Writing Balanced Equations for Redox Reactions Chemistry Tutorial

Key Concepts

• Reactions in which electrons are transferred from one species to another are known as redox reactions, or, oxidation-reduction reactions, or, electron-transfer reactions.
• A redox reaction is made up of two half-reactions:

  (i) A reduction half-reaction in which one species, X, gains a electrons.

  X + ae^- → X^a-

  (ii) An oxidation half-reaction in which one species, M, loses b electrons.

  M → M^b+ + be^-

• In a redox reaction, the number of electrons lost by the species being oxidized must balance the number of electrons gained by the species being reduced.

  (1) Multiply balanced reduction half-equation by b (number of electrons lost in oxidation half-equation)	bX	+	bae-	→	bXa-
  (2) Multiply balanced oxidation half-equation by a (number of electrons gained in reduction half-equation)			aM	→	abe-	+	aMb+
  (3) Add the 2 half-equations together to give the overall redox equation	bX	+	aM	→	bXa-	+	aMb+

• In a balanced redox reaction equation:

  (1) The number of atoms of each element must be balanced
  (see balancing chemical equations).

  (2) The total charge on the ions on the left hand side of the equation will equal the total charge on the ions on the right hand side of the equation.

  (3) Electrons will not appear as either a reactant or a product in the equation
  (they have been "cancelled out" in the process of balancing the redox reaction equation).

• Redox Reactions are very widely used, for example: to

  (i) generate electricity (galvanic cells, fuel cells)

  (ii) recharge rechargable batteries

  (iii) extract of metals from their ores electrolytically (electrowinning, electrorefining, extraction of aluminium, extraction of sodium)

  (iv) plate one metal onto another electrolytically in a process called electroplating

  (v) quantitatively determine the amount of a substance present (redox titrations, ethanol in alcoholic drinks, concentration of hypochlorite in commercial bleach)

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Worked Examples for Writing Balanced Redox Equations

Question 1. Write the balanced redox equation given the two balanced half-equations below:
    (i) Mg → Mg²⁺ + 2e^-
    (ii) ½Cl₂ + e^- → Cl^-

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Write the balanced redox reaction equation

2. What data (information) have you been given in the question?

   Extract the data from the question:

   (i) oxidation reaction: Mg → Mg²⁺ + 2e^-
   Mg has lost two electrons to form Mg²⁺, oxidation is the loss of electons)

   (ii) reduction reaction: ½Cl₂ + e^- → Cl^-
   Cl₂ has gained electrone to form Cl^-, reduciton is the gain of electons)

3. What is the relationship between what you know and what you need to find out?

   For the redox equation, the number of electrons lost by Mg atoms must equal the number of electrons gained by Cl atoms:

   (1) Multiply balanced reduction half-equation by 2
   (number of electrons lost by Mg atoms in the oxidation half-equation is 2)

   2 × (½Cl_2(g) + e^- → Cl^-)
   2 × ½Cl_2(g) + 2 × 1e^- → 2 × 1Cl^-
   reduction equation: Cl_2(g) + 2e^- → 2Cl^-

   (2) Multiply balanced oxidation half-equation by 1
   (number of electrons gained by Cl atoms in the reduction half-equation is 1)

   1 × (Mg_(s) → Mg²⁺ + 2e^-)
   oxidation equation: Mg_(s) → Mg²⁺ + 2e^-

   (3) Add the two half-equations together, making sure there are the same number of electrons on the left hand side of the equation as there are on the right hand side.
   (This is referred to as "cancelling out", or "balancing out", the electrons.)

4. Write the balanced redox reaction equation:

   reduction equation:	Cl2(g)	+	2 e-	→	2Cl-
   oxidation equation:			Mg	→	2 e-	+	Mg2+
   redox equation:	Cl2(g)	+	Mg	→	2Cl-	+	Mg2+

5. Is your answer plausible?

   First, make sure there are no electrons on either side of your redox equation.
   If there are, then the redox equation is not properly balanced.
   Our equation is OK, there are no electrons present on either side of the equation.

   Second, check that the number of atoms of each element present one side of the equation is the same as that on the other side:
   

   redox equation:	Cl2(g)	+	Mg	→	2Cl-	+	Mg2+
   No. Cl "atoms":	2			=	2			balanced
   No. Mg "atoms":			1	=			1	balanced

   
   Our equation is balanced as regards to the number of atoms of each element present.

   Finally, check that the total charge on the left hand side of the equation is the same as that on the right hand side:
   

   redox equation:	Cl2(g)	+	Mg	→	2Cl-	+	Mg2+
   total charge:	0	+	0	=	(2 × -1)	+	2	balanced
   	0			=	-2 + 2 = 0

   We are reasonably confident that our balanced redox equation is plausible.

6. State your solution to the problem "write a balanced redox reaction equation":

   Cl_2(g) + Mg → 2Cl^- + Mg²⁺

Question 2: Write the redox equation given the two balanced half-equations below:
        (i) Cu → Cu²⁺ + 2e^-
        (ii) HNO₃ + H⁺ + e^- → NO₂ + H₂O

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Write the balanced redox reaction equation

2. What data (information) have you been given in the question?

   Extract the data from the question:

   (i) oxidation reaction: Cu → Cu²⁺ + 2e^-
   (oxidation means electons are lost, electrons are a product of the reaction)

   (ii) reduction reaction: HNO₃ + H⁺ + e^- → NO₂ + H₂O
   (reduction means electrons are being gained, electrons are a reactant)

3. What is the relationship between what you know and what you need to find out?

   The number of electrons produced by the oxidation reaction must be equal to the number of electrons consumed by the reduction reaction.

   (1) Multiply balanced reduction half-equation by 2
   (number of electrons produced in the oxidation half-equation is 2)

   2 × (HNO₃ + H⁺ + e^- → NO₂ + H₂O)
   2 × HNO₃ + 2 × 1H⁺ + 2 × 1e^- → 2 × 1NO₂ + 2 × 1H₂O
   reduction reaction: 2HNO₃ + 2H⁺ + 2e^- → 2NO₂ + 2H₂O

   (2) Multiply balanced oxidation half-equation by 1
   (number of electrons gained in the reduction half-equation is 1)

   1 × (Cu_(s) → Cu²⁺ + 2e^-)
   oxidation equation: Cu_(s) → Cu²⁺ + 2e^-

   (3) Add the two half-equations together, making sure there are the same number of electrons on the left hand side of the equation as there are on the right hand side.
   (This is referred to as "cancelling out", or "balancing out", the electrons.)

4. Write the balanced redox reaction equation:

   reduction:	2HNO3 + 2H+ + 2e-	→	2NO2 + 2H2O
   oxidation:	Cu(s)	→	2e- + Cu2+
   redox:	2HNO3 + 2H+ + Cu(s)	→	Cu2+ + 2NO2 + 2H2O

5. Is your answer plausible?

   First, make sure there are no electrons on either side of your redox equation.
   If there are, then the redox equation is not properly balanced.
   Our equation is OK, there are no electrons present on either side of the equation.

   Second, check that the number of atoms of each element present one side of the equation is the same as that on the other side:
   

   redox:

   2HNO3

   +

   2H+

   +

   Cu(s)

   →

   Cu2+

   +

   2NO2

   +

   2H2O

   No. H "atoms":

   2

   +

   2

   =

   2 × 2 = 4

   balanced

   No. N "atoms":

   2

   =

   2

   balanced

   No. O "atoms":

   2×3=6

   =

   2×2=4

   +

   2×1=2

   balanced

   No. Cu "atoms":

   1

   =

   1

   balanced

   
   Atoms in our redox equation are balanced.

   Finally, check that the total charge on the left hand side of the equation is the same as that on the right hand side:
   

   redox:

   2HNO3

   +

   2H+

   +

   Cu(s)

   →

   Cu2+

   +

   2NO2

   +

   2H2O

   total charge:

   2×1+ = 2+

   =

   2+

   balanced

   
   We are reasonably confident that our redox reaction equation is balanced.

6. State your solution to the problem "write a balanced redox reaction equation":

   2HNO₃ + 2H⁺ + Cu → Cu²⁺ + 2NO₂ + 2H₂O