1. Balance the following redox reaction using the method of half-reactions (also called the ion-electron method).
Mn2+ (aq) + NaBiO3 (s) → Bi3+ (aq) + MnO4- (aq) + Na+ (aq)
2. Balance the following reaction is basic media. (remember you should have no H+ ions in your final reaction.)
Pb(OH)42- (aq) + ClO- (aq) → PbO2 (s) + Cl- (aq)
1. Balance the following redox reaction using the method of half-reactions (also called the ion-electron method)....
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 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
More Practice Balancing Redox Reactions in Acid and Basic Solution Balance the following redox reaction in both acidic and basic solution using the half reaction method outlined in Recitation 10, Part III. 3. Unbalanced: PbO2(s) + Mn2+(aq) → Pb2+ (aq) + MnO4- (aq) 4. Unbalanced: SO42-(aq) + Cr3+(aq) → SO2(g) + Cr2O72-(aq)
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)
1. (a) Balance the following half-reactions by the ion-electron half-reaction method: (i) (acid solution) HSO3- (aq) → SO42- (aq) (ii) (acid solution) Pu(OH)4 (8) ► Pu+ (aq) (iii) (base solution) CN- (aq) → OCN- (aq) (iv) (base solution) RhO42- (aq) → Rh,03 (s) (b) Which of the above equations are oxidation half-reactions? (c) Give the formulas of the reactants that become oxidized in the course of the reactic Balance the following redox equations by the ion-electron half-reaction method: (a) (acid...
Balance the following redox-reaction in a basic solution by using the ion-electron method ClO-(s) + Cr2O72-(aq) --> ClO3-(aq) + CrO22-(aq)
How do I solve the following redox reactions? What are the balanced half-reactions? What is the final balanced equation? Problems 1. MnO4 (aq) + SO32- (aq) → MnO2 (s) + SO42- (aq) in basic solution 2. NO2" (aq) + Al(s) NH3(aq) + Al(OH)4 (aq) in basic solution 3. Mn2+ (aq) + NaBiO3 (s) → Bi* (aq) + MnO4 (aq) + Nat (aq) in acidic solution 4. As2O3 (s) + NO3- (aq) → H3ASO4 (aq) + N2O3 (aq) in acidic solution...
Balance the following redox reactions by balancing the half reactions and then combine the half reactions to get the overall balanced redox reaction with the lowest possible whole number coefficients. 1. Consider the following unbalanced redox reaction: MnO2(s) + BrO3−(aq) → MnO4−(aq) + Br−(aq) (a) Balance the corresponding half reactions in acidic conditions using the lowest possible whole number coefficients. (Enter coefficients for one and zero. Blanks will be marked incorrect.) MnO2(s) + H2O(l) + OH−(aq) + H+(aq) + e− → MnO4−(aq) + H2O(l) + OH−(aq) + H+(aq)...
Use the half reaction method for a redox reaction to balance the following skeletal reaction in an acidic solution. (Give step by step) Mn2+ + BiO3– = MnO4– + Bi3+ [acidic]
Balancing Oxidation–Reduction Reactions (Section)Complete and balance the following equations, and identify the oxidizing and reducing agents:(a) Cr2O72-(aq) + I-(aq)→Cr3+(aq) + IO3-(aq) (acidic solution)(b) MnO4-(aq) + CH3OH(aq) →Mn2+(aq) +HCO2H(aq) (acidic solution)(c) I2(s) + OCl-(aq)→IO3-(aq) + Cl-(aq) (acidic solution)(d) As2O3(s) + NO3-(aq)→H3AsO4(aq) + N2O3(aq) (acidic solution)(e) MnO4-(aq) + Br-(aq)→MnO2(s) + BrO3-(aq) (basic solution)(f) Pb(OH)42-(aq) + ClO-(aq)→PbO2(s) + Cl-(aq) (basic solution)