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Chemical transformations · Oxidation and reduction

Oxidation numbers and balancing redox equations

Before you start, make sure you know: Simple substances, ions and inorganic compounds (Chemistry (high school))
Before you start, make sure you know: Balancing chemical equations (Chemistry (high school))

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 with oxidation numbers and half-reactions, the way you will on the AP® Chemistry exam.

Oxidation numbers

Put a clean strip of copper into colorless silver nitrate solution, like the demo in your chemistry lab. 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 permanganate (manganese is $+7$). When writing it on an atom, put the sign first: $+2$, $-1$.

Oxidation numbers in barium peroxide and calcium hydride, showing the exceptions for oxygen (-1) and hydrogen (-1) BaO₂ (peroxide) Ba +2 O −1 O −1 CaH₂ (hydride) Ca +2 H −1 H −1 The O–O bond and the metal reverse the usual rule
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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:

  1. a) Determine the oxidation number of oxygen in barium peroxide, $\mathrm{BaO_2}$, and in oxygen difluoride, $\mathrm{OF_2}$.
  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.
Solve

Oxidation, reduction and agents

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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:

$$\mathrm{Zn(s) + 2\,HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)}$$
  1. a) Calculate the oxidation number of all the atoms in the reactants and products.
  2. b) Identify the species that is oxidized, the one that is reduced, the oxidizing agent and the reducing agent, justifying your answer in terms of electron transfer.
Solve

Balancing using oxidation numbers

One way to balance a redox equation is to use the changes in oxidation numbers. The idea is simple: the electrons lost by the species oxidized must equal the electrons gained by the species reduced.

The half-reaction method in acidic solution

Add purple permanganate 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 help you balance whole equations. The half-reaction method is also what you need for the half-reactions at the electrodes in galvanic cells and electrolysis, so it is worth mastering.

Checking mass and charge for the half-equations of nitrate to nitrogen monoxide and of tin(II) to tin(IV) NO₃⁻ + 4 H⁺ + 3 e⁻ → NO + 2 H₂O mass: 1 N, 3 O, 4 H each side. Charge: (−1+4−3) = 0 = 0 ✓ Sn²⁺ → Sn⁴⁺ + 2 e⁻ mass: 1 Sn each side. Charge: +2 = (+4−2) ✓
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In acidic aqueous solution, 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 solution, the nitrate ion can be converted into nitrogen monoxide, while the tin(II) ion can change to tin(IV).

  1. a) Write and balance the reduction half-reaction of the nitrate ion, $\mathrm{NO_3^-(aq)}$, to nitrogen monoxide, $\mathrm{NO(g)}$, in acidic solution, stating the number of electrons transferred.
  2. b) Write the oxidation half-reaction of the cation $\mathrm{Sn^{2+}(aq)}$ to $\mathrm{Sn^{4+}(aq)}$ and check that both half-reactions balance simultaneously in mass and electric charge.
Solve
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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:

Left half-cell Zn (s) Zn²⁺ (aq) Right half-cell Cu (s) Cu²⁺ (aq) ? e−
  1. 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.
  2. b) Assign, with justification, the roles of oxidizing agent and reducing agent from the changes in oxidation numbers.
Solve

Basic solution

Turning an acidic balance into a basic one by adding hydroxide ions to both sides ... + n H⁺ + n e⁻ → ... add n OH⁻ to both sides n H⁺ + n OH⁻ → n H₂O (cancel with any water already present)
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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, $\mathrm{ClO^-}$.

  1. 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.
  2. b) Deduce the balanced overall net ionic equation and justify quantitatively the concept of disproportionation from the oxidation states of chlorine.
Solve
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In basic solution, the permanganate anion, $\mathrm{MnO_4^-}$, is able to oxidize the sulfite anion, $\mathrm{SO_3^{2-}}$, producing solid manganese dioxide,…

In basic solution, the permanganate anion, $\mathrm{MnO_4^-}$, is able to oxidize the sulfite anion, $\mathrm{SO_3^{2-}}$, producing solid manganese dioxide, $\mathrm{MnO_2(s)}$, and the sulfate anion, $\mathrm{SO_4^{2-}}$.

  1. 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.
  2. b) Obtain the overall net ionic equation and check explicitly that the total charge of the reactants equals that of the products.
Solve

From the ionic to the full equation

On the lab shelf there are no bottles labeled "permanganate 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?

Flow from the net ionic equation to the full equation by adding spectator ions net ionic equation + spectator ions full equation
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In acidic solution with sulfuric acid, $\mathrm{H_2SO_4}$, potassium permanganate, $\mathrm{KMnO_4}$, quantitatively oxidizes oxalic acid (ethanedioic acid),…

In acidic solution with sulfuric acid, $\mathrm{H_2SO_4}$, potassium permanganate, $\mathrm{KMnO_4}$, quantitatively oxidizes oxalic acid (ethanedioic acid), $\mathrm{H_2C_2O_4}$, producing carbon dioxide gas, $\mathrm{CO_2(g)}$, and manganese(II) sulfate, $\mathrm{MnSO_4(aq)}$.

  1. a) Write and balance the corresponding half-equations using the ion-electron method and obtain the complete balanced ionic and molecular equations.
  2. 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 analyzed.
Solve
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In strongly alkaline solution with sodium hydroxide, $\mathrm{NaOH}$, solid chromium(III) hydroxide, $\mathrm{Cr(OH)3}$, is oxidized by the hypochlorite ion,…

In strongly alkaline solution with sodium hydroxide, $\mathrm{NaOH}$, solid chromium(III) hydroxide, $\mathrm{Cr(OH)_3}$, is oxidized by the hypochlorite ion, $\mathrm{ClO^-}$, turning into soluble chromate anions, $\mathrm{CrO_4^{2-}}$, while chlorine is reduced to the chloride ion, $\mathrm{Cl^-}$.

  1. a) Balance the overall net ionic equation using the half-reaction method in basic solution and write the complete molecular reaction considering sodium salts.
  2. 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: $\mathrm{H}=1.0\,\text{u}$; $\mathrm{O}=16.0\,\text{u}$; $\mathrm{Na}=23.0\,\text{u}$; $\mathrm{Cl}=35.5\,\text{u}$; $\mathrm{Cr}=52.0\,\text{u}$.

Solve
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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, $\mathrm{O_2(g)}$.

  1. a) Balance the half-equations using the ion-electron method in acidic medium and obtain the balanced overall ionic equation.
  2. 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 liters, 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.

Solve

Summary

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