Question

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). Include coefficients of 1, as required, for grading purposes.

Answer 1

SO₄^{2-}(aq) + Sn^{2+}(aq) + 4 H^{+} = H₂SO₃(aq) + Sn^{4+}(aq) + H₂O

Reaction stoichiometry
Compound	Coefficient	Molar Mass
SO4{2-}(aq)	1	96.063697159819
Sn{2+}(aq)	1	118.70890284018
H{+}	4	1.0073914200905
H2SO3(aq)	1	82.07908
Sn{4+}(aq)	1	118.70780568036
H2O	1	18.01528

SO4 2- (aq)+ Sn 2+ (aq) --> H2SO3 (aq) + Sn 4+ (aq)

begin by separating into two pieces

So4 2- (aq) --> H2SO3 (aq)

then balance the S (which is already balanced)

after, balance the O by adding H2O where needed

SO4 2- (aq) --> H2SO3 (aq) + H2O (l)

now that the Oxygens are balanced, balance out the Hydrogens by adding H+ wherever needed.

4H+ (aq) + SO4 2- (aq) --> H2SO3 (aq) + H2O (l)


4e- + 4H+ (aq) + SO4 2- (aq) --> H2SO3 (aq) + H2O (l)


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

all you need to do for this equation is add electrons to balance out the charge

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

now, if you multiply the second equation by 2 the electrons will end up cancelling out

4e- + 4H+ (aq) + SO4 2- (aq) --> H2SO3 (aq) + H2O (l)

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

if you add the two together then you get the balanced equation
4H+ (aq) + SO4 2- (aq) + 2Sn 2+ (aq) --> H2SO3 (aq) + 2Sn 4+ (aq) + H2O (l)