Acompound contains 69.9% iron and 30.1% oxygen. what is its empirical formula?

The easiest way to do this is to assume that you have 100g of compound. Then you will have 69.9g of iron and 30.1g of oxygen.

You will need to find how many moles there are of each. Divide these masses by the molar mass of each element (for oxygen, just use the molar mass of O, not O2, as the calculation is easier with atoms than dioxygen):

69.9g / 55.845g/mol = 1.25mol Fe
30.1g / 15.999g/mol = 1.88mol O

Next, find the ratio of these 2 molar amounts. If we have 1 mole of Fe, we will have 1.88 / 1.25 = 1.50 mol O. Make everything whole numbers now: 1.50 is half of 3, so multiply both by 2. We get 2 moles of Fe to every 3 moles of O, so the empirical formula is Fe2O3.

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.