Question

Balance the following redox reaction in acidic solution by inputting the correct coefficients.

Fe²⁺ + O₂ → Fe³⁺ + H₂O

If H⁺ or H₂O is not present as reactant or product, select "0" as the coefficient for the purposes of this question. You must also explicitly label a coefficient as "1" if only one molecule is present.

Important: You MUST simplify all "like terms" on both sides of the chemical equation before selecting your answer: having the same compound on both sides of the equation will be marked incorrect.

* The products are listed below the reactants so the whole equations fits on your computer screens better *

Fe²⁺ + O₂+ H⁺ + H₂O

↓

Fe³⁺ + H⁺ + H₂O

Answer 1

increase of oxidation numbers oxidation Coxidised) decrease of oxidation number Reduction (Reduced) lose of electronsa oxidation gain of electrons = Reduction if both oxidation and Reduction are present in one reaction is called red- of Reaction mical reaction if an atom ioncors. mdocette loses electrons it is said to be oxidised and is called a reducing agent. and if it gain electrons it is said to be reduced and is called an oxidising agent

Rules Free (ory elementary oxidation state of clement in the state is always is zero Ex oxidation state of He to oxidation state of chlorine in clao © oxidation state of the element in the monoatomic ion is equal to the charge on the ion . I-A group element's CH. Li, inq, k, Rb) is a ti ① II-A group element's (Be, mg, ca, sr, Ba, Rq) is at2 - oxidation state. ② oxidation state of vis a group (F,<) Br, I) is =- - - ☺ oxidation state of oxygen is = -2 come of the oxidation state of all aroms in neatral molecule is zero © But in the case of ion: The sum of the oxidation state of all aroms is equal to the charge on the ion.

ASLOW MO22 - 2. Rules i inacidic solution) o increase of oxidation number oxidation .. ☺ decrease of oxidation number Reduction ③ : First separate into two half reactions Balance all atoms except oxygen and hydrogen. Balance oxygen atoms by adding to Balance hydrogen atoms by adding tt : Balance charge by adding electrons : o multiply with suitable integer to eżialise number of elections @ add both.

Feth +ą -> Fet3 + H, 0 tn Ht Feth, 6., Fet3 | H₂O or to | 21201 4=+3 | 2(+) +*20. Xanh oxidation half reaction Reduction half reaction x=12 I x=0 Feth - Fet3 ... ? ? Ho Balance oxygen atoms: by adding Ho molecules fet? Fet3 . q HOTIHO q . >.2 Ho Balance Hydrogen atoms by adding . Htions Fete ? Fet3 q+ 4Ht! 7.2H₂O Balance charge by adding electrons Feth Fet3 tel . Iq+ 4H+ 40 72HO multiply with suitable integer to equalise numbers of electrons 지 O X4 4reth 4 Fets the g+ 447 4401 - 240 add.both 4Feth > 4 Fets + 4e q+ 4Htluye » 2H₂O 4Feth + q + 4H+ -> 4 Fet3+2H0

Feth to - F e +3 + ho Fet2 a l fet | Ho . 1 24301=+32 (+1) +420 H~ Ht Orio 20th 2011 X -L . Reduction half reaction oxidation half reaction Feth - Fet3 . Balance oxygen atoms | by adding Ho molecules Feth Fet3 . 1 > 1 HOTIH 19:-)2H0 i Balance Hydrogen atoms by adding til ions Feth ? Fet3 |q+44' _? 2H2O Balance charge by adding electrons Feth - Fet3 the q+444+ 40 -> 2HO multiply with suitable integer to equalise number of electrons Ox4 4reth 4 Fets the g+ 4H+ + 4€! ~ 240 IDXI add.both 4Feth. q+ 4Htlaqe > 4 Fet3+ 4€ 2 Ho + 4H+ - 4 fets + 2Ho v