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Writing Balanced Equations for Redox Reactions Tutorial

Key Concepts

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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 × (½Cl2(g) + e- → Cl-)
    2 × ½Cl2(g) + 2 × 1e-2 × 1Cl-
    reduction equation: Cl2(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":

    Cl2(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

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