In acidic solution, bromate ion can be used to react with a number of metal ions

Question

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. 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 basic 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) Since this reaction takes place in basic solution, H_2O(l) and OH^- (aq) will be shown in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: 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.

Explanation / Answer

PART-A

Given equation: BrO3 ^1-(aq) + Sn^2+ (aq) -----> Br^1- (aq) + Sn^4+ (aq)

First half equation is: BrO3^1-(aq) ------> Br^1- (aq)

Br^5+ in BrO3^1- (aq) is reduced to form Br^1- (aq)

BrO3^1- (aq) + 6e- -----> Br^1- (aq)

BrO3^1- (aq) + 6e- -----> Br^1-(aq) + 3H2O(l)                            [H2O is added to balance oxygen]

BrO3^1-(aq) + 6H+(aq) +6e- -----> Br^1-(aq) + 3H2O (l)             [H+ is added to balance hydrogen] ------- (1)

Second half equation is Sn^2+ (aq) -----> Sn^4+(aq)

Sn^2+ is oxidized to form Sn^4+

Sn^2+ (aq) - 2e- ------> Sn^4+ (aq)

Mulsiplying both sides of the equation by 3 we get;

3Sn^2+ (aq) -6e- ------> 3Sn^4+ (aq) ---------(2)

Adding equations (1) and (2) we get:

BrO3^1-(aq) + 3Sn^2+(aq) +6H+(aq) -------> Br^1- (aq) + 3Sn^4+ (aq) + 3H2O(l)

Therefore, coefficients of the balanced equation = 1,3,6,1,3,3.

PART-B

Given equation: MnO4^1-(aq) + SO3^2- (aq) ------> MnO2(aq) + SO4^2- (aq)

First half equation is MnO4^1-(aq) ----> MnO2 (aq)

Mn^7+ in MnO4^1- is reduced to Mn^4+ in MnO2

MnO4^1-(aq) + 3e- -----> MnO2(aq)

MnO4^1-(aq) + 3e- -----> MnO2(aq) + 2H2O (l)               [H2O is added to balance oxygen]

MnO4^1-(aq)+ 4H+(aq) + 3e- ------> MnO2(aq) + 2H2O(l)      [H+ s added to balance hydrogen]

Multiplying both sides of the equation by 2 we get:

2MnO4^1-(aq) + 8H+(aq) + 6e- -----> 2MnO2(aq) + 4H2O(l) --------------------------(1)

Second half equation is : SO3^2-(aq) ------> SO4^2-(aq)

S^4+ in SO3^2- is oxidized to S^6+ in SO4^2-

SO3^2-(aq) - 2e- --------> SO4^2-(aq)

SO3^2-(aq) + H2O(l) -2e- -----> SO4^2-(aq)               [Here H2O is added to balance oxygen]

SO3^2-(aq) + H2O(l) -2e- -----> SO4^2- (aq) + 2H+(aq)

Multiplying both sides of the equation by 3 we get;

3SO3^2-(aq) + 3H2O(l) -6e- ------> 3SO4^2- (aq) + 6H+(aq) -----------(2)

Adding (1) and (2) we get;

2MnO4^1-(aq) + 2H+(aq) + 3SO3^2- (aq) -----> 2MnO2 (aq) + 3SO4^2- (aq) + H2O (l)

2OH- ions are added on both sides of the equation because the solution is basic.

2MnO4^1-(aq) + 2H2O(l) + 3SO3^2-(aq) ------> 2MnO2(aq) + 3SO4^2- (aq) + H2O(l) + 2OH-(aq)

Therefore the balanced equartion is :

2MnO4^1-(aq) + 3SO3^2-(aq) + H2O(l) -----> 2MnO2(aq) + 3SO4^2- (aq) + 2OH-(aq)

Therefore, coefficients of the balanced equation = 2,3,1,2,3,2

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