We have to balance the given redox equation by using H2O(l) and H+(aq) and determine their coefficients in the balanced chemical equation.
We will start by splitting the equation into two half-reactions and then determine which halves are reduction and oxidation halves.
Sn2+(aq) → Sn4+(aq)
Recall that the charge on a metal cation is its oxidation state.
The oxidation state of Sn is increasing from +2 to +4 therefore, this is the oxidation half-reaction.
SO42-(aq) → H2SO3(aq)
Since we have already determined the oxidation half-reaction, this one must be the reduction half-reaction.
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)
Sicne 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 balaced equation above? Remember to include H2O (l) and H+ (aq) in the appropriate blanks. Your answer should have six terms.
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