How to Balance a Redox Reaction by the Oxidation State Method – Part 1

Redox, reduction–oxidation or oxidation–reduction is a type of chemical reaction in which the oxidation states (OSs) of the reactants change.

Oxidation (Ox) is the loss of electrons or an increase in the OS, while reduction (Red) is the gain of electrons or a decrease in the OS. The Ox and Red processes occur simultaneously in the chemical reaction; and hence, the name.

Mnemonic: OIL RIG (Oxidation Is Loss, Reduction Is Gain)

Stoichiometry is the relationship between the quantities of reactants and products before, during, and after chemical reactions.

A chemical equation is the representation of a chemical reaction in the form of symbols and chemical formulas. The coefficients next to the symbols and formulas of entities are the absolute values of the stoichiometric numbers.

Question:

Balance the following by the OS method:

K₂Cr₂O₇(s) + HCl(l) → KCl(aq) + CrCl₃(aq) + Cl₂(g) + H₂O(l)

Solution:

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Balancing a redox reaction by the OS method:

1. Write the given unbalanced equation without the states of elements/compounds/ions.
2. Find the OSs of the elements that undergo changes in their OSs. To do that, first find the total OSs of the elements in molecules/ions that may NOT undergo changes in their OSs (for example, O (-2 per atom) or H (+1 per atom). In the elementary states, OS is zero (for instance, H₂ or O₂ or Na or Mg). Look at the formal charges of the molecules/ions, and calculate the OSs of the elements that undergo changes in their OSs. The alkali and alkaline earth metals have only one OS in the compounds/ions (+1 & +2 respectively).
3. Write the half-equations by considering a decrease in OS (Red) of a particular element and an increase in OS (Ox) of another element.
4. Do the mass balance of the elements in the half-equations.
5. Once the mass balance looks good, do the charge balance for the half-equations by adding the required number of electrons (e^-) to the side that needs. You will observe that the Red half-equation has e^- in the left-hand side or LHS (indicating gain of e^-) and the Ox half-equation has e^- in the right-hand side or RHS (indicating loss of e^-).
6. Multiply both equations by appropriate numbers so that the number of e^- becomes the same for both.
7. Add the half-equations (LHS to LHS & RHS to RHS), cancel e^- (since identical in both sides) and excess molecules or ions (like H₂O or H⁺) from one of the sides.
8. Transfer the coefficients to the concerned elements/compounds/ions.
9. Do the mass balance of O followed by H.
10. Do the mass balance for the other substances not undergoing redox.
11. If a stoichiometric balance is achieved (masses are the same in both sides), bring back the states (s, l, g, or aq) to write the final balanced equation.

Looking for the ion-electron method instead? Click HERE and HERE.

Befuddled by balancing a redox reaction ❓ Let Chemaficionado know in the comments below or at mychemistryhomework@gmail.com❗

References:

(1)    Wikipedia Contributors. Redox. Wikipedia. https://en.wikipedia.org/wiki/Redox.

(2)    Wikipedia Contributors. Stoichiometry. Wikipedia. https://en.wikipedia.org/wiki/Stoichiometry.

‌(3)    Wikipedia Contributors. Chemical equation. Wikipedia. https://en.wikipedia.org/wiki/Chemical_equation.

‌‌(4)    Ncert. Chemistry : Textbook for Class XI - Part.II; National Council Of Educational Research And Training: New Delhi.