Why Oxidation Number Is Not Always the Same as Ionic Charge

An acidified permanganate ion, \(\mathrm{MnO_{4}}^{-}\), has an overall charge of \(-1\). Yet the oxidation number of its manganese atom is \(+7\). When permanganate reacts to form \(\mathrm{Mn}^{2+}\), manganese gains five electrons.

How can a negatively charged ion contain an atom with a positive oxidation number of \(+7\)?

The answer is that ionic charge and oxidation number describe different things. Ionic charge belongs to the entire ion. Oxidation number is a formal value assigned to an individual atom to keep track of electrons in chemical reactions. They are equal for monatomic ions, but they are not necessarily equal for atoms inside polyatomic ions or molecules.

This distinction becomes particularly useful when balancing redox reactions and identifying how many electrons have been transferred.

01Two numbers, two different jobs

An ionic charge tells you the overall electrical charge of an ion.

An oxidation number, also called an oxidation state, tells you how electrons are assigned to an individual atom using a set of chemical conventions.

Consider the difference:

• \(\mathrm{Fe}^{3+}\) is a monatomic ion. Its iron atom has an oxidation number of \(+3\), matching the ion’s charge.
• \(\mathrm{MnO_{4}}^{-}\) is a polyatomic ion. Its overall charge is \(-1\), but manganese has an oxidation number of \(+7\).
• \(\ce{H2O}\) is a neutral molecule. Its hydrogen atoms have oxidation numbers of \(+1\), and its oxygen atom has an oxidation number of \(-2\).

The key is to identify exactly which species the charge belongs to.

Think of a polyatomic ion as a team. The ionic charge is the team’s overall score, while the oxidation numbers are the individual bookkeeping entries that add up to that score.

The analogy has a limit: oxidation numbers are formal assignments, not measurements of the actual electrical charge carried by each atom.

The sum rule

The most useful rule is:

The sum of all oxidation numbers in a species must equal the overall charge of that species.

This gives us two cases.

For a neutral molecule:

\[ \sum \text{oxidation numbers}=0 \]

For a polyatomic ion:

\[ \sum \text{oxidation numbers} =\text{ionic charge} \]

This is why manganese can have an oxidation number of \(+7\) inside an ion with an overall charge of \(-1\). The other atoms must also contribute to the total.

02Why oxidation numbers can differ from actual atomic charges

The difference becomes especially important when atoms are joined by covalent bonds.

In an ionic compound, electrons are transferred between atoms, producing positively and negatively charged ions. For a monatomic ion, the oxidation number matches its actual ionic charge.

In a covalent compound, electrons are shared between atoms. However, the electrons are not necessarily shared equally.

To assign oxidation numbers, chemists use a convention: imagine that the bonding electrons are assigned entirely to the more electronegative atom.

Electronegativity describes how strongly an atom attracts bonding electrons.

This hypothetical assignment allows us to track oxidation and reduction consistently, even when electrons are shared rather than completely transferred.

For example, consider hydrogen chloride, \(\ce{HCl}\).

The molecule has an overall charge of zero. However, chlorine is more electronegative than hydrogen, so the bonding electrons are assigned to chlorine when calculating oxidation numbers.

The result is:

• Hydrogen: \(+1\)
• Chlorine: \(-1\)

Their sum is zero.

This does not mean that hydrogen has actually transferred a complete electron to chlorine. The bond is covalent, and the oxidation numbers are formal assignments rather than direct measurements of atomic charge.

Comparing different chemical species

The pattern is clear.

For monatomic ions, the oxidation number equals the ionic charge. For polyatomic ions and neutral molecules, you must calculate the oxidation number of each atom separately.

03The connection to electron transfer

Now we can use oxidation numbers to identify what happens during a redox reaction.

Remember these two rules:

• Oxidation: oxidation number increases, indicating a loss of electrons.
• Reduction: oxidation number decreases, indicating a gain of electrons.

The size of the change tells you how many electrons are transferred per atom.

For example, consider manganese being reduced from permanganate to manganese(II):

\[ \ce{MnO4^- -> Mn^2+} \]

Manganese changes from \(+7\) to \(+2\).

\[ +7 \longrightarrow +2 \]

Its oxidation number decreases by five. Therefore, each manganese atom gains five electrons.

This is why the reduction half-equation contains five electrons:

\[ \ce{MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O} \]

Notice that the electron count comes from the change in manganese’s oxidation number, not from the initial charge of the permanganate ion.

We can check that the equation conserves both atoms and charge.

Atom balance:

• Mn: 1 on each side.
• O: 4 on each side.
• H: 8 on each side.

Charge balance:

Left side:

\[ -1+8-5=+2 \]

Right side:

\[ +2 \]

Both sides have the same total charge.

This is an important distinction: oxidation numbers help us identify electron transfer, while balancing the chemical equation ensures that atoms and electrical charge are conserved.

Connecting manganese reduction to iron oxidation

Suppose the permanganate reacts with iron(II) ions in acidic solution.

Iron(II) becomes iron(III):

\[ \ce{Fe^2+ -> Fe^3+ + e^-} \]

Iron’s oxidation number increases from \(+2\) to \(+3\). Each iron ion loses one electron.

Manganese gains five electrons, so five iron ions must each lose one electron.

Multiplying the iron half-equation by five gives:

\[ \ce{5Fe^2+ -> 5Fe^3+ + 5e^-} \]

Adding the two half-equations produces:

\[ \ce{ MnO4^- + 8H+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+ } \]

The complete reaction is balanced.

The five electrons lost by iron are equal to the five electrons gained by manganese.

This is electron conservation in a redox reaction.

04An experiment that demonstrates the difference

A useful way to see this distinction in practice is through a potassium permanganate titration.

The experiment measures how much iron(II) solution reacts with a known amount of acidified permanganate.

It does not directly measure the oxidation number of an individual atom. Instead, it measures the reacting mole ratio, which can be explained using oxidation numbers and electron conservation.

How the apparatus works

A burette is clamped vertically above a conical flask.

The burette contains a standard solution of acidified potassium permanganate, which is strongly purple.

The conical flask contains a measured volume of iron(II) solution, together with excess dilute sulfuric acid.

The permanganate solution is added gradually from the burette while the flask is swirled to mix the reactants.

As permanganate reacts, manganese is reduced to \(\mathrm{Mn}^{2+}\), which is much less strongly coloured than permanganate.

Eventually, a tiny excess of permanganate produces a faint pink colour that remains after swirling. This indicates the endpoint.

The volume delivered from the burette is recorded.

Sulfuric acid is used to provide the acidic conditions required for the manganese product to be \(\mathrm{Mn}^{2+}\). Hydrochloric acid is avoided because permanganate can oxidise chloride ions, introducing another reaction.

What should the experiment show?

From the oxidation-number changes, we predict:

• Each manganese atom changes from \(+7\) to \(+2\), gaining five electrons.
• Each iron atom changes from \(+2\) to \(+3\), losing one electron.

Therefore, one permanganate ion should react with five iron(II) ions.

The mole ratio is:

\[ \ce{MnO4^- : Fe^2+}=1:5 \]

The titration allows us to test this prediction using measured volumes and known concentrations.

Why the reaction conditions matter

There is another important detail.

The number of electrons accepted by permanganate depends on the manganese-containing product formed.

Under acidic conditions, manganese is reduced from \(+7\) to \(+2\):

\[ +7\longrightarrow+2 \]

It gains five electrons per manganese atom.

Under neutral aqueous conditions, permanganate can instead be reduced to manganese dioxide, \(\ce{MnO2}\).

The oxidation number of manganese in \(\ce{MnO2}\) is \(+4\), because oxygen contributes \(-2\) per atom:

\[ x+2(-2)=0 \]

\[ x=+4 \]

Manganese has therefore changed from \(+7\) to \(+4\), gaining only three electrons per atom.

The starting permanganate ion still has an overall charge of \(-1\). What changes is the reduction product and, consequently, the oxidation-number change.

This is why the reaction medium must be considered when working with redox reactions.

05A reliable method for distinguishing the two

When you encounter an unfamiliar chemical species, follow this sequence.

1. Identify the complete species.

Is it a neutral molecule, a monatomic ion, or a polyatomic ion?

Do not automatically assign the superscript charge to every atom.

2. Assign the oxidation numbers you already know.

Useful rules include:

• Elements in their natural elemental form have an oxidation number of zero.
• Monatomic ions have oxidation numbers equal to their ionic charges.
• Oxygen is usually \(-2\).
• Hydrogen is usually \(+1\) when bonded to non-metals.
• Group 1 metals are \(+1\), and Group 2 metals are \(+2\).

There are exceptions, including peroxides, in which oxygen has an oxidation number of \(-1\).

3. Apply the sum rule.

Add all oxidation numbers and set the result equal to the overall charge of the species.

For a neutral molecule, the sum is zero.

For a polyatomic ion, the sum equals its ionic charge.

4. Compare oxidation numbers before and after the reaction.

An increase indicates oxidation.

A decrease indicates reduction.

5. Calculate the total electron transfer.

Multiply the change in oxidation number per atom by the number of atoms undergoing that change.

The electrons lost must equal the electrons gained.

6. Check conservation.

Make sure the final reaction conserves every type of atom and has the same total electrical charge on both sides.

For a more detailed treatment of assigning oxidation numbers across redox reactions, see Oxidation numbers in redox reactions: HSC Chemistry Guide.

06Questions and solutions

Question 1

Calculate the oxidation number of nitrogen in the nitrate ion, \(\mathrm{NO_{3}}^{-}\).

Explain why its oxidation number is not equal to the overall ionic charge.

Solution 1

The oxidation number of nitrogen is \(+5\), not \(-1\). The \(-1\) charge belongs to the entire nitrate ion, not just its nitrogen atom.

Step 1: Assign oxygen its usual oxidation number of \(-2\).

Three oxygen atoms contribute:

\[ 3(-2)=-6 \]

Step 2: Let the nitrogen oxidation number be \(x\).

The total must equal the ionic charge:

\[ x+3(-2)=-1 \]

Step 3: Solve.

\[ \begin{aligned} x-6&=-1\\ x&=+5 \end{aligned} \]

Therefore, nitrogen has an oxidation number of \(+5\).

The sum of the oxidation numbers is:

\[ +5-6=-1 \]

The result satisfies the overall charge of the nitrate ion.

Question 2

An acidic solution contains dichromate ions, \(\mathrm{Cr_{2}O_{7}}^{2-}\), and iron(II) ions, \(\mathrm{Fe}^{2+}\).

During the reaction:

• Dichromate is reduced to \(\mathrm{Cr}^{3+}\).
• Iron(II) is oxidised to \(\mathrm{Fe}^{3+}\).
• Water is also produced.

(a) Calculate the oxidation number of chromium in \(\mathrm{Cr_{2}O_{7}}^{2-}\).

(b) Determine the number of electrons gained by each dichromate ion.

(c) Write the balanced net ionic equation.

(d) Demonstrate that atoms and electrical charge are conserved.

Solution 2

Chromium has an oxidation number of \(+6\), and each dichromate ion gains six electrons. The balanced reaction requires six iron(II) ions for every dichromate ion.

Step 1: Calculate the oxidation number of chromium.

Oxygen has an oxidation number of \(-2\).

Let the oxidation number of chromium be \(x\).

There are two chromium atoms and seven oxygen atoms:

\[ 2x+7(-2)=-2 \]

Rearranging:

\[ \begin{aligned} 2x-14&=-2\\ 2x&=12\\ x&=+6 \end{aligned} \]

Each chromium atom has an oxidation number of \(+6\).

Step 2: Determine the electron transfer.

Chromium changes from \(+6\) to \(+3\).

Each chromium atom gains three electrons.

However, there are two chromium atoms in each dichromate ion.

Therefore:

\[ 2\times3=6\text{ electrons gained} \]

Each iron(II) ion changes from \(+2\) to \(+3\), losing one electron.

Six iron(II) ions must therefore be oxidised.

Step 3: Balance the reaction.

The balanced net ionic equation is:

\[ \boxed{ \ce{ Cr2O7^2- + 14H+ + 6Fe^2+ -> 2Cr^3+ + 7H2O + 6Fe^3+ } } \]

The acidic conditions are important because the reaction requires hydrogen ions to balance the oxygen and hydrogen atoms.

Step 4: Check atom conservation.

All atoms are conserved.

Step 5: Check charge conservation.

Total charge on the reactant side:

\[ -2+14+12=+24 \]

Total charge on the product side:

\[ 2(+3)+6(+3)=+24 \]

Both sides have a total charge of \(+24\).

The reaction is balanced.

The important distinction is that dichromate has an overall charge of \(-2\), but its chromium atoms have oxidation numbers of \(+6\). The seven oxygen atoms account for a combined oxidation number of \(-14\).

Question 3

Hydrogen peroxide decomposes according to the following balanced reaction:

\[ \ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \]

The decomposition can be accelerated by a manganese dioxide catalyst, which is not consumed.

(a) Determine the oxidation number of oxygen in hydrogen peroxide.

(b) Determine the oxidation numbers of oxygen in both products.

(c) Explain how hydrogen peroxide undergoes both oxidation and reduction in this reaction.

(d) Show that electron transfer is balanced.

Solution 3

Oxygen in hydrogen peroxide has an oxidation number of \(-1\). Some oxygen atoms are oxidised to zero, while others are reduced to \(-2\). Hydrogen peroxide therefore undergoes disproportionation.

Step 1: Calculate the oxidation number of oxygen in hydrogen peroxide.

Hydrogen has an oxidation number of \(+1\).

Let the oxidation number of oxygen be \(x\).

Because hydrogen peroxide is neutral:

\[ 2(+1)+2x=0 \]

\[ \begin{aligned} 2+2x&=0\\ x&=-1 \end{aligned} \]

Each oxygen atom in hydrogen peroxide has an oxidation number of \(-1\).

Step 2: Calculate the oxidation numbers in the products.

In water, oxygen has an oxidation number of \(-2\).

In elemental oxygen, \(\ce{O2}\), the oxidation number is zero because oxygen is in its elemental form.

Therefore:

Step 3: Identify oxidation and reduction.

Some oxygen atoms change from \(-1\) to \(0\).

Their oxidation numbers increase, so they are oxidised and lose electrons.

Other oxygen atoms change from \(-1\) to \(-2\).

Their oxidation numbers decrease, so they are reduced and gain electrons.

The same starting substance supplies both the oxidised and reduced products.

Step 4: Demonstrate electron conservation.

In the balanced equation, two oxygen atoms are oxidised from \(-1\) to \(0\).

They lose:

\[ 2\times1=2\text{ electrons} \]

The other two oxygen atoms are reduced from \(-1\) to \(-2\).

They gain:

\[ 2\times1=2\text{ electrons} \]

Therefore:

\[ \text{Electrons lost}=\text{Electrons gained}=2 \]

The reaction conserves electrons.

It also conserves four oxygen atoms and four hydrogen atoms. Both sides are electrically neutral.

The catalyst changes the reaction rate, not the oxidation-number changes or overall reaction stoichiometry.

This reaction is a useful reminder that oxidation and reduction can occur within the same substance. For more on this pattern, see Disproportionation: When One Substance Is Oxidised and Reduced.

Question 4

Ammonium nitrate, \(\ce{NH4NO3}\), is an ionic compound.

When it dissolves in water, it separates into ammonium ions, \(\ce{NH4+}\), and nitrate ions, \(\mathrm{NO_{3}}^{-}\).

A student calculates the average oxidation number of nitrogen in the neutral formula unit:

\[ \frac{-3+5}{2}=+1 \]

They conclude that both nitrogen atoms have an oxidation number of \(+1\).

Explain why this conclusion is incorrect. Calculate the oxidation number of nitrogen in each ion and demonstrate how the overall compound remains electrically neutral.

Solution 4

The nitrogen atoms have different oxidation numbers: \(-3\) in ammonium and \(+5\) in nitrate. The average of \(+1\) is mathematically correct, but it is not the oxidation number of either nitrogen atom.

The mistake is treating the entire neutral formula unit as though both nitrogen atoms belonged to the same chemical species.

Step 1: Calculate nitrogen’s oxidation number in ammonium.

The ammonium ion is \(\ce{NH4+}\).

Hydrogen has an oxidation number of \(+1\).

Let nitrogen’s oxidation number be \(x\).

\[ x+4(+1)=+1 \]

\[ \begin{aligned} x+4&=1\\ x&=-3 \end{aligned} \]

Nitrogen in ammonium has an oxidation number of \(-3\).

Step 2: Calculate nitrogen’s oxidation number in nitrate.

The nitrate ion is \(\mathrm{NO_{3}}^{-}\).

Oxygen has an oxidation number of \(-2\).

Let nitrogen’s oxidation number be \(y\).

\[ y+3(-2)=-1 \]

\[ \begin{aligned} y-6&=-1\\ y&=+5 \end{aligned} \]

Nitrogen in nitrate has an oxidation number of \(+5\).

Step 3: Explain why the formula unit is neutral.

Ammonium has a charge of \(+1\), while nitrate has a charge of \(-1\).

Therefore:

\[ (+1)+(-1)=0 \]

The complete formula unit is electrically neutral.

We can also check the total oxidation numbers across the entire formula:

\[ -3+5+4(+1)+3(-2)=0 \]

The sum is zero, as required.

The tempting approach was to average the two nitrogen oxidation numbers and assign \(+1\) to both atoms. The problem is that oxidation numbers must be calculated within the actual chemical species.

An average can describe the combined arithmetic, but it cannot replace the distinct oxidation numbers of atoms in different chemical environments.

Question 5

Two separate flasks are prepared with identical starting amounts:

• \(0.00400\ \mathrm{mol}\) of \(\mathrm{MnO_{4}}^{-}\).
• \(0.0150\ \mathrm{mol}\) of \(\mathrm{Fe}^{2+}\).

Both flasks contain sufficient water for the reactions to proceed.

The conditions and observed manganese-containing products are different.

In both flasks, iron(II) is oxidised to iron(III).

Assume that:

• All reacted iron(II) forms \(\mathrm{Fe}^{3+}\).
• All reacted permanganate forms the stated manganese-containing product.
• No other redox reactions occur.
• In flask B, hydroxide ions remain in the solution as reaction products.

(a) Calculate the oxidation number of manganese in the starting permanganate ion and in each product.

(b) Write the balanced net ionic equation for each flask.

(c) For each flask, determine:

• The limiting reactant.
• The amount, in moles, of permanganate consumed.
• The amount, in moles, of iron(II) remaining after the reaction.

(d) Explain why the same starting permanganate ion can accept different numbers of electrons in the two flasks.

Also explain why the neutral charge of \(\ce{MnO2}\) does not mean that manganese has an oxidation number of zero.

Solution 5

Flask A is limited by iron(II), leaving \(0.00100\ \mathrm{mol}\) of permanganate unreacted. Flask B is limited by permanganate, leaving \(0.00300\ \mathrm{mol}\) of iron(II).

The difference comes from the different manganese-containing products formed under the two conditions.

The starting ion is identical in both flasks, but manganese undergoes a different oxidation-number change.

Step 1: Determine the oxidation numbers of manganese.

In the starting permanganate ion:

\[ x+4(-2)=-1 \]

Therefore:

\[ x=+7 \]

In flask A, manganese forms \(\mathrm{Mn}^{2+}\).

Its oxidation number is \(+2\), because it is a monatomic ion.

The change is:

\[ +7\longrightarrow+2 \]

Manganese gains five electrons per atom.

In flask B, manganese forms \(\ce{MnO2}\).

The compound is neutral, and oxygen has an oxidation number of \(-2\).

\[ x+2(-2)=0 \]

Therefore:

\[ x=+4 \]

The change is:

\[ +7\longrightarrow+4 \]

Manganese gains three electrons per atom.

We can summarise the difference:

Step 2: Write the balanced equation for flask A.

The acidic reduction half-equation is:

\[ \ce{ MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O } \]

Iron(II) is oxidised:

\[ \ce{Fe^2+ -> Fe^3+ + e^-} \]

Five iron(II) ions are required to supply the five electrons.

Combining the half-equations gives:

\[ \boxed{ \ce{ MnO4^- + 8H+ + 5Fe^2+ -> Mn^2+ + 4H2O + 5Fe^3+ } } \]

The mole ratio is:

\[ 1\ \mathrm{MnO_{4}}^{-}:5\ \mathrm{Fe}^{2+} \]

Check the charge.

Reactants:

\[ -1+8+10=+17 \]

Products:

\[ +2+15=+17 \]

The atoms are also conserved:

• Mn: 1 on each side.
• O: 4 on each side.
• H: 8 on each side.
• Fe: 5 on each side.

The equation is balanced.

Step 3: Determine the limiting reactant in flask A.

The flask contains:

\[ 0.00400\ \mathrm{mol}\ \mathrm{MnO_{4}}^{-} \]

and

\[ 0.0150\ \mathrm{mol}\ \mathrm{Fe}^{2+} \]

To consume all the permanganate, the amount of iron(II) required would be:

\[ \begin{aligned} n(\mathrm{Fe}^{2+}) &=5(0.00400)\\ &=0.0200\ \mathrm{mol} \end{aligned} \]

However, only \(0.0150\ \mathrm{mol}\) of iron(II) is available.

Therefore, iron(II) is the limiting reactant.

Step 4: Calculate the permanganate consumed and iron(II) remaining in flask A.

Using the \(1:5\) mole ratio:

\[ \begin{aligned} n(\mathrm{MnO_{4}}^{-}) &=\frac{0.0150}{5}\\ &=0.00300\ \mathrm{mol} \end{aligned} \]

The remaining permanganate is:

\[ \begin{aligned} 0.00400-0.00300 &=0.00100\ \mathrm{mol} \end{aligned} \]

All the iron(II) has reacted.

Therefore, flask A contains:

• Permanganate remaining: \(0.00100\ \mathrm{mol}\).
• Iron(II) remaining: \(0\ \mathrm{mol}\).

Step 5: Write the balanced equation for flask B.

In neutral aqueous conditions, manganese is reduced to \(\ce{MnO2}\).

The reduction half-equation is:

\[ \ce{ MnO4^- + 2H2O + 3e^- -> MnO2 + 4OH^- } \]

Iron(II) is oxidised:

\[ \ce{Fe^2+ -> Fe^3+ + e^-} \]

Three iron(II) ions are required to supply three electrons.

Combining the half-equations gives:

\[ \boxed{ \ce{ MnO4^- + 2H2O + 3Fe^2+ -> MnO2 + 4OH^- + 3Fe^3+ } } \]

The mole ratio is now:

\[ 1\ \mathrm{MnO_{4}}^{-}:3\ \mathrm{Fe}^{2+} \]

Notice that the ratio differs from flask A.

Check the charge.

Reactants:

\[ -1+3(+2)=+5 \]

Products:

\[ -4+3(+3)=+5 \]

Check the atoms:

• Mn: 1 on each side.
• Fe: 3 on each side.
• O: 6 on each side.
• H: 4 on each side.

The equation is balanced.

Step 6: Determine the limiting reactant in flask B.

To consume all \(0.00400\ \mathrm{mol}\) of permanganate, the amount of iron(II) required is:

\[ \begin{aligned} n(\mathrm{Fe}^{2+}) &=3(0.00400)\\ &=0.0120\ \mathrm{mol} \end{aligned} \]

There is \(0.0150\ \mathrm{mol}\) of iron(II) available.

This is more than the required amount.

Therefore, permanganate is the limiting reactant.

Step 7: Calculate the remaining iron(II).

All the permanganate reacts, consuming \(0.0120\ \mathrm{mol}\) of iron(II).

The remaining amount is:

\[ \begin{aligned} n(\mathrm{Fe}^{2+})_{\text{remaining}} &=0.0150-0.0120\\ &=0.00300\ \mathrm{mol} \end{aligned} \]

Therefore, flask B contains:

• Permanganate remaining: \(0\ \mathrm{mol}\).
• Iron(II) remaining: \(0.00300\ \mathrm{mol}\).

Step 8: Explain the underlying chemical difference.

The tempting assumption is that permanganate always accepts five electrons because its starting ion has a charge of \(-1\), or that a neutral manganese-containing product must contain manganese with an oxidation number of zero.

Both assumptions are incorrect.

The oxidation number of manganese in the starting permanganate ion is \(+7\), not \(-1\).

In flask A, manganese finishes at \(+2\), so it gains five electrons.

In flask B, manganese finishes at \(+4\), so it gains three electrons.

The product in flask B, \(\ce{MnO2}\), is neutral overall, but its oxygen atoms contribute a combined oxidation number of \(-4\). Manganese must therefore have an oxidation number of \(+4\).

This is the central idea: the ionic charge describes the whole species, while the oxidation number describes an individual atom. The oxidation-number change, not the overall ionic charge, determines the electron transfer.

The reaction medium matters because it can influence which reduction product forms. That changes the electron requirement, the mole ratio, and even which reactant is limiting.

07Where to take this next

Once you can distinguish ionic charge from oxidation number, you can approach redox equations much more reliably.

The next useful skill is balancing half-equations in the correct reaction medium. You use oxidation-number changes to identify electron transfer, then balance atoms and charge to produce a chemically consistent equation.

For a step-by-step treatment of that process, continue with How to Balance a Redox Half-Equation in Acidic Solution.