It's difficult to give you a comprehensive answer without knowing details. Juts general comments. The question is not correctly formulated. I would suggest the following wording:

How Fe(II) can increase the reaction rate but increase the apparent activation energy.

Second. " a difference between an activator and a catalyst" Pease define what is an activator? 

May be 

 Maybe in this case it has a decisive importance entropy of activation, but activation energy?

As mentioned by Yurii Geletti, the word catalyst can be misused. I think that Fe(II) creates a new reaction pathway: K2S2O8 oxidizes Fe(II) and the product of this oxidation oxide E131. The mechanism of the reaction is different in the presence of Fe(II) and there is no contradiction between "faster" and "larger activation energy".

In my first comment I recommended to reformulate the question. A catalyst does NOT change the rate constant. A catalytic reaction proceeds through a different pathway then uncatalyzed. The reaction rate law is commonly different than that of uncatalyzed. Comparing activation energies of catalyzed and uncatalyzed reactions is like comparing apple and orange. I upvoted the answer of Guy Schmitz. I do not understand well the answer of Jamal Alfadhli. 

For the definitions of "observed rate coefficient, partial order of reaction, pseudo-first-order rate coefficient, rate constant, k" 

see

http://goldbook.iupac.org/O04322.html

activation energy (Arrhenius activation energy)

http://goldbook.iupac.org/A00102.html

The answer of Dr Schmitz is more acceptable than of Dr Alfhadli. I think the reaction between Fe(II) and persulfate(KSP)  produces more sulfate radical, which is the principal oxydant in the reaction between persulfate and the food colorant. So increasing sulfate radical by adding Fe(II) to the mixture  (E131 + KSP) will increases the observed rate constant k (pseudo 1st order , C=Cox e-kt ), but did not change the pathway ( or the mechanism). But in this case we would have the same Ea, Experimentally, it is higher in the presence of Fe(II). so may be as Dr Schmitz and Dr Geletii said the reaction occurred  via another mechanism

I Think we give some mechanism and some values since 1991 or 1992 in Fast reaction in solution in Erlangen

I did not understand the answer of K. Yango

I dont know if the topic is over, however I have a several comments to this phenomenon.

You have assumed that the Fe(II) is an catalyst. Why?

Is this reaction of K2S2O8 and the food colorant (E131) accuring?YES.

OK.

Next simple question is this reaction accuring between Fe(II) and E131 probably no.

Is this reaction occurs between Fe(III) and E131?? probably Yes.

In my opinion you have nice case of subsequent and parallel reaction, not catalytic one.

Fe(II)+K2S2O8 -k1-> Fe(III)+products - (1)

Fe(III)+E131-k2-> Fe(II)+ products - (2)

and

E131+K2S2O8 -k3->products - (3)

k3 Fe(II)+ products. Do not give any KS2O8. you should also observe fast reaction.

Note. Fe(III) is also very strong oxydant (in appropriate conditions).

Thx for yr interesting comments. The regeneration of Fe(II) in reaction 2, let me think about a catalytic effect of Fe(II), since by def. a catalyst increases the reaction rate and is regenerated at the end of the reaction. Both are present for Fe(II) in my case. 

I will try the reaction between E131 and Fe(III) but in this case the pH must be below 2 to prevent the precipitation of Fe(III) or the formation of Fe(OH)2+. May be the discoloration will be slower since the acidic form of E131 is more stable than of its basic form. (pka of E131 =2.2).

It is a good idea to test the effect of Fe(III) but you should not draw conclusions too quickly. Fe(IV) is unstable but exists as an intermediate in some reactions. The Fe(II) + K2S2O8 reaction could produce Fe(IV) which quickly oxidizes E131.

Why to avoid the formation of Fe(OH) ++?

Your result is not unexpected at all. You calculated apparent activation energies for lumped equations, thus as already mentioned ignored the new reaction pathway. By using a lumped equation, the catalyst is obviously increasing the reaction rate constant compared to the non-catalyzed reference. So far this is clear. If you further calculate the activation energy of the two cases, the catalyzed case has to appear with an higher activation energy, because of the kinetic compensation effect, which correlates the pre-exponential factor and the acitivation energy. A higher reaction rate constant is therefore always accompagnied by a higher activation energy for these kind of lumped kinetic approaches.

Dennis Jung, In the case of lumped equations with lumped parameters, the term activation energy is unjustified. Personally, then I prefer to treat the parameter E in the Arrhenius equation as a measure of the sensitivity of the reaction rate to a change in temperature, and not as an energy barrier. Apart from that, a matter-of -fact response.

The reaction rate constant of the catalysed reaction should be higher than that of the uncatalysed. The reaction mechanism of the catalysed reaction being different and the higher yield of product will be achieved in the same reaction time (faster oxidation rates) comparative to the uncatalysed case (absence of Fe(II)).

The catalytic reaction mechanism normally proceeds by the catalyst combining with the reactant to form a substrate which decomposes to intermediates (radicals-highly reactive species) via activating collision that further participate in the reaction sequence.

You have to detect the radicals (Electron spectroscopy for chemical analysis studies) and propose a fitting reaction mechanism, then derive the overall rate expression and kinetic model based on the experimental data, The nonlinear regression of the kinetic data to fit the proposed model via suitable data analysis tool- (Excel spreadsheet, SciDavis, Origin) will give the real parameter values of kinetic rate constant and activation energy via Arrhenius equation