How to Check Charge Balance in a Redox Equation

An iodine titration reaches its endpoint, and your calculations show that two thiosulfate ions react with every iodine molecule. You write the equation:

\[ \ce{I2 + 2S2O3^2- -> 2I^- + S4O6^2-} \]

But something looks odd. The total charge on the left is \(-4\). Surely a balanced equation should have zero charge on both sides?

Not quite. To check charge balance in a redox equation, calculate the total charge on each side and confirm that the two totals are equal. They do not have to be zero.

For this reaction, the products also have a total charge of \(-4\).

That distinction is the key to checking ionic redox equations correctly. You need to conserve atoms, conserve total charge, and account for the electrons transferred between reacting species.

Let’s work through why these checks matter, using an actual type of redox experiment.

01Start with the experiment: iodine and thiosulfate

Imagine you want to determine how much iodine is present in an aqueous solution.

You could estimate its concentration from the colour, but that would be imprecise. A redox titration gives you a more reliable measurement.

The reaction involves iodine, \(\ce{I2}\), and thiosulfate ions, \(\mathrm{S_{2}O_{3}}^{2-}\), supplied by sodium thiosulfate solution.

What does the apparatus look like?

A burette containing sodium thiosulfate solution sits above a conical flask containing a measured volume of iodine solution.

You gradually release the thiosulfate into the flask while swirling it.

Initially, the iodine solution is brown. As thiosulfate is added, iodine is converted into iodide ions.

Near the endpoint, the solution becomes pale yellow. You then add starch indicator, which produces a blue-black colour while iodine remains.

Continue adding thiosulfate drop by drop. When all the iodine has reacted, the blue-black colour disappears.

That disappearance tells you that the reaction has reached its endpoint.

What should the reaction ratio be?

Before looking at the equation, predict how many thiosulfate ions should react with one iodine molecule.

There are two competing electron transfers:

• Iodine gains electrons and is reduced to iodide.
• Thiosulfate loses electrons and is oxidised to tetrathionate.

The reduction half-equation is:

\[ \ce{I2 + 2e^- -> 2I^-} \]

One iodine molecule gains two electrons.

The oxidation half-equation is:

\[ \ce{2S2O3^2- -> S4O6^2- + 2e^-} \]

Two thiosulfate ions release two electrons in total.

The electron numbers match, so the overall reaction requires two thiosulfate ions per iodine molecule.

Adding the half-equations cancels the electrons:

\[ \ce{I2 + 2S2O3^2- -> 2I^- + S4O6^2-} \]

This gives us a prediction that can be tested using titration data.

02How to check charge balance in a redox equation

The titration gives us a useful starting point. Now let’s turn the charge check into a method you can use in an exam.

1. Identify the charge of every species

Look at the superscript of each ion.

Remember:

• A neutral atom or molecule has a charge of zero.
• A positive ion has a positive charge.
• A negative ion has a negative charge.
• A polyatomic ion has one overall charge, even though its individual atoms may have different oxidation numbers.

For example:

Be particularly careful with polyatomic ions.

In \(\mathrm{MnO_{4}}^{-}\), manganese has an oxidation number of \(+7\), but the overall charge of the permanganate ion is \(-1\).

For a charge-balance calculation, you use \(-1\), not \(+7\).

2. Multiply each charge by its coefficient

A coefficient tells you how many particles of a particular species are present.

For example, three iron(III) ions have a combined charge of:

\[ 3\times(+3)=+9 \]

Similarly, four sulfate ions contribute:

\[ 4\times(-2)=-8 \]

You must include coefficients when calculating total charge.

3. Add the charges on each side separately

Use:

\[ Q_{\text{total}}=\sum n_iq_i \]

Here:

• \(Q_{\text{total}}\) is the total charge on one side of the equation.
• \(n_i\) is the coefficient of a species.
• \(q_i\) is the charge of that species.

Calculate the reactant-side total and product-side total independently.

Then compare them.

4. Check atoms separately

Charge balance does not automatically mean that the atoms are balanced.

You need both checks:

• Atom conservation: the number of each type of atom must be equal on both sides.
• Charge conservation: the total electrical charge must be equal on both sides.

One cannot replace the other.

5. Check the electron transfer

For a redox reaction, the total number of electrons lost must equal the total number gained.

This is particularly important when checking half-equations.

In a complete redox equation, electrons should cancel when the oxidation and reduction half-equations are combined.

03Why the total charge does not have to be zero

This is one of the most common mistakes in ionic equations.

A student sees a negative charge on the reactant side and assumes something must be missing.

But think of charge balance as an accounting problem. You compare what enters the reaction with what leaves it.

You do not need both sides to have a zero balance. You need the same balance.

The analogy has one important limit: ions really do carry electrical charge. We are not just assigning numbers for convenience.

Consider the iodine-thiosulfate equation again:

\[ \ce{I2 + 2S2O3^2- -> 2I^- + S4O6^2-} \]

Both sides have a charge of \(-4\).

The equation is perfectly valid.

What happens to spectator ions?

You might notice that the experiment uses sodium thiosulfate, but sodium ions do not appear in the net ionic equation.

That’s deliberate.

Sodium ions do not undergo oxidation or reduction. They are spectator ions, meaning they remain unchanged during the reaction.

If we include them, the full ionic equation is:

\[ \ce{ I2 + 4Na+ + 2S2O3^2- -> 4Na+ + 2I^- + S4O6^2- } \]

Both sides now have a total charge of zero.

The four sodium ions appear on both sides and cancel, leaving the net ionic equation.

This explains why net ionic equations can have non-zero total charges even though the complete reaction mixture remains electrically neutral.

04Electrons are the missing piece in an incomplete half-equation

Consider the reduction of iodine:

\[ \ce{I2 -> 2I^-} \]

The iodine atoms balance, but the charges do not.

• Left side: 0
• Right side: -2

Where has the extra negative charge come from?

It comes from two electrons that have not yet been included.

The correct reduction half-equation is:

\[ \ce{I2 + 2e^- -> 2I^-} \]

Now check the charges:

\[ Q_L=0+2(-1)=-2 \]

\[ Q_R=2(-1)=-2 \]

The half-equation conserves charge.

The electrons are essential in the half-equations because each half-reaction describes only one part of the electron transfer.

When the two half-equations are added, the electrons cancel:

\[ \begin{aligned} \ce{I2 + 2e^-}&\ce{-> 2I^-}\\ \ce{2S2O3^2-}&\ce{-> S4O6^2- + 2e^-} \end{aligned} \]

The final equation is:

\[ \ce{I2 + 2S2O3^2- -> 2I^- + S4O6^2-} \]

There are no electrons left in the overall reaction.

This is why checking the charge of each half-equation is such a useful diagnostic. If a half-equation does not balance electrically, it is incomplete or incorrect.

05The reaction medium changes which ions you can use

You cannot check a redox equation properly without knowing the reaction conditions.

The balancing species depend on whether the solution is acidic, alkaline, or approximately neutral.

This matters because hydrogen and oxygen atoms can be balanced differently depending on the medium.

Let’s examine an acidic reaction.

06A charge-balanced equation can still be wrong

Passing the charge check is necessary, but it is not sufficient.

Consider this proposed equation:

\[ \ce{Zn + 2Ag+ -> Zn^2+ + Ag} \]

Check the charge.

Left:

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

Right:

\[ Q_R=+2+0=+2 \]

The charge balances.

However, look at the silver atoms.

There are two silver atoms on the left but only one on the right.

The equation fails atom conservation.

The correct equation is:

\[ \ce{Zn + 2Ag+ -> Zn^2+ + 2Ag} \]

Now both the atoms and charges balance.

There is another important trap: do not confuse oxidation numbers with ionic charges.

For example, manganese in \(\mathrm{MnO_{4}}^{-}\) has oxidation number \(+7\), but the ion’s overall charge is \(-1\).

If you use \(+7\) when calculating the total charge of the reactants, your entire charge audit will be incorrect.

07A quick charge-checking decision process

When you finish balancing a redox equation, use this sequence.

If the charges do not match, don’t randomly add hydrogen or hydroxide ions.

First, identify where the mismatch occurs. Check whether you have omitted electrons from a half-equation, used an incorrect coefficient, or introduced the wrong balancing species.

For a complete walkthrough of constructing redox equations from the beginning, see How to Balance Redox Equations Without Guessing.

08Questions and solutions

The following questions move from straightforward charge calculations to situations where electron transfer, reaction conditions, and electrical neutrality must all be considered together.

Question 1

A magnesium strip is placed into aqueous silver nitrate solution. Magnesium reacts with silver ions to produce magnesium ions and solid silver.

The proposed net ionic equation is:

\[ \ce{Mg + 2Ag+ -> Mg^2+ + 2Ag} \]

Verify that the equation conserves atoms and charge.

Write the two half-equations and identify the total number of electrons transferred per magnesium atom.

Solution 1

The equation is balanced. Both sides have a total charge of \(+2\), and the numbers of magnesium and silver atoms are equal.

Step 1: Check the atoms.

Both elements balance.

Step 2: Calculate the total charge on each side.

Reactants:

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

Products:

\[ Q_R=(+2)+0=+2 \]

The total charges are equal.

Step 3: Write the oxidation half-equation.

Magnesium changes from oxidation number 0 to \(+2\).

It loses two electrons:

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

Step 4: Write the reduction half-equation.

Each silver ion gains one electron:

\[ \ce{Ag+ + e^- -> Ag} \]

There are two silver ions, so multiply the entire half-equation by two:

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

Step 5: Identify the total electron transfer.

Magnesium loses two electrons, and the two silver ions gain two electrons.

Therefore, two electrons are transferred per magnesium atom oxidised.

The electrons cancel when the half-equations are combined, giving the original balanced equation.

Question 2

In acidic aqueous solution, iodate ions react with iodide ions to produce iodine.

A student proposes:

\[ \ce{IO3^- + I^- + H+ -> I2 + H2O} \]

Determine the correct balanced net ionic equation.

Your response must:

• Show the correct coefficients.
• Calculate the total charge on each side.
• Explain how the oxidation-number changes determine the electron ratio.
• Explain why the oxidation number of iodine in \(\mathrm{IO_{3}}^{-}\) is not equal to the overall charge of the ion.

Assume dilute sulfuric acid supplies the hydrogen ions and that iodine is the final iodine-containing product.

Solution 2

The balanced equation is:

\[ \ce{IO3^- + 5I^- + 6H+ -> 3I2 + 3H2O} \]

The key is that iodate and iodide undergo different oxidation-number changes. Their electron transfers must be equal.

Step 1: Determine the oxidation number of iodine in iodate.

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

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

\[ \begin{aligned} x+3(-2)&=-1\\ x-6&=-1\\ x&=+5 \end{aligned} \]

Iodine in \(\mathrm{IO_{3}}^{-}\) has oxidation number \(+5\).

It changes from \(+5\) to 0 in \(\ce{I2}\), gaining five electrons.

Step 2: Determine the oxidation-number change of iodide.

Iodine in \(\mathrm{I}^{-}\) has oxidation number \(-1\).

It changes from \(-1\) to 0 in \(\ce{I2}\), losing one electron per iodide ion.

Therefore, five iodide ions must lose five electrons to match the five electrons gained by one iodate ion.

Step 3: Establish the iodine coefficients.

There is one iodine atom in each iodate ion and five iodine atoms from five iodide ions.

That gives six iodine atoms in total.

Therefore, the product must contain three iodine molecules:

\[ \ce{IO3^- + 5I^- -> 3I2} \]

Step 4: Balance oxygen and hydrogen.

There are three oxygen atoms on the left.

Add three water molecules to the right:

\[ \ce{IO3^- + 5I^- -> 3I2 + 3H2O} \]

The three water molecules contain six hydrogen atoms.

Add six hydrogen ions to the left:

\[ \ce{IO3^- + 5I^- + 6H+ -> 3I2 + 3H2O} \]

Step 5: Check the charge.

Reactants:

\[ Q_L=-1+5(-1)+6(+1)=0 \]

Products:

\[ Q_R=0 \]

Both sides have zero total charge.

Step 6: Check the atoms.

The equation conserves both atoms and charge.

The important misconception is treating the charge of an entire polyatomic ion as the oxidation number of its central atom. Iodate has an overall charge of \(-1\), but iodine has oxidation number \(+5\). These describe different things.

Question 3

In an aqueous redox reaction, iron(III) ions react with tin(II) ions.

The products are iron(II) ions and tin(IV) ions.

A student proposes:

\[ \ce{Fe^3+ + Sn^2+ -> Fe^2+ + Sn^4+} \]

The student argues that the atoms are balanced, so the equation must be correct.

Determine the correct coefficients and verify the total charge.

Explain how the electron transfer determines the coefficient ratio.

Assume the reaction occurs in aqueous solution without additional acidic or alkaline balancing species.

Solution 3

The correct equation is:

\[ \ce{2Fe^3+ + Sn^2+ -> 2Fe^2+ + Sn^4+} \]

Two iron(III) ions are needed for every tin(II) ion because tin loses two electrons while each iron(III) ion gains one.

Step 1: Identify oxidation and reduction.

Iron:

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

Each iron(III) ion gains one electron.

Tin:

\[ \ce{Sn^2+ -> Sn^4+ + 2e^-} \]

Each tin(II) ion loses two electrons.

Step 2: Match the electron transfer.

One tin ion releases two electrons.

Therefore, two iron(III) ions are required to accept those electrons.

The coefficient ratio is:

\[ \ce{Fe^3+ : Sn^2+}=2:1 \]

Step 3: Write the combined equation.

\[ \ce{2Fe^3+ + Sn^2+ -> 2Fe^2+ + Sn^4+} \]

The electrons have cancelled.

Step 4: Check the total charge.

Reactants:

\[ \begin{aligned} Q_L&=2(+3)+(+2)\\ &=+8 \end{aligned} \]

Products:

\[ \begin{aligned} Q_R&=2(+2)+(+4)\\ &=+8 \end{aligned} \]

Both sides have a total charge of \(+8\).

Step 5: Check the atoms.

There are two iron atoms and one tin atom on each side.

The equation is balanced.

The original student’s mistake was assuming that matching the number of atoms was enough. The proposed equation actually had a charge of \(+5\) on the left and \(+6\) on the right.

The electron-transfer ratio fixes that mismatch.

Question 4

In strongly alkaline aqueous solution, permanganate ions react with sulfite ions.

The products are manganese dioxide and sulfate ions.

A student proposes:

\[ \ce{2MnO4^- + 3SO3^2- -> 2MnO2 + 3SO4^2-} \]

The student has chosen coefficients that match the electron transfer but has not included any hydrogen- or oxygen-balancing species.

Complete the equation.

Then:

• Identify the oxidation and reduction processes.
• Show the number of electrons transferred.
• Verify the atom counts and total charge.
• Explain why hydroxide ions appear in the final equation.

Assume manganese dioxide is the manganese-containing product and no other reactions occur.

Solution 4

The balanced equation in alkaline solution is:

\[ \ce{2MnO4^- + 3SO3^2- + H2O -> 2MnO2 + 3SO4^2- + 2OH^-} \]

The electron transfer determines the coefficients of the redox species. Water and hydroxide then complete the atom and charge balance.

Step 1: Determine the oxidation-number changes.

In permanganate, manganese has oxidation number \(+7\).

In manganese dioxide, oxygen has oxidation number \(-2\), so manganese has oxidation number \(+4\).

Manganese is reduced:

\[ +7\rightarrow+4 \]

Each manganese atom gains three electrons.

There are two manganese atoms, so six electrons are gained in total.

In sulfite, sulfur has oxidation number \(+4\).

In sulfate, sulfur has oxidation number \(+6\).

Sulfur is oxidised:

\[ +4\rightarrow+6 \]

Each sulfur atom loses two electrons.

Three sulfite ions therefore release six electrons.

Step 2: Match the electron transfer.

Two permanganate ions gain six electrons.

Three sulfite ions lose six electrons.

This gives the initial ratio:

\[ \ce{MnO4^- : SO3^2-}=2:3 \]

Step 3: Write the reduction half-equation.

In alkaline solution:

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

There are three electrons on the left.

Multiply the entire equation by two:

\[ \ce{2MnO4^- + 4H2O + 6e^- -> 2MnO2 + 8OH^-} \]

Step 4: Write the oxidation half-equation.

\[ \ce{SO3^2- + 2OH^- -> SO4^2- + H2O + 2e^-} \]

Multiply the entire equation by three:

\[ \ce{3SO3^2- + 6OH^- -> 3SO4^2- + 3H2O + 6e^-} \]

Step 5: Combine the half-equations.

Add the two equations and cancel the six electrons.

You initially obtain:

\[ \ce{ 2MnO4^- + 4H2O + 3SO3^2- + 6OH^- -> 2MnO2 + 8OH^- + 3SO4^2- + 3H2O } \]

Cancel six hydroxide ions from both sides.

Cancel three water molecules from both sides.

The simplified equation is:

\[ \ce{2MnO4^- + 3SO3^2- + H2O -> 2MnO2 + 3SO4^2- + 2OH^-} \]

Step 6: Check the total charge.

Reactants:

\[ Q_L=2(-1)+3(-2)=-8 \]

Products:

\[ Q_R=2(0)+3(-2)+2(-1)=-8 \]

The charges match.

Step 7: Check the atoms.

All atoms balance.

The hydroxide ions appear because the reaction occurs in alkaline solution. They are produced when the two balanced half-equations are combined and simplified.

The tempting approach is to assume that matching the six electrons is enough. It isn’t. The electron ratio establishes the redox coefficients, but the final equation must still conserve every atom and the total charge.

Question 5

A galvanic cell is constructed using two half-cells:

• An aluminium electrode immersed in aqueous aluminium nitrate, \(\ce{Al(NO3)3}\).
• A silver electrode immersed in aqueous silver nitrate, \(\ce{AgNO3}\).

The half-cells are connected by an external wire and a potassium nitrate, \(\ce{KNO3}\), salt bridge.

During operation, aluminium dissolves from its electrode and silver deposits onto the other electrode.

A student proposes the following net ionic equation:

\[ \ce{Al + 3Ag+ -> Al^3+ + 3Ag} \]

Another student argues that this cannot be balanced because both sides have a total charge of \(+3\), rather than zero.

During one run, 1.50 mmol of aluminium is consumed, and 0.485 g of silver is deposited.

Assume:

• The only electrode reactions are aluminium oxidation and silver-ion reduction.
• The solution remains aqueous, with no precipitation or other side reactions.
• The salt bridge contains only \(\mathrm{K}^{+}\) and \(\mathrm{NO_{3}}^{-}\) as its mobile ions.
• Salt bridge migration maintains electrical neutrality in each half-cell.
• Use a molar mass of silver of 107.87 g mol\(^{-1}\).

Answer the following:

(a) Write the oxidation and reduction half-equations.

(b) Verify the net ionic equation using atom conservation, charge conservation, and electron transfer. Explain which student’s argument is correct.

(c) Calculate the amount of electrons transferred during the run.

(d) Determine the approximate amounts of \(\mathrm{K}^{+}\) and \(\mathrm{NO_{3}}^{-}\) that migrate through the salt bridge, identifying the direction of movement for each ion.

(e) Write the complete molecular equation using nitrate as the spectator ion. Explain why the salt bridge ions do not need to appear in the net ionic redox equation.

Solution 5

The first student is correct: the net ionic equation is balanced, even though both sides have a total charge of \(+3\).

The second student’s mistake is confusing charge conservation in a net ionic equation with electrical neutrality in each half-cell.

Step 1: Write the oxidation half-equation.

Aluminium loses three electrons:

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

This is oxidation because aluminium’s oxidation number increases from 0 to \(+3\).

Step 2: Write the reduction half-equation.

Each silver ion gains one electron:

\[ \ce{Ag+ + e^- -> Ag} \]

Three silver ions are needed to accept the three electrons released by aluminium.

Therefore:

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

Step 3: Combine the half-equations.

The three electrons cancel:

\[ \ce{Al + 3Ag+ -> Al^3+ + 3Ag} \]

This is the balanced net ionic equation.

Step 4: Verify atom conservation.

Both elements balance.

Step 5: Verify charge conservation.

Reactants:

\[ Q_L=0+3(+1)=+3 \]

Products:

\[ Q_R=(+3)+0=+3 \]

The total charges are equal.

There is no requirement for the two totals to be zero.

The net ionic equation is therefore balanced.

Step 6: Check the experimental evidence.

Convert the silver mass into moles:

\[ \begin{aligned} n(\ce{Ag})&=\frac{m}{M}\\ &=\frac{0.485}{107.87}\\ &=0.004497\text{ mol} \end{aligned} \]

Therefore:

\[ n(\ce{Ag})\approx4.50\times10^{-3}\text{ mol} \]

The amount of aluminium consumed is:

\[ n(\ce{Al})=1.50\times10^{-3}\text{ mol} \]

Calculate the mole ratio:

\[ \frac{n(\ce{Ag})}{n(\ce{Al})} = \frac{4.50\times10^{-3}}{1.50\times10^{-3}} =3.00 \]

The experimental ratio is approximately three silver atoms deposited per aluminium atom consumed.

This agrees with the balanced equation.

Step 7: Calculate the amount of electrons transferred.

Each aluminium atom releases three electrons.

Therefore:

\[ \begin{aligned} n(e^-)&=3n(\ce{Al})\\ &=3(1.50\times10^{-3})\\ &=4.50\times10^{-3}\text{ mol} \end{aligned} \]

The reaction transfers approximately \(4.50\times10^{-3}\) mol of electrons.

The same amount of electrons is accepted by the silver ions.

Step 8: Determine the movement of salt bridge ions.

This is where the distinction between net ionic charge and electrical neutrality becomes important.

At the aluminium electrode, aluminium atoms become aluminium ions:

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

The concentration of positive ions in that half-cell increases as aluminium dissolves.

To maintain electrical neutrality, negative ions from the salt bridge migrate into the aluminium half-cell.

Therefore, \(\mathrm{NO_{3}}^{-}\) migrates towards the aluminium electrode.

The amount required is three moles of nitrate ions per mole of aluminium oxidised:

\[ \begin{aligned} n(\mathrm{NO_{3}}^{-})&=3(1.50\times10^{-3})\\ &=4.50\times10^{-3}\text{ mol} \end{aligned} \]

At the silver electrode, silver ions are removed from solution:

\[ \ce{Ag+ + e^- -> Ag} \]

This leaves an excess of negative nitrate ions in the silver half-cell.

To maintain electrical neutrality, positive ions from the salt bridge migrate into this compartment.

Therefore, \(\mathrm{K}^{+}\) migrates towards the silver electrode.

The amount required is:

\[ n(\mathrm{K}^{+})=4.50\times10^{-3}\text{ mol} \]

The approximate ion migration is:

The salt bridge maintains charge neutrality within the separate compartments.

It does not change the electron-transfer ratio or the net ionic redox equation.

Step 9: Write the complete molecular equation.

The nitrate ions can be included in the molecular equation:

\[ \ce{Al + 3AgNO3 -> Al(NO3)3 + 3Ag} \]

All the formula units are electrically neutral.

The potassium nitrate in the salt bridge is not consumed by the overall reaction. Its ions migrate to maintain electrical neutrality, rather than participating in the electron-transfer reaction.

This is the subtle distinction that makes the question difficult:

• The net ionic equation conserves total charge, even though its two sides have a non-zero charge.
• Each half-cell remains approximately electrically neutral because ions migrate through the salt bridge.
• The complete molecular equation includes spectator ions and can be written using neutral formula units.

These are three related ideas, but they describe different levels of the same electrochemical system.

Once your redox equation passes the atom, charge, and electron checks, you have established that its bookkeeping is consistent. The next question is whether the reaction can actually drive an electrochemical cell.

That requires comparing standard reduction potentials to determine the cell potential and predict the direction of electron flow.