Oxidation-Reduction Reactions

Oxidation States

Oxidation state is the hypothetical charge of an atom if all of its bonds to other atoms were completely ionic.

Learning Objectives

Predict the oxidation states of common elements by their group number.

Key Takeaways

Key Points

  • The oxidation state of a pure element is always zero.
  • The oxidation state for a pure ion is equivalent to its ionic charge.
  • In general, hydrogen has an oxidation state of +1, while oxygen has an oxidation state of -2.
  • The sum of the oxidation states for all atoms of a neutral molecule must add up to zero.


Key Terms

  • reduction: the gain of electrons, which causes a decrease in oxidation state
  • oxidation: the loss of electrons, which causes an increase in oxidation state


Oxidation state indicates the degree of oxidation for an atom in a chemical compound; it is the hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic. Oxidation states are typically represented by integers, which can be positive, negative, or zero. In some cases, the average oxidation state of an element is a fraction, such as 8/3 for iron in magnetite (Fe3O4).

The highest known oxidation state is +8 in the tetroxides of ruthenium, xenon, osmium, iridium, hassium, and some complexes involving plutonium; the lowest known oxidation state is −4 for some elements in the carbon group.

image

Oxidation states of plutonium: Here, plutonium varies in color with oxidation state.

An atom's increase in oxidation state through a chemical reaction is called oxidation, and it involves a loss of electrons; an decrease in an atom's oxidation state is called reduction, and it involves the gain of electrons.

General Rules Regarding Oxidation States

  1. The oxidation state of a free element (uncombined element) is zero.
  2. For a simple (monoatomic) ion, the oxidation state is equal to the net charge on the ion. For example, Cl- has an oxidation state of -1.
  3. When present in most compounds, hydrogen has an oxidation state of +1 and oxygen an oxidation state of −2. The exceptions to this are that hydrogen has an oxidation state of −1 in hydrides of active metals (such as LiH), and an oxidation state of −1 in peroxides (such as H2O2) or -1/2 in superoxides (such as KO).
  4. The algebraic sum of oxidation states for all atoms in a neutral molecule must be zero. In ions, the algebraic sum of the oxidation states of the constituent atoms must be equal to the charge on the ion.


Predicting Oxidation States

Generally, the oxidation state for most common elements can be determined from their group number on the periodic table. This is summarized in the following chart:

image

Typical oxidation states of the most common elements by group: Transition metals are not included, as they tend to exhibit a variety of oxidation states.

The above table can be used to conclude that boron (a Group III element) will typically have an oxidation state of +3, and nitrogen (a group V element) an oxidation state of -3. Keep in mind that oxidation states can change, and this prediction method should only be used as a general guideline; for example, transition metals do not adhere to any fixed rules and tend to exhibit a wide range of oxidation states.

As stated in rule number four above, the sum of the oxidation states for all atoms in a molecule or polyatomic ion is equal to the charge of the molecule or ion. This helps determine the oxidation state of any one element in a given molecule or ion, assuming that we know the common oxidation states of all of the other elements. For example, in a sulfite ion (SO32-), the total charge of the ion is 2-, and each oxygen is assumed to be in its usual oxidation state of -2. Because there are three oxygen atoms in sulfite, oxygen contributes
3×−2=−63\times-2=-6
to the total charge. Therefore, sulfur must have an oxidation state of +4 for the overall charge on sulfite to be 2-: 
(+4−6=−2).(+4-6=-2).


Do not confuse the formal charge on an atom with its formal oxidation state, as these may be different (and often are different, in polyatomic ions). For example, the charge on the nitrogen atom in ammonium ion NH4+ is 1+, but the formal oxidation state is -3—the same as it is for nitrogen in ammonia. In the case between ammonium and ammonia, the formal charge on the N atom changes, but its oxidation state does not.

Types of Redox Reactions

The five main types of redox reactions are combination, decomposition, displacement, combustion, and disproportionation.

Learning Objectives

Explain the processes involved in a redox reaction and describe what happens to their various components.

Key Takeaways

Key Points

  • In combination reactions, two elements are combined.
    A+B→ABA + B \rightarrow AB
    .
  • In decomposition reactions, a compound is broken down into its constituent parts.
    AB→A+BAB \rightarrow A + B
    .
  • In displacement reactions, one or more atoms is swapped out for another.
    AB+C→A+CBAB + C \rightarrow A + CB
    .
  • In combustion reactions, a compound reacts with oxygen to produce carbon dioxide, water, and heat.
  • In disproportionation reactions, a molecule is both reduced and oxidized; these types of reactions are rare.


Key Terms

  • redox: a shorthand term for "reduction-oxidation," two methods of electron transfer that always occur together
  • combustion: a process in which a fuel combines with oxygen, usually at high temperature, releasing CO2, H2O, and heat


Redox reactions are all around us. In fact, much of our technology, from fire to laptop batteries, is largely based on redox reactions. Redox ( reduction - oxidation ) reactions are those in which the oxidation states of the reactants change. This occurs because in such reactions, electrons are always transferred between species. Redox reactions take place through either a simple process, such as the burning of carbon in oxygen to yield carbon dioxide (CO2), or a more complex process such as the oxidation of glucose (C6H12O6) in the human body through a series of electron transfer processes.

The term "redox" comes from two concepts involved with electron transfer: reduction and oxidation. These processes are defined as follows:

  • Oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom, or ion.
  • Reduction is the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.


A simple mnemonic for remembering these processes is "OIL RIG"—Oxidation Is Losing (electrons), Reduction Is Gaining (electrons).

Redox reactions are matched sets: if one species is oxidized in a reaction, another must be reduced. Keep this in mind as we look at the five main types of redox reactions: combination, decomposition, displacement, combustion, and disproportion.

Combination

Combination reactions "combine" elements to form a chemical compound. As usual, oxidation and reduction occur together.

General equation:  
A+B→ABA + B \rightarrow AB


Sample 1. equation: 2 H2 + O2 → 2 H2O

The sum of oxidation states in the reactants is equal to that in the products: 0 + 0 → (2)(+1) + (-2)

In this equation, both H2 and O2 are the molecular forms of their respective elements and therefore their oxidation states are 0. The product is H2O: the oxidation state is -2 for oxygen and +1 for hydrogen.

Decomposition

Decomposition reactions are the reverse of combination reactions, meaning they are the breakdown of a chemical compound into its component elements.

General equation: AB → A + B

Sample 2. equation: 2 H2O → 2 H2 + O2

Calculation: (2)(+1) + (-2) = 0 → 0 + 0

In this equation, the water is "decomposed" into hydrogen and oxygen, both of which are neutral. Similar to the previous example, H2O has a total oxidation state of 0, with each H taking on a +1 state and the O a -2; thus, decomposition oxidizes oxygen from -2 to 0 and reduces hydrogen from +1 to 0.

Displacement

Displacement reactions, also known as replacement reactions, involve compounds and the "replacing" of elements. They occur as single and double replacement reactions.

General equation (single displacement): A + BC → AB + CA

A single replacement reaction "replaces" an element in the reactants with another element in the products.

Sample 3. equation: Cl2 + 2 NaBr → 2 NaCl + Br2

Calculation: 0 + [(+1) + (-1) = 0]
→\rightarrow
 [(+1) + (-1) = 0] + 0

In this equation, Cl is reduced and replaces Br, while Br is oxidized.

General equation (double displacement): AB + CD → AD + CB

A double replacement reaction is similar to a single replacement reaction, but involves "replacing" two elements in the reactants with two in the products.

Sample 4. equation: Fe2O3 + 6 HCl → 2 FeCl3 + 3 H2O

In this equation, Fe and H as well as O and Cl trade places.

Combustion

Combustion reactions always involve oxygen and an organic fuel. In the following image, we see methane combusting to release energy.

image

Combustion reaction of methane: This is an example of a combustion reaction, a redox process. Methane (

CH4\text{CH}_4
), reacts with oxygen (
O2\text{O}_2
)  to form carbon dioxide (
CO2\text{CO}_2
) and two water molecules (
2H2O2\text{H}_2\text{O}
).

The general equation of a combustion reaction is:

CxHy+(x+y4)O2→xCO2+y2H2O\text{C}_x\text{H}_y +\left( x + \dfrac{y}{4} \right) \text{O}_2 \rightarrow x \text{CO}_2 + \dfrac{y}{2}\text{H}_2\text{O}


Disproportionation

In some redox reactions, substances can be both oxidized and reduced. These are known as disproportionation reactions. One real-life example of such a process is the reaction of hydrogen peroxide, H2O2, when it is poured over a wound. At first, this might look like a simple decomposition reaction, because hydrogen peroxide breaks down to produce oxygen and water:

2 H2O2(aq) → 2 H2O(l) + O2(g)

The key to this reaction lies in the oxidation states of oxygen, however. Notice that oxygen is present in the reactant and both products. In H2O2, oxygen has an oxidation state of -1. In H2O, its oxidation state is -2, and it has been reduced. In O2 however, its oxidation state is 0, and it has been oxidized. Oxygen has been both oxidized and reduced in the reaction, making this a disproportionation reaction. The general form for this reaction is as follows:

2A → A' + A"

Balancing Redox Equations

Balancing redox reactions involves splitting the reaction into two half-reactions.

Learning Objectives

Formulate a balanced redox reaction from two half-reactions.

Key Takeaways

Key Points

  • Reducing agents get oxidized, and therefore lose electrons.
  • Oxidizing agents get reduced, and therefore gain electrons.
  • Remember the mnemonic device OIL RIG—" Oxidation Involves Loss" and " Reduction Involves Gain" to distinguish between oxidizing and reducing agents.


Key Terms

  • oxidation: a reaction in which an element's atoms lose electrons and its oxidation state increases
  • reduction: a reaction in which electrons are gained and oxidation state is reduced, often by the removal of oxygen or the addition of hydrogen
  • half-reactions: one of the two constituent parts of any redox reaction in which only oxidation or reduction is shown


Every balanced redox reaction is composed of two half-reactions: the oxidation half-reaction, and the reduction half-reaction. For example, look at the following redox reaction between iron and copper:

Fe+Cu+2→Fe+2+Cu\text{Fe}+\text{Cu}^{+2}\rightarrow\text{Fe}^{+2}+\text{Cu}


In this reaction, iron is oxidized, and copper is reduced (or, iron is the reducing agent, and copper is the oxidizing agent.) We can split this reaction into two half-reactions. The oxidation half-reaction looks as follows:

Fe→Fe+2+2e−\text{Fe}\rightarrow\text{Fe}^{+2}+2e^-


This shows the oxidation of iron and the loss of two electrons. Notice that this equation is balanced in both mass and charge: we have one atom of iron on each side of the equation (mass is balanced), and the net charge on each side of the equation is equal to zero (charge is balanced).

Next, look at the reduction half-reaction:

Cu+2+2e−→Cu\text{Cu}^{+2}+2e^-\rightarrow\text{Cu}


This half-reaction explicitly shows the copper (II) ion gaining two electrons. Note again that the equation is balanced in mass and charge. Now that we have our two balanced half-reactions, we can combine them to get the full redox reaction:

Fe→Fe2++2e−Cu2++2e−→CuFe+Cu2++2e−‾→Cu+Fe2++2e−‾\begin{array}{rcl}\text{Fe}&\rightarrow&\text{Fe}^{2+}+2e^{-}\\\text{Cu}^{2+}+2e^{-}&\rightarrow &\text{Cu}\\\overline{\text{Fe}+\text{Cu}^{2+}+2e^{-}}&\rightarrow&\overline{\text{Cu}+\text{Fe}^{2+}+2e^{-}}\end{array}


Adding the two halves of a redox reaction: These two halves of the reaction can be added like any other chemical equation. Once the equations are added, the electrons on each side cancel out.

Note that the two electrons on each side of the equation cancel out. This is very important, because the final balanced equation for any redox reaction should never contain any electrons.

Electrons moving from one pole of a battery through a circuit and back through the battery's other pole is an example of applied redox reaction.

Balancing Redox Equations in Acidic Solution: Basic Rules

If a reaction occurs in an acidic environment, you can balance the redox equation as follows:

  1. Write the oxidation and reduction half-reactions, including the whole compound involved in the reaction—not just the element that is being reduced or oxidized.
  2. Balance both reactions for all elements except oxygen and hydrogen.
  3. If the oxygen atoms are not balanced in either reaction, add water molecules to the side missing the oxygen.
  4. If the hydrogen atoms are not balanced, add hydrogen ions (H+) until the hydrogen atoms are balanced.
  5. Multiply the half-reactions by the appropriate numbers so that they both have equal numbers of electrons.
  6. Add the two equations to cancel out the electrons to balance the equation.


Example: Balancing Redox Equations in Acidic Solution

The following is an unbalanced redox equation that takes place in acidic solution:

Cr2O72−(aq)+NO2−(aq)→Cr3+(aq)+NO3−(aq)\text{Cr}_2\text{O}_7^{2-}(aq)+\text{NO}_2^-(aq)\rightarrow\text{Cr}^{3+}(aq)+\text{NO}_3^-(aq)


First, we need to split this reaction into its two half-reactions. Let's start with the oxidation half-reaction:

NO2−(aq)→NO3−(aq)\text{NO}_2^-(aq)\rightarrow\text{NO}_3^-(aq)


We need to balance this equation by mass. The equation is already balanced in nitrogens, but not oxygens. You can balance oxygen by adding the appropriate number of water molecules:

H2O(l)+NO2−(aq)→NO3−(aq)\text{H}_2\text{O}(l)+\text{NO}_2^-(aq)\rightarrow\text{NO}_3^-(aq)


Now the equation is balanced in oxygens, but not hydrogens. To balance hydrogens in acidic solution, we add H+(aq):

H2O(l)+NO2−(aq)→NO3−(aq)+2H+(aq)\text{H}_2\text{O}(l)+\text{NO}_2^-(aq)\rightarrow\text{NO}_3^-(aq)+2\text{H}^+(aq)


The equation is now balanced in mass, but not charge. To balance the charge, we will add two electrons to the right side of the equation:

H2O(l)NO2−(aq)→NO3−(aq)+2H+(aq)+2e−\text{H}_2{O}(l)\text{NO}_2^-(aq)\rightarrow\text{NO}_3^-(aq)+2\text{H}^+(aq)+2e^-


The charge is now balanced; each side of the equation has an equal net charge of -1.

Let's now move on to the reduction half-reaction:

Cr2O72−→Cr3+\text{Cr}_2\text{O}_7^{2-}\rightarrow\text{Cr}^{3+}


Once again, we will balance mass first, balancing oxygens by adding in water molecules:

Cr2O72−→2Cr3++7H2O(l)\text{Cr}_2\text{O}_7^{2-}\rightarrow{2}\text{Cr}^{3+}+7\text{H}_2\text{O}(l)


Again, since we are in acidic solution, we balance hydrogens by adding H+(aq):

14H+(aq)+Cr2O72−→2Cr3++7H2O(l)14\text{H}^+(aq)+\text{Cr}_2\text{O}_7^{2-}\rightarrow{2}\text{Cr}^{3+}+7\text{H}_2\text{O}(l)


The equation is now balanced for mass, and we need only balance for charge. As it stands now, the left side of the equation has a net charge of +14 - 2 = +12. The right side of the equation has a net charge of +6 (2 x +3 = +6). We therefore need to add 6 electrons to the right side of the equation to balance the charges:

6e−+14H+(aq)+Cr2O72−→2Cr3++7H2O(l)6e^{-}+14\text{H}^+(aq)+\text{Cr}_2\text{O}_7^{2-}\rightarrow{2}\text{Cr}^{3+}+7\text{H}_2\text{O}(l)


Now we have balanced both of our half-reactions. To sum up so far:

Balanced oxidation half-reaction: 
H2O(l)+NO2−(aq)→NO3−(aq)+2H+(aq)+2e−\text{H}_2\text{O}(l)+\text{NO}_2^-(aq)\rightarrow\text{NO}_3^-(aq) + 2\text{H}^+(aq) + 2e^-


Balanced reduction half-reaction: 
6e−+14H+(aq)+Cr2O72−→2Cr3++7H2O(l)6e^{-}+14\text{H}^+(aq)+\text{Cr}_2\text{O}_7^{2-}\rightarrow{2}\text{Cr}^{3+}+7\text{H}_2\text{O}(l)


Lastly, in order to get our full balanced redox equation, we need to add our half-reactions so that all the electrons cancel out. For this reaction, we can multiply the first half-reaction by 3:

3×(H2O(l)+NO2−(aq)→NO3−(aq)+2H+(aq)+2e−)=3H2O(l)+3NO2−(aq)→3NO3−(aq)+6H+(aq)+6e−\begin{array}{l}3\times(\text{H}_2\text{O}(l)+\text{NO}_2^-(aq)\rightarrow\text{NO}_3^-(aq)+2\text{H}^+(aq)+2e^-)=\\3\text{H}_2\text{O}(l)+3\text{NO}_2^-(aq)\rightarrow{3}\text{NO}_3^-(aq)+6\text{H}^+(aq) + 6e^-\end{array}


Now we can add our half-reactions together:

3H2O(l)+3NO2−(aq)→3NO3−(aq)+6H+(aq)+6e−6e−+14H+(aq)+Cr2O72−→2Cr3++7H2O(l)6e−+3H2O(l)+3NO2−(aq)+14H+(aq)+Cr2O72−‾→2Cr3++7H2O(l)+3NO3−(aq)+6H+(aq)+6e−‾\begin{array}{rcl}3\text{H}_2\text{O}(l)+3\text{NO}_2^-(aq)&\rightarrow&3\text{NO}_3^-(aq)+6\text{H}^+(aq)+6e^-\\6e^{-}+14\text{H}^+(aq)+\text{Cr}_2\text{O}_7^{2-}&\rightarrow&{2}\text{Cr}^{3+}+7\text{H}_2\text{O}(l)\\\overline{6e^-+3\text{H}_2\text{O}(l)+3\text{NO}_2^-(aq)+14\text{H}^+(aq)+\text{Cr}_2\text{O}_7^{2-}}&\rightarrow&\overline{2\text{Cr}^{3+}+7\text{H}_2\text{O}(l)+3\text{NO}_3^-(aq)+6\text{H}^+(aq)+6e^-}\end{array}


Canceling out all common terms, including the electrons, we get our final balanced equation:

3NO2−(aq)+8H+(aq)+Cr2O72−→2Cr3++4H2O(l)+3NO3−(aq)3\text{NO}_2^-(aq)+8\text{H}^+(aq)+\text{Cr}_2\text{O}_7^{2-}\rightarrow{2}\text{Cr}^{3+} + 4 H_2O(l) + 3NO_3^-(aq)


Notice that the final equation is balanced in mass as well as charge (each side of the equation has a net charge of +3).

Although this example seems intimidating, balancing redox reactions in acidic solution becomes much easier with careful practice.

Example: Balancing Redox Reactions in Basic Solution

If a redox reaction occurs in basic solution, we proceed as we did above, with one minor difference: once we have added H+(aq) to balance hydrogens, we simply add the same number of hydroxides to both sides of the equation. For instance:

H2O(l)+NO2−(aq)→NO3−(aq)+2H+(aq)\text{H}_2\text{O}(l)+\text{NO}_2^-(aq)\rightarrow\text{NO}_3^-(aq)+2\text{H}^+(aq)


The half-reaction above is balanced for mass in acidic solution. If we are in basic solution, however, we would need to add 2 hydroxides to both sides of the equation:

2OH−(aq)+H2O(l)+NO2−(aq)→NO3−(aq)+2H+(aq)+2OH−(aq)2\text{OH}^-(aq)+\text{H}_2\text{O}(l)+\text{NO}_2^-(aq)\rightarrow\text{NO}_3^-(aq)+2\text{H}^+(aq)+2\text{OH}^-(aq)


Notice that we have both H+(aq) and OH-(aq) on the right side of the equation. These species will neutralize each other to form water, so we can rewrite this as follows:

2OH−(aq)+H2O(l)+NO2−(aq)→NO3−(aq)+2H2O(l)2\text{OH}^-(aq)+\text{H}_2\text{O}(l)+\text{NO}_2^-(aq) \rightarrow\text{NO}_3^-(aq) + 2\text{H}_2\text{O}(l)


Lastly, because we have water molecules on both sides of the equation, we cancel out like terms to give us:

2OH−(aq)+NO2−(aq)→NO3−(aq)+H2O(l)2\text{OH}^-(aq)+\text{NO}_2^-(aq)\rightarrow\text{NO}_3^-(aq)+\text{H}_2\text{O}(l)


This half-reaction is now balanced for mass in basic solution. From here, we proceed just as we did above in acidic solution: balance the charge by adding the appropriate number of electrons.

Balancing redox equations can certainly be complicated and time-consuming, so it is wise to practice them extensively.



Balance a Redox Reaction (ACIDIC solution): A great walkthrough on how to balance a redox reaction in acidic solution.



Balance a Redox Reaction (BASIC solution): A great walkthrough on how to balance a redox reaction in basic solution.

Redox Titrations

Redox titration determines the concentration of an analyte containing either an oxidizing or a reducing agent.

Learning Objectives

Calculate the concentration of an unknown analyte by performing a redox titration.

Key Takeaways

Key Points

  • The titrant is the standardized solution; the analyte is the analyzed substance.
  • Redox titration determines the concentration of an unknown solution (analyte) that contains an oxidizing or reducing agent.
  • Not all titrations require an external indicator. Some titrants can serve as their own indicators, such as when potassium permanganate is titrated against a colorless analyte.


Key Terms

  • titration: a method in which known amounts of the titrant are added to the analyte until the reaction reaches the endpoint
  • analyte: any substance undergoing analysis
  • titrant: the standardized solution used in titrations; the solution of known concentration


Determining the Concentration of an Analyte

As with acid-base titrations, a redox titration (also called an oxidation- reduction titration) can accurately determine the concentration of an unknown analyte by measuring it against a standardized titrant. A common example is the redox titration of a standardized solution of potassium permanganate (KMnO4) against an analyte containing an unknown concentration of iron (II) ions (Fe2+). The balanced reaction in acidic solution is as follows:

MnO4−(aq)+5Fe2+(aq)+8H+(aq)→5Fe3+(aq)+Mn2+(aq)+4H2O(l)\text{MnO}_4^-(aq)+5\text{Fe}^{2+}(aq)+8\text{H}^+(aq)\rightarrow{5}\text{Fe}^{3+}(aq)+\text{Mn}^{2+}(aq)+4\text{H}_2\text{O}(l)


In this case, the use of KMnO4 as a titrant is particularly useful, because it can act as its own indicator; this is due to the fact that the KMnO4 solution is bright purple, while the Fe2+ solution is colorless. It is therefore possible to see when the titration has reached its endpoint, because the solution will remain slightly purple from the unreacted KMnO4.



Permanganate Titration Endpoint: A redox titration using potassium permanganate as the titrant. Because of its bright purple color, KMnO4 serves as its own indicator. Note how the endpoint is reached when the solution remains just slightly purple.

Example: Sample Calculation from Experimental Data

A standardized 4 M solution of KMnO4 is titrated against a 100 mL sample of an unknown analyte containing Fe2+. A student conducts the redox titration and reaches the endpoint after adding 25 mL of the titrant. What is the concentration of the analyte?

We know from our balanced equation above that permanganate and iron react in a 1:5 mole ratio. We can therefore perform the following calculation:

25 mL KMnO4(1 L1000 mL)(4 mol KMnO41 L)(5 mol Fe2+1 mol KMnO4)=0.5 mol Fe2+25\text{ mL KMnO}_4\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\left(\frac{4\text{ mol KMnO}_4}{1\text{ L}}\right)\left(\frac{5\text{ mol Fe}^{2+}}{1\text{ mol KMnO}_4}\right)=0.5\text{ mol Fe}^{2+}


Now that we know the number of moles of iron present in the sample, we can calculate the concentration of the analyte:

M=molL=0.5 mol0.100 L=5 M\text{M}=\frac{\text{mol}}{\text{L}}=\frac{0.5\text{ mol}}{0.100\text{ L}}=5\text{ M}


Other Types of Redox Titrations

There are various other types of redox titrations that can be very useful. For example, wines can be analyzed for sulfur dioxide using a standardized iodine solution as the titrant. In this case, starch is used as an indicator; a blue starch-iodine complex is formed in the presence of excess iodine, signaling the endpoint.

Another example is the reduction of iodine (I2) to iodide (I−) by thiosulphate (S2O32−), again using starch as the indicator. This is essentially the reverse titration of what was just described; here, when all the iodine has been reduced, the blue color disappears. This is called an iodometric titration.

Most often, the reduction of iodine to iodide is the last step in a series of reactions in which the initial reactions are used to convert an unknown amount of the analyte to an equivalent amount of iodine, which can then be titrated. Sometimes halogens (or organic compounds containing halogens) other than iodine are used in the intermediate reactions because they are available in better-measurable standard solutions or they react more readily with the analyte. While these extra steps make an iodometric titration much more involved, they are often worthwhile, because the equivalence point involving the bright blue iodine-starch complex is more precise than various other analytical methods.

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