Balance the following redox equations by the ion-electron method: (a) H2O2 + Fe2+ → Fe3+ + H2O (in acidic solution) (b) Cu + HNO3 → Cu2+ + NO + H2O (in acidic solution) (c) CN− + MnO4− → CNO− + MnO2 (in basic solution)
Here is the solution of your question. If you have doubt please comment in comment box and will definitely resolve your query, sequence in the solution is get disturbed but your whole question is answered.. Thanks in advance.
Balance the following redox equations by the ion-electron method: (a) H2O2 + Fe2+ → Fe3+ +...
Write the balanced redox reaction for each of the following: a. H2O2(aq) + Fe2+(aq) → H2O(l) + Fe3+(aq) in acidic solution b. Cu(s) + HNO3(aq) → Cu2+(aq) + NO(g) + H2O(l) in acidic solution c. Bi(OH)3(s) + SnO22–(aq) → Bi(s) + SnO32–(aq) in basic solution
Balance the following redox equations by the half-reaction method: (a) Mn2+ + H2O2 → MnO2 + H2O (in basic solution) (b) Bi(OH)3 + SnO22− → SnO32− + Bi (in basic solution) (c) Cr2O72− + C2O42− → Cr3+ + CO2 (in acidic solution) (d) ClO3− + Cl− → Cl2 + ClO2 (in acidic solution) (e) Mn2+ + BiO3− → Bi3+ + MnO4− (in acidic solution)
Balance the following redox reactions: a) CN− + MnO4− → CNO− + MnO2 (basic solution) b) O2 + As → HAsO2 + H2O (acidic solution) c) Br− + MnO4− → MnO2 + BrO3− (basic solution) d) NO2 → NO3− + NO (acidic solution) e) ClO4− + Cl− → ClO− + Cl2 (acidic solution) f) AlH4− + H2CO → Al3+ + CH3OH (basic solution)
(10 pts) Balance the following redox reactions by first separating the oxidation and reduction half-reactions. a. Cut (aq) + Fe (s) Fe3+ (aq) + Cu(s) b. Cu(s) + HNO3 (aq) Cu2+ (aq) + NO (g) (basic solution) c. NH(aq) + O2(g) → N03 (aq) + H2O(l) (acidic solution) d. Cd(s) + NiO(OH)(s) + Ca(OH)2(s) + Ni(OH)2(s) (Nicad battery) e. The oxidation of iodide ion (1) by permanganate ion (MnO4) in basic solution to yield molecular iodine (12) and manganese(IV) oxide...
2. Balance the following redox equations by the ion-electron half-reaction method: (a) (acid solution): MnO (s) + PbO2 (s) → MnO,- + Pb2+ (b) (acid solution): MnO4 + Mn2+ → MnO2 (s) (c) (basic solution): Al (8) + OH → Al(OH) - + H2(g)
balance pls show work its just to check HW12 Balance the following redox in acidic/basic medium MnO- + Fe2+ → Mn2+ + Fe3+ (acidic) Cr2O72- + C,H-OH → Cr3+ + CO2 (acidic) Ag + CN- + 02 → Ag(CN)2- (basic) Cr2O72- + S032– → Cr3+ + SO42- (acidic) C1,02 + H2O2 → C102 +02 (basic) MnO4 + C H204 → Mn2+ + CO2 (acidic) Sn + NO3- → SnC162- + NO2 (basic)
please help! thanks so much! 10.) Balance the following redox reactions by the ion-electron method: Show your step by step Work, do not just give your answer. a) S2O32- + 12 = 1 +S4062" (in acidic solution) b) Mn2+ + H2O2 → MnO2 + H20 (in basic solution) c) Bi(OH)3 + SnO22- → Sn032 + Bi (in basic solution) d) CIO3 +Cl → Cl2 + ClO2 (in acidic solution)
Balance the following redox equations by the ion-electron half-reaction method: (a) (acid solution): VŽ+ + V(OH). - VO2 (b) (acid solution): C 02 + MnO - CO2(g) + Mn (c) (basic solution): Cr(OH), (s) + H2O2 → CrO
Balance the following redox equations by the ion-electron half-reaction method: (a) (acid solution): MnO2 (s) + Cl → Mn2+ + Cl2 (g) (b) (acid solution): NaBiO; (s) + Ce3+ BiO+ + Ce4+ + Na+ (c) (basic solution): Fe(OH)2 (s) + CrO42- → Fe,0, (s) + Cr(OH)4
please show work so i understand, thank you Balance the following redox equations by the ion-electron half-reaction method: (a) (acid solution): U4+ + MnO4 → UO2+ + Mn2 (b) (acid solution): Zn (s) + NO, Zn2+ + N,O (g) (c) (basic solution): HPO; + MnO4 → P04 + MnO42