Part A In acidic solution, the sulfate ion can be used to react with a number of
ID: 835205 • Letter: P
Question
Part A In acidic solution, the sulfate ion can be used to react with a number of metal ions. One such reaction is
SO42?(aq)+Sn2+(aq)?H2SO3(aq)+Sn4+(aq)
Since this reaction takes place in acidic solution, H2O(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:
SO42?(aq)+Sn2+(aq)+ ????H2SO3(aq)+Sn4+(aq)+ ???
What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and H+(aq) in the appropriate blanks. Your answer should have six terms.
Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3).
Part B Basic solution
Potassium permanganate, KMnO4, 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 fluoride:
MnO4?(aq)+F?(aq)?MnO2(s)+F2(aq)
Since this reaction takes place in basic solution, H2O(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:
MnO4?(aq)+F?(aq)+ ????MnO2(s)+F2(aq)+ ???
What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l) and OH?(aq) in the blanks where appropriate. Your answer should have six terms.
Enter the equation coefficients in order separated by commas (e.g., 2,2,1,4,4,3).
Explanation / Answer
reduction: (SO4^2- (aq) with sulfur at +6) & 2 electrons taken ----> H2SO3 (aq) (with sulfur at +4) oxidation: Sn2+ (aq) ----> Sn^4+ (aq) & 2 electrons lost
simplifies to, & has electrons balanced as well:
(SO4)-2 & 2 e- ----> H2SO3 (aq)
oxidation: Sn2+ (aq) ----> Sn^4+ (aq) & 2 e-
giving us:
(SO4)-2 & Sn+2 (aq) (with a total ion charge of zero) ----> H2SO3 & Sn+4 (w/ total ion charge of +4)
adding 4 H+'s to balance the charges at +4 <+> +4
(SO4)-2 & Sn+2 (aq) & 4 H+ ----> H2SO3 & Sn+4
adding H2O to balance H's & O's
(SO4)-2 & Sn+2 (aq) & 4 H+ ----> H2SO3 & Sn+4 & H2O
(SO4)-2 + Sn+2 (aq) + 4 H+ ----> H2SO3 + Sn+4 + H2O
done
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reduction: (MnO4)-1 with Mn at +7 takes 3 electrons ----> MnO2 with its Mn at +4
oxidation: 2 F-1 (aq) ----> F2 & 2 electrons lost
they simplify as:
reduction: (MnO4)-1 & 3 e- ----> MnO2
oxidation: 2 F-1 ----> F2 & 2 e-
they balance electrons as:
reduction: 2 (MnO4)-1 & 6 e- ----> 2 MnO2
oxidation: 6 F-1 ----> 3 F2 & 6 e-
giving us:
2 (MnO4)-1 & 6F- (with total ion charges at -8) ----> 2 MnO2 & 3F2 (with total ion charges at zero)
adding (OH)-1 's to balance charges as -8 <=> -8:
2 (MnO4)-1 & 6F- ----> 2 MnO2 & 3F2 & 8 (OH)-1
adding water, to balance H's & O's
2 (MnO4)-1 & 6F- & 4 H2O ----> 2 MnO2 & 3F2 & 8 (OH)-1
2 (MnO4)-1 + 6F- + 4 H2O ----> 2 MnO2 + 3F2 + 8 (OH)-1