Electrode potentials question help

Tip: The more positive E value will be spontaneous.

PART (I)

Two reactions:

1. S2O8(2-) + 2e- ---> 2SO4(2-); E = +2.01V

2. 2Ag+ + 2e- ---> 2Ag; E = + 0.80V

First equation more positive, so spontaneous (forward). The second equation will go backward, and electrons cancel:

S2O8(2-) + 2Ag ---> 2SO4(2-) + 2Ag+

Emf = 2.01 - 0.8 = 1.21 V

PART (II)

Increase in SO4(2-) shifts the equilbrium to the left - backwards. So the electrode potential should decrease.

thanks a lot . So the reaction which is less positive always goes backwards? and hence this is the way to work out the half equation and thus the full ionic equations?