Calculate E°(cell) for the reaction

Question

Calculate E°(cell) for the reaction, 2 IO3-(aq) + 10 Fe2+(aq) <=> 10 Fe3+(aq) + I2(aq) given the reduction potentials:

Fe3+(aq) + e- <=> Fe2+(aq) , E° = 0.87 V

2 IO3-(aq) + 10 e- <=> I2(aq), E° = 1.10 V

I am confused why the answer is +0.23

I thought the answer would be -0.23 because Fe³⁺ is on the opposite side or does that not matter?

Answer 1

2IO₃- + 10e- ==> I₂ should be the reduction half reaction (cathode)

Fe²⁺ ===> Fe³⁺ + e- should be the oxidation half reaction (anode) x 10 to equalize electrons

______________________

2IO₃- + 10Fe²⁺ ==> I₂ + 10Fe³⁺ overall reaction

Eºcell = cathode - anode = 1.10 v - 0.87 V

Eºcell = +0.23 V