[Moz]

Sn2+ + Mn --> Sn + Mn2+

the question was to find ecell at standard conditions

Sn2+ + 2e- --> Sn    e=-.014V
Mn2+ + 2e- --> Mn   e=-1.18V

So i did e cathode- e anode

-.14-1.18=-1.32

since the half reactions are

Sn2+ + 2e- --> Sn    e=-.014V
Mn --> Mn2+ + 2e-   e=1.18V (was i suppose to switch the sign or leave it for e?)

the answer is 1.04

[Schrödinger]

Your half reactions are correct and you have identified cathode and anode correctly.

One small problem : Once you change the sign of E^o after reversing the reaction, it no more remains E^o. This is not the standard E^o as prescribed by IUPAC. Let's call this reaction potential 'R'. It's not the standard electrode potential, since the reduction reaction with which it was associated (by definition) has been changed to oxidation.
 
So, once you reverse the sign, you just add the values. You don't perform E_cathode - E_anode. This formula is valid only if you have with you the actual values of E_cathode and E_anode.

The advantage of these R's is that you don't have to follow any formula. You just add the R's to get the R for the net reaction.