Question

Balance the redox reaction given

7. H → NO2+ Fe + SO4 + H2O FeS2 + NO3 + (pyrite) Oxidation Half Reaction: Reduction Half Reaction: Oxidizing agent: Reducing agent

Answer 1

In the equation given above Fe in pyrite is in +2 oxidation state and S in pyrite is in -1 oxidation state. The common oxidation state of sulphur i.e., -2 will not work because then Fe would have to assume +4 oxidation state which is highly unstable.

We start by calculating oxidation numbers of the atoms in the given equation. Next we identify the species undergoing oxidation and reduction and based on that, we separate this equation into two half-equations(the oxidation state of the underlined atom is assumed to be x):-

x+31-2) =-| x=+5 x + 2(-2) x =0 NO2 = +4 x = +6 +41-2)=- 2 x so 2 - Reduction +3 +6 +6 +4 2- 2- I +2 -1 FeS Therefore, -+5 + Noz &H + 50,2- → NO2 + Fe NO₂+ Fe oxidation + H₂O oxidation Half-Equation : + FeS₂ 3 Fe + Total loss of electrons is 1 + 7*2 = 15 (fe, 23) (s. 16) So, Fes t et + 2 sou + 150 '0' is balanced by adding Ho molecules as- 2- 4 Fes + IS e- fet + 8 H₂O And 't is balanced by adding H. + 8 H₂0 Fe3+ + 2 5042 + 16 H Fes + 15 é OXIDATION HALF EQUATION REDUCING AGENT Fes, Reduction NO2 Non Hall-Equation : Gain Total loss of election = 1 (N + 5 4 )

So Noz & œ NO₂ Noz tê NO₂ + H₂O Balancing oxygen Balancing H NO3 + 2H+ + NO + H₂O REDUCTION HALF EQUATION OXIDISING AGENT: NO3 & Now, Multiplying Reduction half equation by 15 adding to oxidation half-equation: 14 H+ Fe + 8 +220 + 15 NO3 + 30 H + 150 – feet + 2 sou + 16 + 15/0 + 15 NO₂ + 15 ₂0 7 H₂O So, the balanced equation: 1 FeS₂ + 15 No3 + 14 H - 15NO₂ + 1 Fest + 2504 2 + 740