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When charge has to balance as well as atoms, and what the electrons in a half equation are counting.
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By the end of this lesson you will be able to balance an ionic equation so that the total charge matches on both sides as well as every element, and to write and balance a half equation with the right number of electrons on the right side of the arrow. You will be able to say why a spectator ion is left out, why the two totals have to be equal rather than zero, and how two half equations combine into one redox equation with the electrons cancelled.
You can balance an equation by counting atoms, in order, and finish it in lowest terms. This lesson adds one more thing that has to come out equal — charge — and one more kind of species that can appear in an equation: the electron.
| Term | What it means |
|---|---|
| Ion | An atom or group with a charge, written as a superscript: $\mathrm{Na^+}$, $\mathrm{SO_4^{2-}}$. |
| Ionic equation | An equation showing only the ions that take part, without the spectators. |
| Spectator ion | An ion that is dissolved before and after, unchanged. |
| Half equation | One side of an electron transfer, with the electrons written in. |
| Oxidation | Losing electrons. |
| Reduction | Gaining electrons. |
| Precipitate | A solid that appears when two solutions are mixed. |
Atoms are not the only thing a reaction cannot create. Charge is conserved as well, and for the same reason: the electrons that carry it go somewhere, they do not stop existing.
So an ionic equation has two conditions rather than one.
A half equation makes the electrons visible. When zinc dissolves it becomes $\mathrm{Zn^{2+}}$, and the two electrons it shed are written as a species:
$$\mathrm{Zn \rightarrow Zn^{2+} + 2e^-}.$$
Left: charge 0. Right: $+2$ from the ion and $-2$ from the two electrons, so 0 as well. The electron count is not decoration — it is the answer to the question the half equation asks.
Which side the electrons go on is decided by the chemistry. Lost electrons are written with the products, gained electrons with the reactants. Put them on the other side and the same numbers describe the reverse reaction.
Another way: picture
Think of two ledgers, one for atoms and one for charge. Balancing the atoms is the first ledger; the second is a separate column that also has to come out level, and the electrons are the entry you are allowed to add to make it do so.
Another way: steps
To balance a half equation:
For a precipitation. Write the two ions that meet and the solid they make. Balance the atoms of the solid's formula, which fixes how many of each ion you need. Then add up the charge on the left: it must come to zero, because the solid is neutral.
For a half equation. Balance the atoms of the element that is changing — this is usually one coefficient, sometimes a 2 for a diatomic element like iodine or bromine. Total the charge on each side without electrons. Add electrons to the more positive side until the totals agree: on the right for an oxidation, on the left for a reduction.
For a whole redox equation. Write the two half equations. Multiply each so that the electrons given equal the electrons taken — the smallest common multiple of the two electron counts. Add the halves together and cancel the electrons.
Check the work. Count every element on each side. Then add up the charge on each side, coefficient times charge, and confirm the two totals are equal. Finally, ask the direction question: did the charge on the changing species go up? Then electrons belong on the right.
Leaving out spectator ions is allowed because they appear identically on both sides; cancelling them is the same move as subtracting the same thing from both sides of an algebraic equation.
Adding electrons to a half equation is allowed because electrons are real particles that really moved. Nothing else may be added to fix a charge, because nothing else took part.
Putting lost electrons with the products is required, not chosen, because the equation reads left to right as a description of what happens. A species that leaves during the reaction is something produced.
Multiplying half equations before adding them is allowed for the same reason scaling any balanced equation is allowed: every count on both sides changes by the same factor. It is needed because electrons are never left over in a real reaction — every one given is taken.
Checking charge separately from atoms is necessary because the two conditions are independent. An equation can balance in atoms and fail in charge, and the atom count alone will not show it.
Mixing silver nitrate solution with sodium chloride solution gives a white solid. Written in full:
$$\mathrm{AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_3(aq)}.$$
But the sodium ions and the nitrate ions were dissolved before and are dissolved after. Nothing happened to them. They are spectator ions, and leaving them in hides what the reaction is:
$$\mathrm{Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)}.$$
That is the whole reaction, and it is the same equation whichever soluble silver salt and whichever soluble chloride you started from. The ionic equation is shorter because it is more general, not because it is an abbreviation.
Charge: $+1$ and $-1$ on the left, so zero; the solid is neutral, so zero. Balanced on both counts.
A metal displacing another from solution is two half equations happening at once. Copper in silver nitrate solution:
$$\mathrm{Cu \rightarrow Cu^{2+} + 2e^-} \qquad \mathrm{Ag^+ + e^- \rightarrow Ag}.$$
The copper sheds two electrons; each silver ion can take only one. So the second equation has to happen twice for every time the first one does, and doubling it lets the electrons cancel:
$$\mathrm{Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag}.$$
Charge: $+2$ on the left, $+2$ on the right. The electrons are gone from the finished equation, and the coefficient 2 is the trace they left behind. That is why a full redox equation balances for charge as well as atoms — it was built out of two equations that each did.
Aluminum dipped in copper sulfate solution slowly turns pink-brown as copper coats it. The two halves are
$$\mathrm{Al \rightarrow Al^{3+} + 3e^-} \qquad \mathrm{Cu^{2+} + 2e^- \rightarrow Cu}.$$
Aluminum gives three electrons; copper takes two. Neither number divides the other, so look for the smallest number both go into evenly: six. Six electrons means two aluminum atoms giving three each, and three copper ions taking two each:
$$\mathrm{2Al + 3Cu^{2+} \rightarrow 2Al^{3+} + 3Cu}.$$
Check the charge: $3 \times (+2) = +6$ on the left, $2 \times (+3) = +6$ on the right. Equal, as they must be, because the six electrons that moved from aluminum to copper were counted once on each side and then cancelled.
This is the same arithmetic as finding a common denominator, and it is the step learners most often skip — writing one of each and wondering why the charge will not balance. If the totals disagree after you have added the halves, the electron counts were never matched.
An ordinary AA alkaline battery, the kind sold by the billion in American stores every year, is a redox reaction split into its two halves and kept apart. At the negative terminal, zinc powder is oxidized:
$$\mathrm{Zn \rightarrow Zn^{2+} + 2e^-}.$$
At the positive terminal, manganese dioxide is reduced, taking those electrons back. The electrons cannot cross through the paste inside the battery, so the only way from one half to the other is out through the wire, through the remote, and back in at the other end. That trip is the electric current.
The half equation says exactly how much charge each zinc atom can supply: two electrons. A fresh AA cell holds a few grams of zinc — roughly $3 \times 10^{22}$ atoms in each gram, which is $6 \times 10^{22}$ electrons per gram. At the small current a remote draws, about 10 milliamps while a button is pressed, that is enough for many thousands of presses. When the zinc runs out, so do the electrons, and the battery is dead.
If the half equations were written with the electrons on the wrong side, they would describe a battery charging rather than discharging — the same numbers, the opposite chemistry. Engineers who design rechargeable batteries use exactly that reversal: forcing electrons back in drives each half equation the other way.
Water in much of the Midwest is hard, carrying dissolved calcium ions. A testing lab adds carbonate ions and watches for $\mathrm{Ca^{2+} + CO_3^{2-} \rightarrow CaCO_3}$: one of each, charges cancelling to zero, and the same white solid that builds up as scale inside a water heater. Weighing the solid tells the lab how much calcium the water carried, and the balanced one-to-one equation is what turns that mass into a count of ions.
Electrons on the wrong side. The commonest error in this lesson, and it changes nothing numerically, which is exactly why it survives. Ask whether the charge went up or down: up means electrons left, so they belong with the products.
Balancing charge with an invented ion. Adding a stray $\mathrm{H^+}$ or $\mathrm{Na^+}$ to make the totals agree balances the arithmetic and describes a reaction that did not happen. The only species you may add to fix a charge is the electron, because it is the thing that actually moved.
Expecting each side to come to zero. They have to be equal, not zero. $\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+}}$ has $+2$ on each side and is perfectly balanced.
Balancing charge before atoms. Atoms first, always. Changing a coefficient to fix the charge will usually break an element, and then you are chasing both ledgers at once.
Leaving electrons in a whole equation. If electrons survive in the finished redox equation, the two halves were not scaled to match. Electrons given and taken must be equal, so they always cancel completely.
Write the ions and solid.
$\mathrm{Pb^{2+} + I^- \rightarrow PbI_2}$
Spectators already left out.
Balance the iodine first.
$\mathrm{2I^-}$
Two iodines in the solid.
Check the lead atoms.
$1 = 1$
Every element now agrees.
Total the left charge.
$+2 + 2 \times (-1) = 0$
Charge checked as a separate question.
Write the equation.
$\mathrm{Pb^{2+} + 2I^- \rightarrow PbI_2}$
The solid is neutral, so both sides are zero.
Write the change.
$\mathrm{I^- \rightarrow I_2}$
Iodide becomes iodine.
Balance the iodine atoms.
$\mathrm{2I^- \rightarrow I_2}$
Two iodines in the molecule.
Total the left charge.
$2 \times (-1) = -2$
Coefficient times charge.
Total the right charge.
$0$
Iodine is a neutral molecule.
Read the direction.
$\text{left more negative: electrons left}$
The sign of the mismatch decides the side.
Add the electrons.
$\mathrm{2I^- \rightarrow I_2 + 2e^-}$
An oxidation, with $-2$ on each side.
Write the oxidation half.
$\mathrm{Zn \rightarrow Zn^{2+} + 2e^-}$
Zinc's charge goes up.
Write the reduction half.
$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$
Copper's charge comes down.
Compare the electron counts.
$2 = 2$
No scaling needed this time.
Add the two halves.
$\mathrm{Zn + Cu^{2+} + 2e^- \rightarrow Zn^{2+} + Cu + 2e^-}$
Everything from both sides.
Cancel the electrons.
$\mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu}$
Given equals taken.
Check the atoms.
$\text{one Zn, one Cu each side}$
Atoms balance.
Check the charge.
$+2 = +2$
Equal, and not zero.
Balance the atoms first.
$\mathrm{Al \rightarrow Al^{3+}}$
One aluminum on each side already.
Compare the charges.
$0 \text{ against } +3$
The charge went up, so electrons left.
Add the electrons.
In aluminum at the anode of an electrolysis cell, $\mathrm{Al}$ turns into $\mathrm{Al^{3+}}$. Its charge has gone up. What has to appear in the half equation, and where?
Complete the worked solution: an aluminum strip is dipped in copper sulfate solution. Each aluminum atom gives up three electrons, and each copper ion takes two. Find the smallest number of electrons both halves can share, then the aluminum atoms and the copper ions that exchange it.
Find the shared electron count.
$\text{the smallest multiple of three and of two} =$ e
Electrons given must equal electrons taken.
Find the aluminum atoms.
$(\text{shared electrons}) \div \text{three} =$ a
Each atom gives three.
Find the copper ions.
$(\text{shared electrons}) \div \text{two} =$ c
Each ion takes two.
Two solutions are mixed and silver chloride settles out as a solid. Only the ions that take part are written. Balance the equation so that both the atoms and the total charge match.
This task has no paper form; do it on a device.
Here is bromide ions in seawater: $\mathrm{Br^-}$ loses electrons and becomes $\mathrm{Br_2}$. The electrons are written as a species of their own, on the right, because they are leaving. Balance the half equation.
This task has no paper form; do it on a device.
A plating shop in Ohio coats steel parts with copper: $\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$. If $9$ billion copper atoms are deposited on a part, how many billion electrons did the power supply deliver?
The answer: a.
Now both halves at once: bromine water added to potassium iodide solution. The electrons never appear, because they pass straight from one species to the other — but they are still what decides the coefficients. Balance the equation so that both the atoms and the total charge match.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
This is the other half: bromine taken out of solution. Electrons are being gained, so they are written on the left, among the reactants. Balance it.
This task has no paper form; do it on a device.
You can balance atoms and charge together, and you know which side the electrons go on and why. Say what the half equation for zinc dissolving looks like, and what would change about the chemistry if the electrons were written on the other side. Next: the mole, which is what turns any of these equations into a mass you can weigh.
16. Your turn: write the half equation for aluminum becoming $\mathrm{Al^{3+}}$., step 3
$\mathrm{Al \rightarrow Al^{3+} + 3e^-}$
Both sides now come to zero.