How to Balance a Redox Reaction by the Ion-electron 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
ion-electron method (in basic medium):
Cr(OH)3(s) + IO3─(aq) → I(aq)─ + CrO42─(aq)
Solution:
Balancing a redox reaction by the ion-electron or half reaction method:
- Write the given unbalanced equation without the
states of elements/compounds/ions.
- 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.
- Write the half-equations by considering a decrease
in OS (Red) of a particular element and an increase in OS
(Ox) of another element.
- Do the mass balance of the elements other than O
& H in the half-equations.
- Do the mass balance of O followed by H
for the half-equations.
Acidic or neutral medium: To the
other side, add the same number of H2O molecules as the
number of excess O atoms in one side. Then balance the number of H
atoms by introducing H+ to the side that requires.
Basic medium: Find the O
excess in one side, add the same number of H2O molecules as
excess and to the same side as excess. For each H2O, add two OH-
ions in the other side. If H is still unbalanced, it is balanced by
adding one OH- for every excess of H on the same side
as excess and one H2O on the other side.
- 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-).
- Multiply both equations by appropriate numbers so
that the number of e- becomes the same for
both.
- 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.
- If a stoichiometric balance is achieved (masses and charges are the same in both sides), bring back the states (s, l, g, or aq) to write the final balanced equation.
Stoichiometry settled❓ 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.
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