Oxidation numbers and balancing redox equations
When a piece of zinc is dropped into hydrochloric acid, the gas that fizzes off is hydrogen. Each zinc atom hands over two electrons, and those electrons end up on hydrogen ions. In a reaction with many atoms, counting electrons one by one would be exhausting. Oxidation numbers give you a bookkeeping system to track them, and half-equations let you balance the equation step by step. By the end of this page you will be able to balance equations like these in the Leaving Cert style.
Oxidation numbers
Put a clean strip of copper into colourless silver nitrate solution. The liquid slowly turns blue and shiny crystals of silver grow on the metal. No wires, no sparks, yet electrons have moved. How can we track the electrons gained or lost by each element in any compound?
When naming compounds, the oxidation number is written in Roman numerals, as in iron(II) sulfate or potassium manganate(VII). When writing it on an atom, put the sign first: $+2$, $-1$.
The concept of oxidation number presents special cases in certain binary compounds such as peroxides and metal hydrides. Using the general assignment rules: a) Determine…
The concept of oxidation number presents special cases in certain binary compounds such as peroxides and metal hydrides. Using the general assignment rules:
- a) Determine the oxidation number of oxygen in barium peroxide, $\mathrm{BaO_2}$, and in oxygen difluoride, $\mathrm{OF_2}$.
- b) Determine the oxidation number of hydrogen in calcium hydride, $\mathrm{CaH_2}$, and in water, $\mathrm{H_2O}$. Briefly justify each value in terms of the relative electronegativities of the combined elements.
Oxidation, reduction and agents
When a sheet of metallic zinc is placed in an aqueous solution of hydrochloric acid, vigorous bubbling of hydrogen gas occurs according to the following process:…
When a sheet of metallic zinc is placed in an aqueous solution of hydrochloric acid, vigorous bubbling of hydrogen gas occurs according to the following process:
$$\mathrm{Zn(s) + 2\,HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)}$$- a) Calculate the oxidation number of all the atoms in the reactants and products.
- b) Identify the species that is oxidised, the one that is reduced, the oxidising agent and the reducing agent, justifying your answer in terms of electron transfer.
Balancing using oxidation numbers
The Leaving Cert method is to balance redox equations using the changes in oxidation numbers. The idea is simple: the electrons lost by the species oxidised must equal the electrons gained by the species reduced.
The ion-electron method in acidic solution
Add purple manganate(VII) to iron(II) sulfate in sulfuric acid and the purple disappears instantly. The ion-electron method splits the change into two half-equations. Oxidation numbers are your main method for balancing whole equations. You need the ion-electron method for the half-equations at the electrodes in cells and electrolysis, so it is worth mastering.
In acidic aqueous medium, the nitrate ion can be converted into nitrogen monoxide, while the tin(II) ion can change to tin(IV). a) Write and balance the reduction…
In acidic aqueous medium, the nitrate ion can be converted into nitrogen monoxide, while the tin(II) ion can change to tin(IV).
- a) Write and balance the reduction half-equation of the nitrate ion, $\mathrm{NO_3^-(aq)}$, to nitrogen monoxide, $\mathrm{NO(g)}$, in acidic medium, stating the number of electrons transferred.
- b) Write the oxidation half-equation of the cation $\mathrm{Sn^{2+}(aq)}$ to $\mathrm{Sn^{4+}(aq)}$ and check that both half-equations balance simultaneously in mass and electric charge.
The diagram below illustrates the flow of electrons that occurs spontaneously when two redox half-cells interact through a conducting circuit. Observe the arrangement of…
The diagram below illustrates the flow of electrons that occurs spontaneously when two redox half-cells interact through a conducting circuit. Observe the arrangement of the chemical species and the direction of the electron flow:
- a) Write the balanced oxidation and reduction half-equations, deducing the numerical value that must replace the question mark in the number of electrons transferred, and write the overall net ionic equation.
- b) Assign, with justification, the roles of oxidising agent and reducing agent from the changes in oxidation numbers.
Basic solution
Chlorine gas, $\mathrm{Cl_2(g)}$, bubbled into a cold dilute aqueous solution of sodium hydroxide, undergoes a disproportionation process leading to the formation of…
Chlorine gas, $\mathrm{Cl_2(g)}$, bubbled into a cold dilute aqueous solution of sodium hydroxide, undergoes a disproportionation process leading to the formation of chloride anions, $\mathrm{Cl^-}$, and hypochlorite anions, $\mathrm{ClO^-}$.
- a) Write and balance, using the ion-electron method in basic medium, the separate oxidation and reduction half-equations, indicating the auxiliary reactants and products ($\mathrm{H_2O}$ and $\mathrm{OH^-}$) and the electrons exchanged.
- b) Deduce the balanced overall net ionic equation and justify quantitatively the concept of disproportionation from the oxidation states of chlorine.
In basic solution, the permanganate anion, $\mathrm{MnO_4^-}$, is able to oxidise the sulfite anion, $\mathrm{SO_3^{2-}}$, producing solid manganese dioxide,…
In basic solution, the permanganate anion, $\mathrm{MnO_4^-}$, is able to oxidise the sulfite anion, $\mathrm{SO_3^{2-}}$, producing solid manganese dioxide, $\mathrm{MnO_2(s)}$, and the sulfate anion, $\mathrm{SO_4^{2-}}$.
- a) Write and balance, using the ion-electron method in basic medium, the oxidation and reduction half-equations, detailing the balancing of atoms using $\mathrm{H_2O}$ and $\mathrm{OH^-}$, as well as the balancing of charges using electrons.
- b) Obtain the overall net ionic equation and check explicitly that the total charge of the reactants equals that of the products.
From the ionic to the full equation
On the lab shelf there are no bottles labelled "manganate(VII) ions" or "iron(II) ions". There are real salts such as $\mathrm{KMnO_4}$ and $\mathrm{FeSO_4}$. How do we turn an ionic equation into one with the real formulae?
In acidic solution with sulfuric acid, $\mathrm{H_2SO_4}$, potassium permanganate, $\mathrm{KMnO_4}$, quantitatively oxidises oxalic acid (ethanedioic acid),…
In acidic solution with sulfuric acid, $\mathrm{H_2SO_4}$, potassium permanganate, $\mathrm{KMnO_4}$, quantitatively oxidises oxalic acid (ethanedioic acid), $\mathrm{H_2C_2O_4}$, producing carbon dioxide gas, $\mathrm{CO_2(g)}$, and manganese(II) sulfate, $\mathrm{MnSO_4(aq)}$.
- a) Write and balance the corresponding half-equations using the ion-electron method and obtain the complete balanced ionic and molecular equations.
- b) $25.0\,\text{mL}$ of an aqueous solution of oxalic acid acidified with $\mathrm{H_2SO_4}$ is taken. To reach the faint persistent pink end point of the permanganate itself in the redox titration, exactly $18.4\,\text{mL}$ of a standard solution of $\mathrm{KMnO_4}$ of concentration $0.0250\,\text{M}$ is used. Calculate the molar concentration of the oxalic acid analysed.
In strongly alkaline solution with sodium hydroxide, $\mathrm{NaOH}$, solid chromium(III) hydroxide, $\mathrm{Cr(OH)3}$, is oxidised by the hypochlorite ion,…
In strongly alkaline solution with sodium hydroxide, $\mathrm{NaOH}$, solid chromium(III) hydroxide, $\mathrm{Cr(OH)_3}$, is oxidised by the hypochlorite ion, $\mathrm{ClO^-}$, turning into soluble chromate anions, $\mathrm{CrO_4^{2-}}$, while chlorine is reduced to the chloride ion, $\mathrm{Cl^-}$.
- a) Balance the overall net ionic equation using the ion-electron method in basic medium and write the complete molecular reaction considering sodium salts.
- b) The starting point is $5.15\,\text{g}$ of $\mathrm{Cr(OH)_3}$ and an excess of $\mathrm{NaClO}$ solution in the presence of $\mathrm{NaOH}$. Calculate the mass in grams of sodium chromate, $\mathrm{Na_2CrO_4}$, obtained if the overall yield of the process is $82.0\,\%$.
Atomic mass data: $\mathrm{H}=1.0\,\text{u}$; $\mathrm{O}=16.0\,\text{u}$; $\mathrm{Na}=23.0\,\text{u}$; $\mathrm{Cl}=35.5\,\text{u}$; $\mathrm{Cr}=52.0\,\text{u}$.
In acidic solution with sulfuric acid, potassium dichromate, $\mathrm{K_2Cr_2O_7}$, reacts with hydrogen peroxide, $\mathrm{H_2O_2}$. In this reaction, chromium is…
In acidic solution with sulfuric acid, potassium dichromate, $\mathrm{K_2Cr_2O_7}$, reacts with hydrogen peroxide, $\mathrm{H_2O_2}$. In this reaction, chromium is reduced to the chromium(III) cation, $\mathrm{Cr^{3+}}$, while the peroxide acts as a reducing agent, releasing oxygen gas, $\mathrm{O_2(g)}$.
- a) Balance the half-equations using the ion-electron method in acidic medium and obtain the balanced overall ionic equation.
- b) $40.0\,\text{mL}$ of an aqueous solution of $\mathrm{K_2Cr_2O_7}$ $0.125\,\text{M}$ is mixed with an excess of acidified $\mathrm{H_2O_2}$ solution. Calculate the volume of $\mathrm{O_2(g)}$, expressed in litres, that will be collected at a temperature of $295\,\text{K}$ and a pressure of $1.02\times 10^5\,\text{Pa}$.
Data: $R = 8,314\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1} = 8,314\times 10^{-2}\,\text{bar}\times\text{L}\times\text{mol}^{-1}\times\text{K}^{-1}$; take $1\,\text{Pa} = 10^{-5}\,\text{bar}$. Assume an ideal dry gas and no secondary decomposition of the peroxide.