Question

H₂O₂(aq) + FeSO₄(aq)

1. The changes you observed involves two sets of half reactions. For each set, write the oxidation and reduction half reactions, the overall reaction, and calculate ..   
   1. Set 1
   • Reduction Half Reaction
   • Oxidation Half Reaction
   • Overall Balanced Redox Reaction
   • Calculation
   2. Set 2

1. • Reduction Half Reaction
   • Oxidation Half Reaction
   • Overall Balanced Redox Reaction
   • Calculation
2. Overall Reaction

Can you help to answer these?

Answer 1

2 FeSO4 (aq) + H2O2 (aq) + H2SO4 (aq) → Fe2(SO4)3(s)+ 2 H2O When one member of the redox couple is oxygen with an oxidation state of -2 or hydrogen with an oxidation state of +1, it is best to replace it with a water molecule. 0: Fe+2S+60-24 → Fe+3,0-23 (Fe) H+1,0-12 + H+120-2 (0) R: one for the oxidation, and one for the reduction (see: Divide the redox reaction into two half- reactions). Fe+2S+60-24 → Fe+3,0-23 0: R: H+1,0-12 + H+120 half cell method -- is essentially the same as a redox reaction. there are always 2 reactions in a redox reaction, namely the reduction reaction and oxidation reaction. reduction can be defined as the gain of electrons in an atom, or the decrease in oxidation state of an atom. oxidation can be defined as the loss of electrons in an atom, or the increase in oxidation state of an atom an example is the oxidation of iron in iron (II) sulphate by hydrogen peroxide. the equation at first glance, is H202 + FeSO4 --> H2O + Fe2(SO4)3 (1) if u seperate the reduction and oxidation reactions u'll get H202 --> H2O Fe2+ --> Fe3+ note that SO4- is not counted. it is a spectator ion and hence is not reduced or oxidised. (2) now to balance them. first consider that all the atoms are balanced. (except for hydrogen and oxygen) no balancing needed here, but u know what to do next time.

(3) balance oxygen by adding H2O H202 --> 2H20 (4) balance hydrogen by adding H+ 2H+ + H202 --> 2H20 (5) balance charges by adding electrons. 2H+ + 2e- + H202 --> 2H2O Fe2+ --> Fe3+ + e- now to combine them. remember, charge conversation requires charges on both sides to be equal. this can be done by ensuring the number of electrons on both sides of the equation are equal; overall electron gain = overall electron loss so we get 2H+ + 2e- + H202 --> 2H2O 2Fe2+ --> 2Fe3+ + 2e- combining now... 2H+ + 2e- + 2Fe2+ + H202 --> 2H20 + 2Fe3+ + 2e- removing electrons... 2H+ + 2Fe2+ + H202 --> 2H2O + 2Fe3+ if u want u can add the SO42- which will give us.. H2SO4 + 2FeSO4 + H202 --> 2H2O + Fe2(SO4)3 H202 + 2 H+ + 2 e- → 2 H20 E°= 1.763 Fe/Fe2+ || Fe3+ / Fe2+ E l = 0.44 + 0.77 = 1.21 V

So - E cell for 2 FeSO4 (aq) + H2O2 (aq) + H2SO4 (aq) → Fe2(SO4)3(s)+ 2 H2O E cathode- Eanode So= E°= 1.763-1.21 =0.553 v