
So the conditions for the reduction are the equation is below the oxidation equation on the standard reduction potentials sheet (because we want to get a positive number). that is your answer in which you pick now from potassium or dichromate, the dichromate because its reaction is below the oxidation of the so2, and thus will be a reduction. Ereduction = 1.36 volts.
Add the two giving Ecell = Ereduction + Eoxidation = 1.36-0.16 = 1.20 which is A