Given the redox reaction: Sn + HNO3 --> SnO2 + NO2+ H2O, identify the following: (see details)?
a.) Identify the substance oxidized and the substance reduced.
b.) Identify the oxidizing and reducing agents.
c.) Identify the oxidation and reduction half-reactions and determine the number of electrons lost and gained.
d.) Please balance the equation.
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- 1 decade agoFavorite Answer
Sn in Sn: Oxid State 0
Sn in SnO2: Oxid State +4 (since O is -2 and there are 2 of them)
During oxidation, oxidation number increases, so Sn is being oxidised.
N in HNO3: +5 (since H is +1 and O is -2 and there are 3 O's)
N in NO2: +4 (since O is -2 and there are 2 O's)
During reduction, oxidation number decreases, so N is being reduced.
b) The Sn is being oxidised by the HNO3, so the HNO3 is the oxidising agent.
The N is being reduced by the Sn, so the Sn is the reducing agent.
c) Sn + 2H2O --> SnO2 + 4H+ + 4e
NO3- + 2H+ + e --> NO2 + H2O
The Sn loses 4e's (0 to +4) and the N gains an e (+5 to +4)
d) In a redox reaction, as many electrons are lost as gained. So the second half-equation needs to be multiplied by 4
Sn + 2H2O --> SnO2 + 4H+ + 4e
4NO3- + 8H+ + 4e --> 4NO2 + 4H2O
Summing up, and noticing the 4e on right cancel 4e on left, 4 of the H+ on the left cancel out the 4H+ on the right (leaving 4H+ from 8H+ on the left), and 2 H2O on the left cancel out 2 of the H2O on the right (leaving 2 from 4 H2O on the right)
Sn + 4NO3- + 4H+ --> SnO2 + 4NO2 + 2H2O
Sn + 4HNO3 --> SnO2 + 4NO2 + 2H2O