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Balancing OxidationReduction Equations
There are different methods and several steps to balancing oxidationreduction
equations (also known as redox reactions). These include:
The Oxidation Number Method:
Step 1: Write the unbalanced equation, showing the reactants and the products of the reaction (if it is not given).
Step 2: Determine the change in oxidation number which certain elements in both the oxidizing and reducing agents undergo  express these changes as electronic equations .
Step 3: Balance these electronic equations by ensuring that the number of electrons lost by the reducing agent equals those gained by the oxidizing agent.
Combine the two electronic equations that have been balanced.
Step 4: Hence, substitute the coefficients of the balanced electronic equations in the original equation.
Step 5: Finally, balance all remaining groups and elements that have not been balanced.
Example:
1.
Balance the following oxidationreduction equation:
MnO_{2} + HCl → MnCl_{2} + Cl_{2} + H_{2}O
Step 2: The oxidation state of manganese has changed from +4 in MnO_{2} to +2 in MnCl_{2} (reduction); the oxidation state of Cl in HCl has changed from –1 to zero in Cl_{2} (oxidation).
Electronic equations:
Mn^{4+} + 2e^{} → Mn^{2+} ; Cl^{}  1e^{}
→ Cl^{0}
Step 3: Balancing the electronic equations: number of electrons gained must equal number of electrons lost, therefore 2 Cl^{} ion are required to produce the 2 electrons gained by the 1 Mn^{4+} ion.
Mn^{4+} +2e^{} → Mn^{2+}
2Cl^{}  2e^{} → 2Cl^{0} or Cl_{2}
Net: Mn^{4+} + 2Cl^{} → Mn^{2+} + Cl_{2}
Step 4: Substitute the coefficients (i.e. the number multiplying each ion) of the net electronic equation into the original equation. (the equation given in the question) I.e.
MnO_{2} + 2HCl → MnCl_{2} + Cl_{2} + H_{2}O.
Step 5: The equation above is still not completely balanced. Water has not been balanced  two molecules of water are needed to balance the one molecule of MnO_{2}.
Hence, four molecules of HCl will be required to also balance the two molecules of water and chlorine.
I.e.
MnO_{2} + 4HCl → MnCl_{2} + Cl_{2} + 2H_{2}O
The equation is now completely balanced.
By following the above steps, you will be able to quickly balance any oxidation – reduction equation.
2. Balance the equation:
KMnO_{4} + H_{2}SO_{4} + H_{2}S → K_{2}SO_{4} + MnSO_{4} + H_{2}O + S
Oxidation state of manganese has changed from +7 (in KMnO_{4}) to +2 (in MnSO_{4}) – reduction; while oxidation state of sulphur has changed from –2 (in H_{2}S) to 0 (in S)  oxidation.
Therefore, electronic equations:
Mn^{7+} + 5e^{} → Mn^{2+}
S^{2}  2e^{} → S^{0}
To balance the equations, multiply the upper equation by the coefficient of electron of the lower equation (i.e. 2). Also, multiply the lower equation by the coefficient of electron of the upper equation (this is similar to what you do in solving simultaneous equations by elimination method).
I.e.
2Mn^{7+} + 10e^{} → 2Mn^{2+}
5S^{2}  10e^{} → 5S^{0}
Net: 2Mn^{7+} + 5S^{2} → 2Mn^{2+} + 5S^{0}
Now substitute the coefficients of the net electronic equation into the original equation. Notice that the electrons have cancelled out and 2KMnO_{4}, 5H_{2}S, 2MnSO_{4}, and 5S have been balanced.
I.e.
2KMnO_{4} + H_{2}SO_{4} + 5H_{2}S → K_{2}SO_{4} + 2MnSO_{4} + H_{2}O +5S
This is still not completely balanced (H_{2}SO_{4} and H_{2}O are unbalanced). Notice that for equations like this, the oxygen of the oxidizing agents are used in producing water.
Since there are 8 atoms of oxygen in the oxidizing agent, KMnO_{4}, and there is complete reduction of the oxidizing agent, 16 atoms of hydrogen are required to form 8 molecules of water always try balancing from this point, it makes it easier to balance any equation of this type.
The number of SO_{4}^{2} is 3 on the right, and is balanced by increasing H_{2}SO_{4} to 3 molecules. The balanced equation is therefore:
2KMnO_{4} + 3H_{2}SO_{4} + 5H_{2}S → K_{2}SO_{4} + 2MnSO_{4} + 8H_{2}O + 5S
3. Balance the equation:
Cu + HNO_{3} → Cu(NO_{3})_{2} + H_{2}O + NO
Solution:
Cu^{0} – 2e^{} → Cu^{2+}
Copper of oxidation state of zero changes to +2 in Cu(NO_{3})_{2}.
N^{5+} +3e^{} → N^{2+} Nitrogen changes from oxidation state of +5 (in HNO_{3})to +2 in (NO).
Balancing the electronic equations:
3Cu^{0} – 6e^{} → 3Cu^{2+}
2N^{5+} + 6e^{} → 2N^{2+}
Net: 3Cu^{0} + 2N^{5+} → 3Cu^{2+} + 2N^{2+}
Hence, 3Cu +2HNO_{3} → 3Cu(NO_{3})_{2} + H_{2}O + 2NO
This is not yet completely balanced 3Cu(NO_{3})_{2}, 3Cu and 2NO are balanced but 2HNO_{3} and H_{2}O are not balanced).
There are 8 nitrogen on the right; hence there must be 8 molecules of HNO_{3}. The number of hydrogen becomes 8, and these form 4 molecules of water.
The completely balanced equation is:
3Cu + 8HNO_{3} → 3Cu(NO_{3})_{2} + 4H_{2}O + 2NO
The balanced equation shows that not all the nitrate are reduced by Cu. Out of 8 molecules of nitrate, only 2 are reduced (Cu being a weak reducing agent).
This reaction occurs when copper reacts with
dilute trioxonitrate(V) acid – this is the recognized laboratory preparation of nitrogen monoxide, NO.
4. Balance the equation: Cu + HNO_{3} → Cu(NO_{3})_{2} + H_{2}O
+ NO_{2}
Solution:
Notice that this is the main reaction, which occurs when copper reacts with conc. trioxonitrate(V) acid (NO_{2} is produced)
Cu^{0} – 2e^{} → Cu^{2+}  oxidation state of Cu changes from zero (in Cu) to +2 (in Cu(NO_{3})_{2})
N^{5+} + e^{} → N^{4+}  oxidation state of N changes from +5 (in HNO_{3}) to +4 (in NO_{2})
Balancing the electronic equations:
Cu^{0}  2e^{} → Cu^{2+}
2N^{5+} + 2e^{} → 2N^{4+}
Net: Cu^{0} + 2N^{5+} → Cu^{2+} + 2N^{4+}
Hence, Cu + 2HNO_{3} → Cu(NO_{3})_{2} + H_{2}O
+ 2NO_{2}
This equation is not yet completely balanced (Cu, Cu(NO_{3})_{2} and 2NO_{2} are balanced, but HNO_{3} and H_{2}O are not balanced).
The number of nitrogen on the right is 4, therefore, number of HNO_{3} must be 4 – this makes number of water to be 2 (4 hydrogens).
Notice that the reason why HNO_{3} is not considered balanced from the electronic equation is that Cu, being a weak reducing agent will not reduce all of the nitrates present – the electronic equation shows only the nitrates that are reduced.
The completely balanced equation is: Cu + 4HNO_{3}
→ Cu(NO_{3})_{2} + 2H_{2}O + 2NO_{2}
5. Balance the equation:
HNO_{3} + I_{2} → HIO_{3} + NO_{2} + H_{2}O
Solution:
N^{5+} + e^{} → N^{4+}  oxidation state of N changes from +5 (in HNO_{3}) to +4 (in NO_{2}).
I^{0}  5e^{} → I^{5+}  oxidation state of I changes from zero (in I_{2}) to +5 (in HIO_{3}).
Balancing the electronic equations:
5N^{5+} + 5e^{} → 5N^{4+}
I^{0}  5e^{} → I^{5+}
Net: 5N^{5+} + I^{0} → 5N^{4+} + I^{5+}
Since in the original equation we have two atoms of iodine, we now modify the net electronic equation by multiplying by 2. i.e.
10N^{5+} + 2I^{0} → 10N^{4+} + 2I^{5+}.
Therefore;
10HNO_{3} + I_{2} → 2HIO_{3} + 10NO_{2} + H_{2}O
This is not yet completely balanced. 10HNO_{3}, I_{2}, 2HIO_{3} and 10NO_{2} are balanced, only H_{2}O is not balanced.
Therefore, on the left there are 10 hydrogen atoms, this must be balanced on the right. I.e. water molecules must be 4 (i.e. 8 hydrogen atoms).
Also, oxygen atoms are 30 on the left, this must also be balanced on the right. I.e. 4 oxygen atoms must make up water.
Therefore, the balanced equation is:
10HNO_{3} + I_{2} → 2HIO_{3} + 10NO_{2} + 4H_{2}O
6. Balance the equation:
K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + H_{2}S
→ KHSO_{4} + Cr_{2}(SO_{4})_{3} + H_{2}O + S
Solution:
Cr^{6+} + 3e^{} → Cr^{3+}  oxidation state of Cr changes from +6 (in K_{2}Cr_{2}O_{7}) to +3 (in Cr_{2}(SO_{4})_{3}).
S^{2}  2e^{} → S^{0} – oxidation state of sulphur changes from –2 (in H_{2}S) to zero (in S)
Balancing the equations:
2Cr^{6+} + 6e^{} → 2Cr^{3+}
3S^{2}  6e^{} → 3S^{0}
Net: 2Cr^{6+} + 3S^{2} → 2Cr^{3+} + 3S^{0}
Substitute coefficients of the balanced electronic equation into the original equation that is given, we have:
K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + 3H_{2}S
→ KHSO_{4} + Cr_{2}(SO_{4})_{3} + H_{2}O + 3S
This equation is not yet completely balanced (K_{2}Cr_{2}O_{7}, 3H_{2}S, Cr_{2}(SO_{4})_{3} and 3S are balanced but H_{2}SO_{4}, KHSO_{4} and H_{2}O are not) .
On the left side, there are two atoms of K, hence there must be 2 KHSO_{4} on the right  this makes the number of sulphate on the right 5, hence H_{2}SO_{4} must be 5.
There are 7 oxygen atoms in the oxidizing agent, K_{2}Cr_{2}O_{7}), these are used to make water, i.e. number of water molecules is 7. The balanced equation is therefore:
K_{2}Cr_{2}O_{7} + 5H_{2}SO_{4} + 3H_{2}S
→ 2KHSO_{4} + Cr_{2}(SO_{4})_{3} + 7H_{2}O + 3S
Note: equations in which there is complete reduction of the oxidizing agent (e.g. questions 1, 2 and 6 above) can easily be balanced after going through the process of change in oxidation state. All you need to do is determine the number of oxygen atoms in the oxidizing agent, which forms water. This will give you the key to balancing the hydrogens, and then other species in the reaction.
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