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Module 1: Oxidation and Reduction Reactions

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Balancing Redox Equations

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We would learn how to balance equations for redox reactions in acidic solution using half-equations in six steps.Step 1We establish the species containing the atom that is reduced and the product of reduction.We write the change in the form of a half equation.The dichromate VI anion is reduced to chromate III ions.The same thing is repeated for the oxidized species.Free chlorine is obtained by oxidation of chloride ion.Step 2.We balance the atoms that have changed their oxidation state.In the reduction reaction we double the number of chromium III ions.In the oxidation reaction we double the number of chloride ions.Step 3.We balance oxygen and hydrogen atoms.If an oxygen atom is missing in the equation, we add a water molecule.If on the other hand a hydrogen atom is missing, we add a hydrogen ion.Thus in the equation of the reduction of the dichromate ion, we add 7 H2O molecules on the R.H.S and 14 H+ ions on the L.H.S of the equation.There is no need to add H2O molecules or H+ ions to the other half equation.Now both half equations are balanced in terms of atoms.Step 4.We balance the electrical charge by adding electrons to which-ever side that is deficient of electrons.The total charge on the L.H.S of the reduction equation is 14(+1) for H+ plus the charge of the Cr2O72- ions, -2 which gives a total of +12.On the R.H.S of the reaction the only electrically charged species are the chromium III ions.So the total charge on the R.H.S. is two times +3 which equals +6.To balance the charges we add 6 electrons on the L.H.S.In the oxidation half equation we add two electrons on the R.H.S.Step 5.Now we have to make sure that the numbers of electrons lost and the electrons gained are the same.To achieve this we multiply the oxidation half equation by 3.Step 6.We add the half equations together and eliminate any species that occur on both sides of it.Thus we obtain a balanced overall redox equation.