How to Find Electron Transfer from Oxidation-State Changes
Learn a reliable method for calculating electrons transferred from oxidation-state changes, including how to account for coefficients, subscripts, and reaction conditions. Includes worked examples, redox titration calculations, and challenging HSC-style practice questions with solutions.
A reaction can be balanced for atoms and still leave you wondering how many electrons have actually been transferred. Consider the reaction between iron(III) oxide and carbon monoxide:
\[ \ce{Fe2O3 + 3CO -> 2Fe + 3CO2} \]
Iron changes from an oxidation state of +3 to 0. Carbon changes from +2 to +4. But how many electrons are transferred? Is it three, two, six, or something else?
The answer is six electrons transferred per reaction as written. To find this, you need to calculate the total oxidation-state change for every atom of each element that changes oxidation state, taking the coefficients and subscripts into account.
This gives you a reliable method for finding electron transfer from oxidation-state changes, even when a reaction contains several reacting substances or different numbers of atoms.
01The method: turning oxidation-state changes into electron numbers
An oxidation state tells you how many electrons an atom is formally assigned to have gained or lost compared with its elemental state. A rise in oxidation state represents oxidation (electron loss), while a fall represents reduction (electron gain).
For a balanced redox equation:
- An increase of one oxidation-state unit represents the loss of one electron per affected atom.
- A decrease of one oxidation-state unit represents the gain of one electron per affected atom.
- The total electrons lost must equal the total electrons gained.
The important detail is that you must count every affected atom, not just compare the oxidation states of one atom on each side.
A reliable four-step method
Step 1: Write the balanced reaction and identify the reaction conditions.
Check that the chemical formulas are correct and that the atoms are conserved. For ionic equations, also check that the total charge is conserved.
State whether the reaction occurs in acidic, alkaline, or neutral conditions when this affects the products or the way the equation is balanced.
Step 2: Assign oxidation states to the elements that change.
Work out the oxidation state before and after the reaction. You do not need to calculate every element if some remain unchanged.
Step 3: Calculate the total oxidation-state change.
For each element that changes oxidation state, multiply:
\[ \text{Total change} = \text{number of atoms} \times \text{change per atom} \]
The number of atoms is determined by multiplying the coefficient by the subscript of that element in its chemical formula.
Step 4: Match the electrons lost and gained.
An increase in oxidation state represents electrons lost. A decrease represents electrons gained.
The total number of electrons lost must equal the total number gained in a correctly balanced redox reaction.
Worked example: iron(III) oxide reacting with carbon monoxide
Question: In the reaction below, calculate the number of electrons transferred per reaction as written.
\[ \ce{Fe2O3 + 3CO -> 2Fe + 3CO2} \]
Step 1
Iron and carbon change oxidation states. Oxygen remains at -2 throughout.
Step 2
For iron:
- In \(\ce{Fe2O3}\), iron has an oxidation state of +3.
- In elemental iron, \(\ce{Fe}\), its oxidation state is 0.
Therefore:
\[ +3 \rightarrow 0 \]
Each iron atom gains three electrons.
For carbon:
- In \(\ce{CO}\), oxygen has an oxidation state of -2, so carbon must be +2.
- In \(\ce{CO2}\), two oxygen atoms contribute -4, so carbon must be +4.
Therefore:
\[ +2 \rightarrow +4 \]
Each carbon atom loses two electrons.
Step 3
There are two iron atoms in the products and three carbon atoms in the reactants.
| Element | Atoms involved | Change per atom | Total electron change |
|---|---|---|---|
| Fe | 2 | Gains 3 | Gains 6 |
| C | 3 | Loses 2 | Loses 6 |
Step 4
\[ \begin{aligned} \text{Electrons gained} &= 2 \times 3 = 6\\ \text{Electrons lost} &= 3 \times 2 = 6 \end{aligned} \]
Final answer: Six electrons are transferred per reaction as written. Iron gains six electrons in total, while carbon loses six electrons in total.
Notice why counting just the change from +3 to 0 would give the wrong answer. There are two iron atoms, so the total electron gain is six, not three.
02Why coefficients and subscripts matter
A common mistake is to look only at the oxidation-state change and ignore how many atoms are involved.
The coefficient tells you how many formula units or molecules are present. The subscript tells you how many atoms of an element are in each formula unit or molecule.
For example, in \(\ce{2Cr2O7^2-}\), there are four chromium atoms, not two. The coefficient multiplies the subscript.
You can organise your calculations in an electron-accounting table:
| Quantity | What to calculate |
|---|---|
| Oxidation-state change | Final oxidation state minus initial oxidation state |
| Number of affected atoms | Coefficient × subscript |
| Total change | Number of atoms × magnitude of oxidation-state change |
| Electron loss | Total increase in oxidation state |
| Electron gain | Total decrease in oxidation state |
The sign of the oxidation-state change tells you the direction of electron transfer. Its magnitude, multiplied by the number of affected atoms, tells you the total number of electrons involved.
03The reaction medium can change the electron count
There is one more condition to check before trusting your calculation: the reaction medium.
Acidic, alkaline, and neutral conditions can produce different products from the same starting substance. This matters because the final oxidation state determines how many electrons an atom has gained or lost.
Consider dichromate ions reacting with iron(II) ions in acidic solution:
\[ \ce{Cr2O7^2- + 6Fe^2+ + 14H+ -> 2Cr^3+ + 6Fe^3+ + 7H2O} \]
Step 1: Find the oxidation-state changes.
For chromium in dichromate:
\[ 2x+7(-2)=-2 \]
\[ x=+6 \]
Chromium changes from +6 to +3. Each chromium atom gains three electrons.
For iron, the change is from +2 to +3. Each iron atom loses one electron.
Step 2: Include the coefficients.
There are two chromium atoms and six iron atoms.
| Element | Number of atoms | Change per atom | Total |
|---|---|---|---|
| Cr | 2 | Gains 3 electrons | Gains 6 |
| Fe | 6 | Loses 1 electron | Loses 6 |
Step 3: Check conservation of charge.
On the reactant side:
\[ -2+6(+2)+14(+1)=+24 \]
On the product side:
\[ 2(+3)+6(+3)=+24 \]
Both sides have a total charge of +24. The atoms are also balanced, including the hydrogen and oxygen atoms supplied by the acidic medium.
Final answer: Six electrons are transferred per dichromate ion, or six electrons per reaction as written.
Notice that the hydrogen ions help balance the equation, but they do not undergo an oxidation-state change. They are not included in the electron-transfer calculation.
Worked example: using electron transfer in a redox titration
A useful application of this method is determining the concentration of an iron(II) solution.
Imagine a titration in which:
- A burette contains standard potassium dichromate solution.
- A conical flask contains 25.00 mL of an iron(II) solution and excess dilute sulfuric acid.
- A suitable redox indicator is added to the flask.
- Dichromate solution is added gradually while the flask is swirled until the indicator reaches its endpoint.
The balanced reaction is:
\[ \ce{Cr2O7^2- + 6Fe^2+ + 14H+ -> 2Cr^3+ + 6Fe^3+ + 7H2O} \]
What should the experiment show?
The oxidation-state calculation predicts that one mole of dichromate ions reacts with six moles of iron(II) ions. The titration measures the volume of dichromate needed to reach the endpoint, allowing this predicted mole ratio to be tested.
The indicator does not count electrons. It signals that the reaction has reached its endpoint. The electron count comes from the balanced reaction and oxidation-state changes.
Suppose the following illustrative data are obtained:
| Measurement | Value |
|---|---|
| Iron(II) solution volume | 25.00 mL |
| Dichromate concentration | 0.00500 mol L\(^{-1}\) |
| Dichromate volume at endpoint | 20.00 mL |
Calculate the iron(II) concentration and the amount of electron transfer.
Step 1
Use:
\[ n=cV \]
where \(n\) is amount in moles, \(c\) is concentration in mol L\(^{-1}\), and \(V\) is volume in litres.
Convert the volume:
\[ 20.00\text{ mL}=0.02000\text{ L} \]
Then:
\[ \begin{aligned} n(\mathrm{Cr_{2}O_{7}}^{2-}) &=0.00500\times0.02000\\ &=1.00\times10^{-4}\text{ mol} \end{aligned} \]
Step 2
One dichromate ion reacts with six iron(II) ions.
\[ \begin{aligned} n(\mathrm{Fe}^{2+}) &=6\times1.00\times10^{-4}\\ &=6.00\times10^{-4}\text{ mol} \end{aligned} \]
Step 3
\[ \begin{aligned} c&=\frac{n}{V}\\ &=\frac{6.00\times10^{-4}}{0.02500}\\ &=0.0240\text{ mol L}^{-1} \end{aligned} \]
Step 4
Each iron(II) ion loses one electron. Therefore:
\[ n(e^-)=6.00\times10^{-4}\text{ mol} \]
The dichromate accepts exactly the same amount of electrons.
Final answer: The iron(II) concentration is 0.0240 mol L\(^{-1}\), and the reaction transfers \(6.00\times10^{-4}\) mol of electrons at the measured endpoint.
The experimental result is consistent with the predicted 1:6 mole ratio. It does not directly measure individual electrons, but it provides evidence that the reaction follows the expected stoichiometry.
A single rapid titration can overshoot the endpoint. A better design uses an initial rough titration to locate the endpoint, followed by repeat trials with slower, dropwise additions near that point. Concordant results make the measured volume more reliable.
04A common trap: formal electron transfer is not always literal electron movement
Consider the reaction between hydrogen and oxygen:
\[ \ce{2H2 + O2 -> 2H2O} \]
How many electrons are transferred according to oxidation-state changes?
Hydrogen changes from 0 to +1. There are four hydrogen atoms, so each loses one electron:
\[ 4\times1=4\text{ electrons lost} \]
Oxygen changes from 0 to -2. There are two oxygen atoms, so each gains two electrons:
\[ 2\times2=4\text{ electrons gained} \]
The electron accounting gives four electrons transferred.
But there is an important distinction.
In water, hydrogen and oxygen form covalent bonds. The electrons are shared, although the bonding electrons are assigned to oxygen when calculating oxidation states because oxygen is more electronegative.
So the calculation does not mean four free electrons physically travel from hydrogen molecules to oxygen molecules.
Oxidation states provide a formal electron-accounting model. In a reaction involving ions, such as magnesium reacting with chlorine, the model also describes the formation of oppositely charged ions:
\[ \ce{Mg + Cl2 -> MgCl2} \]
Magnesium loses two electrons, while the two chlorine atoms gain one each.
The useful distinction is this:
- Oxidation-state changes tell you the number of formal electron equivalents exchanged.
- For ionic reactions, this often corresponds directly to electrons transferred between species.
- For covalent reactions, it is an electron-accounting convention, not necessarily a description of complete physical electron movement.
For HSC calculations, use the oxidation-state method consistently. Just be precise when explaining what the calculated electron number represents.
05Questions and solutions
The following questions move from straightforward electron counting to situations where reaction conditions, multiple oxidation states, and conservation of atoms and charge affect the answer.
Question 1
Consider the balanced reaction:
\[ \ce{2Na(s) + Cl2(g) -> 2NaCl(s)} \]
Calculate the total number of electrons transferred per reaction as written.
Identify which element loses electrons and which element gains electrons.
Solution 1
Two electrons are transferred per reaction as written. Sodium loses two electrons in total, and chlorine gains two electrons in total.
Start by assigning oxidation states:
- Sodium changes from 0 to +1.
- Chlorine changes from 0 to -1.
There are two sodium atoms, each losing one electron:
\[ 2\times1=2\text{ electrons lost} \]
There are two chlorine atoms in \(\ce{Cl2}\), each gaining one electron:
\[ 2\times1=2\text{ electrons gained} \]
The totals agree, so the electron transfer is balanced.
Sodium is oxidised, while chlorine is reduced.
The important point is that the subscript in \(\ce{Cl2}\) represents two chlorine atoms. Counting only one chlorine atom would incorrectly give one electron gained.
Question 2
Nitrite ions react with iodide ions in acidic solution according to the balanced net ionic equation:
\[ \ce{2NO2^- + 2I^- + 4H+ -> 2NO + I2 + 2H2O} \]
Determine:
a. The oxidation state of nitrogen before and after the reaction.
b. The oxidation state of iodine before and after the reaction.
c. The total number of electrons transferred per reaction as written.
d. Whether atoms and total ionic charge are conserved.
Solution 2
Two electrons are transferred per reaction as written. Nitrogen gains two electrons in total, while iodine loses two electrons in total.
Step 1: Calculate the oxidation states of nitrogen.
In \(\mathrm{NO_{2}}^{-}\), oxygen has an oxidation state of -2.
\[ x+2(-2)=-1 \]
\[ x=+3 \]
In \(\ce{NO}\):
\[ x+(-2)=0 \]
\[ x=+2 \]
Nitrogen changes from +3 to +2. Each nitrogen atom gains one electron.
Step 2: Calculate the oxidation states of iodine.
Iodide has an oxidation state of -1.
Elemental iodine, \(\ce{I2}\), has an oxidation state of 0.
Each iodine atom therefore loses one electron.
Step 3: Include the number of atoms.
There are two nitrogen atoms and two iodine atoms involved.
\[ \begin{aligned} \text{Electrons gained} &=2\times1=2\\ \text{Electrons lost} &=2\times1=2 \end{aligned} \]
Step 4: Check conservation of atoms and charge.
| Element | Reactants | Products |
|---|---|---|
| N | 2 | 2 |
| O | 4 | 4 |
| I | 2 | 2 |
| H | 4 | 4 |
The total charge on the reactant side is:
\[ 2(-1)+2(-1)+4(+1)=0 \]
The products are all neutral, so their total charge is also zero.
The equation conserves both atoms and charge.
Nitrogen is reduced, and iodide is oxidised. The hydrogen ions provide the acidic conditions needed for the reaction, but hydrogen does not change oxidation state.
Question 3
The following equation represents the overall Claus reaction, used to produce elemental sulfur from hydrogen sulfide and sulfur dioxide:
\[ \ce{2H2S(g) + SO2(g) -> 3S(s) + 2H2O(g)} \]
Calculate the total number of electrons transferred per reaction as written.
Explain why two different sulfur-containing reactants are needed to account for the electron transfer.
Solution 3
Four electrons are transferred per reaction as written. Sulfur in hydrogen sulfide loses four electrons in total, while sulfur in sulfur dioxide gains four electrons.
Step 1: Find the initial oxidation states of sulfur.
In \(\ce{H2S}\), hydrogen has an oxidation state of +1.
\[ 2(+1)+x=0 \]
\[ x=-2 \]
In \(\ce{SO2}\), oxygen has an oxidation state of -2.
\[ x+2(-2)=0 \]
\[ x=+4 \]
Step 2: Find the final oxidation states.
Elemental sulfur has an oxidation state of 0.
Therefore:
- Sulfur from \(\ce{H2S}\) changes from -2 to 0.
- Sulfur from \(\ce{SO2}\) changes from +4 to 0.
Step 3: Calculate the electron changes.
There are two sulfur atoms in the two \(\ce{H2S}\) molecules.
Each changes from -2 to 0, losing two electrons:
\[ 2\times2=4\text{ electrons lost} \]
There is one sulfur atom in \(\ce{SO2}\).
It changes from +4 to 0, gaining four electrons:
\[ 1\times4=4\text{ electrons gained} \]
The electron totals match.
Step 4: Interpret the two sulfur-containing reactants.
This reaction involves two sulfur species with different initial oxidation states.
- Hydrogen sulfide supplies sulfur that is oxidised.
- Sulfur dioxide supplies sulfur that is reduced.
Both end up as elemental sulfur, but they reach that final state through opposite redox changes.
This is why simply looking at the three sulfur atoms in the products is not enough. You must trace their starting oxidation states to determine the electron transfer.
Question 4
Permanganate ions react with sulfite ions in weakly alkaline solution:
\[ \ce{2MnO4^- + 3SO3^2- + H2O -> 2MnO2(s) + 3SO4^2- + 2OH^-} \]
Determine:
a. The oxidation-state change of manganese.
b. The oxidation-state change of sulfur.
c. The total number of electrons transferred per reaction as written.
d. Whether the equation conserves atoms and charge.
A student claims that each permanganate ion must gain five electrons. Explain why this reasoning is incorrect.
Solution 4
Six electrons are transferred per reaction as written. Each manganese atom gains three electrons, while each sulfur atom loses two electrons.
Step 1: Calculate the oxidation state of manganese.
In \(\mathrm{MnO_{4}}^{-}\):
\[ x+4(-2)=-1 \]
\[ x=+7 \]
In \(\ce{MnO2}\):
\[ x+2(-2)=0 \]
\[ x=+4 \]
Each manganese atom changes from +7 to +4, gaining three electrons.
There are two manganese atoms:
\[ 2\times3=6\text{ electrons gained} \]
Step 2: Calculate the oxidation state of sulfur.
In \(\mathrm{SO_{3}}^{2-}\):
\[ x+3(-2)=-2 \]
\[ x=+4 \]
In \(\mathrm{SO_{4}}^{2-}\):
\[ x+4(-2)=-2 \]
\[ x=+6 \]
Each sulfur atom changes from +4 to +6, losing two electrons.
There are three sulfur atoms:
\[ 3\times2=6\text{ electrons lost} \]
The electron totals match.
Step 3: Check conservation of atoms.
| Element | Reactants | Products |
|---|---|---|
| Mn | 2 | 2 |
| S | 3 | 3 |
| O | 18 | 18 |
| H | 2 | 2 |
Step 4: Check conservation of charge.
Reactant side:
\[ 2(-1)+3(-2)=-8 \]
Product side:
\[ 3(-2)+2(-1)=-8 \]
Both sides have a total charge of -8.
Step 5: Explain the student’s mistake.
The claim that permanganate always gains five electrons assumes that manganese finishes at an oxidation state of +2.
That happens in acidic solution when permanganate is reduced to \(\mathrm{Mn}^{2+}\).
Here, however, the solution is weakly alkaline, and the product is \(\ce{MnO2}\), in which manganese has an oxidation state of +4.
The change is therefore +7 to +4, not +7 to +2.
The reaction medium determines the reduction product, and the reduction product determines the electron count.
Question 5
A student investigates the reaction between permanganate ions and sulfite ions under two different conditions.
Both reaction mixtures initially contain:
- 0.0100 mol of \(\mathrm{MnO_{4}}^{-}\)
- 0.0250 mol of \(\mathrm{SO_{3}}^{2-}\)
The student obtains the following results:
| Reaction | Conditions | Manganese-containing product |
|---|---|---|
| A | Excess acid | \(\mathrm{Mn}^{2+}\) ions |
| B | Weakly alkaline solution | \(\ce{MnO2(s)}\), a brown-black solid |
Assume that:
- All permanganate ions react.
- Sulfite is oxidised completely to sulfate ions in both reactions.
- The products listed are the only manganese-containing products.
- There are no competing reactions.
- Acid is in excess in reaction A, and water is available in excess in reaction B.
A student argues:
“Both mixtures start with the same amount of permanganate, so both must transfer the same amount of electrons. Each permanganate ion gains five electrons, regardless of the reaction conditions.”
Evaluate this claim.
For each reaction:
a. Determine the oxidation-state changes.
b. Write the balanced net ionic equation, including the appropriate balancing species.
c. Calculate the amount of electrons transferred in moles.
d. Determine whether any sulfite remains after the permanganate has completely reacted.
Solution 5
The student’s claim is incorrect. Reaction A transfers 0.0500 mol of electrons, while reaction B transfers only 0.0300 mol of electrons. The difference comes from the different reduction products of manganese.
The tempting approach is to assume that permanganate always changes from +7 to +2. That is only correct in reaction A.
In reaction B, manganese finishes at +4 in manganese dioxide. It therefore gains fewer electrons per atom.
Step 1: Calculate the oxidation-state changes common to both reactions.
For permanganate, manganese begins at +7.
For sulfite, sulfur begins at +4:
\[ x+3(-2)=-2 \]
\[ x=+4 \]
Sulfur finishes at +6 in sulfate:
\[ x+4(-2)=-2 \]
\[ x=+6 \]
Therefore, sulfur loses two electrons per atom in both reactions.
Step 2: Analyse reaction A in acidic solution.
Manganese changes from +7 in \(\mathrm{MnO_{4}}^{-}\) to +2 in \(\mathrm{Mn}^{2+}\).
Each manganese atom gains five electrons.
Sulfur changes from +4 to +6, losing two electrons per atom.
To make the total electron gain and loss equal, use the lowest common multiple of five and two, which is ten.
- Two permanganate ions gain ten electrons.
- Five sulfite ions lose ten electrons.
The balanced net ionic equation is:
\[ \ce{2MnO4^- + 5SO3^2- + 6H+ -> 2Mn^2+ + 5SO4^2- + 3H2O} \]
Step 3: Check the equation for reaction A.
The charge on the reactant side is:
\[ -2-10+6=-6 \]
The charge on the product side is:
\[ +4-10=-6 \]
The atoms are also conserved:
| Element | Reactants | Products |
|---|---|---|
| Mn | 2 | 2 |
| S | 5 | 5 |
| O | 23 | 23 |
| H | 6 | 6 |
The equation is balanced.
Step 4: Calculate the amount of sulfite consumed in reaction A.
The mole ratio is:
\[ \ce{MnO4^- : SO3^2-}=2:5 \]
Therefore:
\[ \begin{aligned} n(\mathrm{SO_{3}}^{2-}) &=0.0100\times\frac{5}{2}\\ &=0.0250\text{ mol} \end{aligned} \]
The mixture initially contained 0.0250 mol of sulfite.
All the sulfite is consumed.
The amount of electron transfer is:
\[ \begin{aligned} n(e^-) &=0.0100\times5\\ &=0.0500\text{ mol} \end{aligned} \]
Reaction A transfers 0.0500 mol of electrons.
Step 5: Analyse reaction B in weakly alkaline solution.
Manganese now changes from +7 to +4 in \(\ce{MnO2}\).
Each manganese atom gains only three electrons.
Sulfur still changes from +4 to +6, losing two electrons per atom.
The lowest common multiple of three and two is six.
- Two permanganate ions gain six electrons.
- Three sulfite ions lose six electrons.
The balanced net ionic equation is:
\[ \ce{2MnO4^- + 3SO3^2- + H2O -> 2MnO2 + 3SO4^2- + 2OH^-} \]
Step 6: Check the equation for reaction B.
The reactant-side charge is:
\[ -2-6=-8 \]
The product-side charge is:
\[ -6-2=-8 \]
The atoms are conserved:
| Element | Reactants | Products |
|---|---|---|
| Mn | 2 | 2 |
| S | 3 | 3 |
| O | 18 | 18 |
| H | 2 | 2 |
The equation is balanced.
Step 7: Calculate the amount of sulfite consumed in reaction B.
The mole ratio is now:
\[ \ce{MnO4^- : SO3^2-}=2:3 \]
Therefore:
\[ \begin{aligned} n(\mathrm{SO_{3}}^{2-}) &=0.0100\times\frac{3}{2}\\ &=0.0150\text{ mol} \end{aligned} \]
The initial amount was 0.0250 mol.
The remaining sulfite is:
\[ \begin{aligned} n_{\text{remaining}} &=0.0250-0.0150\\ &=0.0100\text{ mol} \end{aligned} \]
The amount of electron transfer is:
\[ \begin{aligned} n(e^-) &=0.0100\times3\\ &=0.0300\text{ mol} \end{aligned} \]
Reaction B transfers 0.0300 mol of electrons, leaving 0.0100 mol of sulfite unreacted.
Step 8: Compare the two results.
| Quantity | Reaction A: acidic | Reaction B: weakly alkaline |
|---|---|---|
| Initial \(\mathrm{MnO_{4}}^{-}\) | 0.0100 mol | 0.0100 mol |
| Mn final oxidation state | +2 | +4 |
| Electrons gained per Mn | 5 | 3 |
| Sulfite consumed | 0.0250 mol | 0.0150 mol |
| Electrons transferred | 0.0500 mol | 0.0300 mol |
| Sulfite remaining | 0 mol | 0.0100 mol |
The crucial constraint is the final oxidation state of manganese.
The same starting amount of permanganate does not guarantee the same electron transfer. You must know what product forms under the stated conditions before calculating the oxidation-state change.
The brown-black solid in reaction B is consistent with the formation of \(\ce{MnO2}\). It provides an observable clue that the manganese has been reduced to +4 rather than +2.
This is a useful exam habit: never assume a familiar oxidising agent always accepts the same number of electrons. Check the reaction medium and the actual product.
06The final check: have you counted the electrons correctly?
Before submitting an electron-transfer calculation, work through these checks:
- Correct formulas: Are the chemical formulas and ionic charges correct?
- Balanced atoms: Is every element present in equal numbers on both sides?
- Balanced charge: Do the reactant and product sides have equal total charge?
- Correct oxidation states: Have you calculated the oxidation states before and after the reaction?
- Correct atom count: Have you included both coefficients and subscripts?
- Equal electron totals: Do the electrons lost equal the electrons gained?
Remember the central calculation:
\[ \boxed{ \text{Electrons transferred} = \text{number of affected atoms} \times \text{oxidation-state change} } \]
Calculate this separately for oxidation and reduction. The two totals must agree.
For a deeper explanation of why electron loss and gain define oxidation and reduction, see Oxidation and Reduction by Electron Transfer: HSC Chemistry.
If your oxidation-state calculations are correct but your ionic equation does not balance, work through How to Check Charge Balance in a Redox Equation.
Once you can calculate the electron transfer reliably, the next step is using that number to balance redox half-equations. That is where the electron counts become coefficients, allowing oxidation and reduction to combine into one conserved overall reaction.