Transcript Balancing Oxidation Reduction Equations

Balancing Oxidation
Reduction Equations
Oxidation and Reduction
•
Oxidation – the atom loses electrons
– The charge becomes more positive
•
Reduction – the atom gains electrons
– The charge becomes more negative
•
Mass and charge must be conserved for
all chemical equations.
Steps for Balancing Redox
reactions
A) Acidic Solutions
Step 1: Write the half reactions for the
oxidation and reduction.
MnO4- + H2SO3 SO42- + Mn2+ + H2O
(Mn goes from Mn7+ to Mn2+ so this is
reduction)
MnO4-  Mn2+
Steps for Balancing Redox
reactions
• S goes from S4+ to S6+, so this is oxidation
H2SO3  SO42Step 2: Balance the half rxns for atoms other
than H and O.
– Already balanced for the example.
Step 3: Balance the O’s by adding H2O.
MnO4-  Mn2+ + 4H2O
H2O + H2SO3  SO42-
Steps for Balancing Redox
Reactions
Step 4: Balance H atoms with H+
MnO4- + 8H+  Mn2+ + 4H2O
H2O + H2SO3  SO42- + 4H+
Step 5: Balance the charge on each side by
adding electrons.
MnO4- + 8H+  Mn2+ + 4H2O
total charge = +7  total charge = +2
So, add 5e- to the reactants side to balance the
charge.
Steps for Balancing Redox
Reactions
5 e- + MnO4- + 8H+  Mn2+ + 4H2O
H2O + H2SO3  SO42- + 4H+
charge = 0
charge = 2+
So,
H2O + H2SO3  SO42- + 4H+ + 2eElectrons must be added to opposite sides
for each half reaction.
Steps for Balancing Redox
Reactions
Step 6: Balance the electrons between both
half reactions.
(5 e- + MnO4- + 8H+  Mn2+ + 4H2O) x 2
10 e- + 2MnO4- + 16H+  2Mn2+ + 8H2O
(H2O + H2SO3  SO42- + 4H+ + 2e-) x 5
5H2O + 5H2SO3  5SO42- + 20H+ + 10eNow the electrons will cancel out
Steps for Balancing Redox
Reactions
Step 7: Cross cancel the electrons and
other substances.
3
10 e- + 2MnO4- + 16H+  2Mn2+ + 8H2O
5H2O + 5H2SO3  5SO42- + 20H+ + 10e4
Step 8: Add the half reactions together
5H2SO4 + 2MnO4-  2Mn2+ + 3H2O +
5SO42- + 4H+
Basic Solutions
Follow the same steps as for acidic
solutions, and add the following steps.
Step 9: Add OH- to both sides to equal the
H+
5H2SO3 + 2MnO4-  2Mn2+ + 3H2O +
5SO42- + 4H+
4 OH
4OH-
Basic Solutions
Step 10: Combine the H+ and OH- to make
water.
4 OH- + 5H2SO3 + 2MnO4-  2Mn2+ + 3H2O
+ 5SO42- + 4H2O
Step 11: Cancel the H2O if they are on both
sides.
4 OH- + 5H2SO3 + 2MnO4-  2Mn2+ + 7H2O
+ 5SO42-