In addition to mass balance oxidation-reduction reactions must be balanced such

Question

In addition to mass balance oxidation-reduction reactions must be balanced such that the number of elections lost m the oxidation equals the number of electrons gamed in the reduction This balancing can be done by two methods. The oxidation number method balances the net increase in oxidation of the substance oxidized with the net decrease in the oxidation number of the substance reduced The half-reaction method balances the electrons lost in the oxidation half-reaction with the electrons gained w the reduction half-reaction. In both methods H_2;O(l), OH^-(aq), and H^+ (aq) may be added to complete the mass balance. Which substances are used depends on the reaction conditions. In acidic solution, bromate ion can be used to react with a number of metal ions One such reaction is BrO_3^- (aq) + Sn^2+ (aq) rightarrow Br^- (aq) + Sn^4+ (aq) Since this reaction takes place in acidic solution, H_2O(l) and H^+ (aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: BrO_3^- (aq) + Sn^2+ (aq) + rightarrow Br^- (aq) + Sn^4+ (aq) + What are the coefficients of the six species in the balanced equation above? Remember to include coefficients for H_2O(l) and H^+ (aq) in the appropriate blanks. Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3). Potassium permanganate, KMnO_4, is a powerful oxidizing agent. The products of a given redox reaction with the permanganate ion depend on the reaction conditions used. In have solution, the following equation represents the reaction of this ion with a solution containing sodium sulfite MnO_4^- (aq) + SO_3^2- (aq) + rightarrow MnO_2(s) + SO_4^2- (aq) + What are the coefficients of the six species in the balanced equation above? Remember to include coefficients for H_2O(l) and OH^- (aq) in the blanks where appropriate. Enter the equation coefficient in order separated by commas (eg.,2,2,1,4,4,3).

Explanation / Answer

acidic solution

split in half cells

BrO3- --> Br-

Sn+2 --> Sn+4

balance Oxygen adding H2O

BrO3- --> Br- + 3H2O

Sn+2 --> Sn+4

balance H adding H+ (acidic)

6H+ + BrO3- --> Br- + 3H2O

Sn+2 --> Sn+4

balance charges

6H+ + BrO3- --> Br- + 3H2O + 4e-

2e- + Sn+2 --> Sn+4

balance electrons

6H+ + BrO3- --> Br- + 3H2O + 4e-

4e- + 2Sn+2 --> 2Sn+4

badd both equaitons

6H+ + BrO3- + 4e- + 2Sn+2 --> Br- + 3H2O + 4e- +  2Sn+4

cancel common terms

6H+ + BrO3- + 2Sn+2 --> Br- + 3H2O + 2Sn+4

for basic solutoin

MnO4 - --> MnO2

SO3-2 --> SO4-2

balance O adding H2O

MnO4 - --> MnO2

H2O + SO3-2 --> SO4-2

balance H adding H

MnO4 - --> MnO2

H2O + SO3-2 --> SO4-2 + 2H+

balance charges

MnO4 - --> MnO2 + e-

2e- + H2O + SO3-2 --> SO4-2 + 2H+

balance electorns

2MnO4 - --> 2MnO2 + 2e-

2e- + H2O + SO3-2 --> SO4-2 + 2H+

add both

2e- + H2O + SO3-2 + 2MnO4 - --> 2MnO2 + 2e- +  SO4-2 + 2H+

Cancel terms

H2O + SO3-2 + 2MnO4 - --> 2MnO2 +SO4-2 + 2H+

add OH- ions to balance H+ (basic)

2Oh- + H2O + SO3-2 + 2MnO4 - --> 2MnO2 +SO4-2 + 2H+ + 2Oh-

2Oh- + H2O + SO3-2 + 2MnO4 - --> 2MnO2 +SO4-2 + 2H2O

cnacle common terms

2OH-+ SO3-2 + 2MnO4 - --> 2MnO2 +SO4-2

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.