Redox Reaction Balancer: Acidic and Basic | StatsUnlock

To balance a redox reaction, split it into an oxidation half-reaction and a reduction half-reaction. In each half, balance the atoms other than O and H, then O with H₂O, then H with H⁺, then charge with electrons. Multiply the halves so the electrons lost equal the electrons gained, add them and cancel. Permanganate and iron(II) in acidic solution:

Rule of thumb: electrons never appear in the final equation, and both atoms and total charge must match (+17 on each side here). In basic solution, add one OH⁻ to both sides for every H⁺, combine them into water and cancel: 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻.

A redox reaction is a reaction in which electrons move from one species to another. The species that loses electrons is oxidised and acts as the reducing agent; the species that gains them is reduced and acts as the oxidising agent. A balanced redox equation must conserve three things at once: every kind of atom, the total charge, and the electrons (lost = gained).

A redox reaction balancer does that bookkeeping for you. It reads the skeleton equation, assigns oxidation numbers to every atom, works out which element is oxidised and which is reduced, splits the reaction into two half-reactions, balances each one with H₂O, H⁺ (or OH⁻) and electrons, equalises the electrons, adds the halves and cancels. Every answer is then checked three ways: atoms, charge and an independent electron count.

A worked one-liner: copper dissolves in nitric acid as Cu + NO₃⁻ → Cu²⁺ + NO. Copper loses 2 e⁻ and nitrogen (+5 → +2) gains 3 e⁻; the lowest common multiple is 6, so 3 Cu and 2 NO₃⁻ react: 3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO + 4H₂O.

Who uses it: balancing redox reactions is taught in NCERT Class 11 (“Redox Reactions”, including the oxidation number method and the half reaction method), tested in JEE, NEET, AP Chemistry, A level, IB and the MCAT, and used every day in titrations (permanganate, dichromate, iodine–thiosulfate), electrochemistry, corrosion, batteries and water treatment.

What this free redox reaction balancer solves

One tool covers every kind of redox balancing in a typical course: balancing redox reactions in acidic solution, balancing redox reactions in basic solution, neutral solution, the oxidation number method, the ion electron method, disproportionation, reactions with three reactants, single half-reactions, and checking an answer you already wrote. Press Try it to load an example, then press the button to see the working.

Free, with no sign-up. Everything runs in your browser: nothing you type is sent to a server or stored.

En español: ¿cómo se balancea una reacción redox? Separa la oxidación y la reducción, balancea en cada semirreacción los átomos, el oxígeno con H₂O, el hidrógeno con H⁺ y la carga con electrones; iguala los electrones y suma: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.

How accurate is this redox reaction balancer?

Coefficients are exact whole numbers from integer arithmetic, so there is no rounding. Oxidation numbers follow the IUPAC rules (Karen, McArdle & Takats, 2016). Every answer must pass three checks before it gets the Verified badge: atoms and charge are equal on both sides, the electrons lost equal the electrons gained, and an independent electron count from the oxidation numbers gives the same number. Where a plain matrix balance also has a single answer, it is compared as well (Alberty, 1991). Standard potentials are tabulated values at 25 °C (Bratsch, 1989); different tables can differ by a few hundredths of a volt. The R and Python scripts in Sections 10 and 11 return the same MnO₄⁻ + 5Fe²⁺ + 8H⁺ balance and E°cell = +0.739 V.

Studies of how students learn redox find that most errors come from the conventions, not the arithmetic: adding H⁺ in basic solution, forgetting to balance charge, or confusing the oxidising agent with the species that is oxidised (Garnett & Treagust, 1992; Brandriet & Bretz, 2014). That is why the results label each half and each agent.

4.1 Oxidation, reduction and OIL RIG

Oxidation Is Loss of electrons and Reduction Is Gain. In terms of oxidation numbers, an element that is oxidised goes up (Fe +2 → +3) and an element that is reduced goes down (Mn +7 → +2). Oxidation and reduction always happen together: the electrons lost by one species are gained by another, which is why a redox equation is split into two half-reactions and then put back together.

4.2 Oxidising agent and reducing agent

The oxidising agent is the reactant that takes electrons, so it is itself reduced (MnO₄⁻, Cr₂O₇²⁻, Cl₂, H₂O₂, NO₃⁻). The reducing agent gives electrons away and is oxidised (Fe²⁺, I⁻, C₂O₄²⁻, metals such as Zn and Cu). Name the whole species, not just the element: the oxidising agent is MnO₄⁻, not Mn.

4.3 Electrons transferred (n)

The number n is the electrons that move in the balanced equation as written: the lowest common multiple of the electrons in the two half-reactions, divided by any common factor at the end. It is 5 for MnO₄⁻ + 5Fe²⁺, 6 for dichromate and iron(II), and 10 for 2MnO₄⁻ + 5C₂O₄²⁻. You need n for ΔG° = −nFE°, the Nernst equation and Faraday’s laws of electrolysis.

4.4 Acidic, basic and neutral solution

The same skeleton can give a different balanced equation in a different medium. Permanganate with iodide gives MnO₂ in basic solution (2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻) but Mn²⁺ in acid, so the products themselves depend on the medium.

4.5 Disproportionation and comproportionation

In a disproportionation one species is both oxidised and reduced: chlorine in hot base gives chloride (−1) and chlorate (+5), 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O, and hydrogen peroxide decomposes as 2H₂O₂ → O₂ + 2H₂O. In a comproportionation two forms meet in the middle: IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O. The balancer puts the shared species in both half-reactions and adds the amounts.

4.6 E°cell, ΔG° and the equilibrium constant

When both couples are in the built-in table of standard reduction potentials, the results add E°cell = E°cathode − E°anode, where the cathode is the reduction half. A positive E°cell means the reaction is spontaneous under standard conditions; ΔG° = −nFE° and log K = nE° ÷ 0.05916 at 25 °C. For permanganate and iron(II), 1.51 − 0.771 = +0.739 V, ΔG° = −356.5 kJ/mol and K ≈ 10⁶²: the reaction goes to completion, which is why permanganate titrations of iron work.

4.7 Sum of coefficients

Multiple-choice questions often ask for the sum of the coefficients in the balanced equation. The results table gives it, counting H⁺, OH⁻ and H₂O but not electrons: for MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O it is 14 + 10 = 24.

4.8 A field note on where redox answers go wrong

Marking redox equations, the same slips come up: putting electrons on the wrong side, balancing atoms but not charge, writing H⁺ in an answer for basic solution, multiplying only one species of a half-reaction instead of the whole half, and leaving electrons in the final equation. Some textbooks add to the confusion by mixing the oxygen, hydrogen and electron definitions of redox (Österlund, Berg & Ekborg, 2010), so the step list always says which rule is being used.

▶ Balance a reaction above to fill all five cards with your own answer.

Balancing redox reactions rests on three conservation rules and a handful of equations. The cards below give each one with a worked number from the permanganate and iron(II) example.

This flowchart follows the half reaction method on the same problem used in the Excel, R, Python and by-hand sections: balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic solution.

Check any box yourself: the Excel sheet in Section 9, the R script in Section 10 and the Python script in Section 11 all return this equation and E°cell = +0.739 V.

1. Choose what you want to do. Balance a full redox equation, balance one half-reaction, or check a balanced equation you already wrote.
2. Type the skeleton equation. Write MnO4- + Fe2+ -> Mn2+ + Fe3+ with charges after each formula; the preview shows every species with its oxidation numbers.
3. Leave out H⁺, OH⁻ and H₂O. The balancer adds them in the right amounts; if you type them, they are used to choose the medium.
4. Pick the medium. Choose acidic, basic or neutral solution, or leave it on automatic (acidic unless you typed OH⁻; molecular equations are balanced as written).
5. Pick the method. The half-reaction (ion-electron) method or the oxidation-number method; both give the same equation.
6. Or load a sample. Sixteen samples cover acidic, basic and neutral solution, disproportionation, three reactants, the oxidation-number method and answer checking.
7. Press the button. Nothing is calculated before you press it, and changing the input clears the old answer.
8. Read the half-reactions and the steps. The Verified badge means atoms, charge and electrons all balance.
9. Read E°cell when it is shown. It appears when both couples are in the table of standard reduction potentials.
10. Copy, share or batch. Copy the equation, share a link, download a report, or balance a whole worksheet on the Batch tab.

Excel cannot read a chemical equation, but it is ideal for the electron bookkeeping of the oxidation number method: the change in oxidation number, the electrons per formula, the lowest common multiple, the multipliers, an atom-and-charge check with SUMPRODUCT, and E°cell, ΔG° and K. The sheet uses the headline problem, so it must return 5 Fe²⁺ per MnO₄⁻ and E°cell = +0.739 V, the same as the redox reaction balancer above.

9.1 The functions you need

9.2 Step by step

9.3 Awkward cases

• Two elements oxidised in one formula (As₂S₃, FeS₂): add one row per element and SUM them for the electrons per formula: As₂S₃ loses 2 × 2 + 3 × 8 = 28.
• Fractional oxidation numbers (Fe₃O₄, S₄O₆²⁻): work with the total for the formula, not per atom, so the electrons come out whole.
• Disproportionation: count the electrons per product (ClO₃⁻ 5, Cl⁻ 1) rather than per reactant.

9.4 Why your Excel answer might differ

9.5 Charting in Excel

1. Next to the atom-check table, keep the reactant totals in one column and the product totals in the next.
2. Insert → Clustered Column; every pair of bars must be equal, as in chart 1 above.

9.6 Excel errors you will probably hit

This base R script writes the composition of every species (including H⁺ and H₂O for acidic solution) as a matrix with one row per element and one for charge, finds the null space with svd, scales it to whole numbers, counts the electrons with the lowest common multiple, and gives E°cell, ΔG° and log K. For the headline problem it prints MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O and +0.739 V, matching the redox reaction balancer on this page.

10.1 The complete script

R, base only, saves redox_balance.png

# ============================================================
# Redox reaction balancer in R (base R only):
# balance atoms + charge, count electrons, E°cell, ΔG° and K
# StatsUnlock - https://statsunlock.com/redox-reaction-balancer/
# ============================================================

# ---- 1. Species: element counts, charge and side (MnO4- + Fe2+ in acid) ----
sp <- data.frame(
  name   = c("MnO4-", "Fe2+", "H+", "Mn2+", "Fe3+", "H2O"),
  Mn     = c(1, 0, 0, 1, 0, 0),
  Fe     = c(0, 1, 0, 0, 1, 0),
  O      = c(4, 0, 0, 0, 0, 1),
  H      = c(0, 0, 1, 0, 0, 2),
  charge = c(-1, 2, 1, 2, 3, 0),
  side   = c(1, 1, 1, -1, -1, -1)       # 1 = reactant, -1 = product
)

# ---- 2. Matrix: one row per element plus charge; its null space is the balance
A <- t(as.matrix(sp[, c("Mn", "Fe", "O", "H", "charge")]) * sp$side)
v <- svd(A, nv = ncol(A))$v[, ncol(A)]           # vector with A %*% v = 0
v <- v / min(abs(v[abs(v) > 1e-9]))               # smallest coefficient = 1
for (k in 1:12) if (all(abs(v * k - round(v * k)) < 1e-6)) { v <- round(v * k); break }
if (v[1] < 0) v <- -v
coef <- v
side_txt <- function(s) paste(ifelse(coef[sp$side == s] == 1, "", coef[sp$side == s]), sp$name[sp$side == s], sep = "", collapse = " + ")
cat("Balanced:", side_txt(1), "->", side_txt(-1), "\n")

# ---- 3. Electrons: Fe +2 -> +3 loses 1, Mn +7 -> +2 gains 5 ----------------
lost <- 1; gained <- 5
gcd <- function(a, b) if (b == 0) a else gcd(b, a %% b)
n <- lost * gained / gcd(lost, gained)             # lowest common multiple
cat(sprintf("Oxidation x %d, reduction x %d, n = %d electrons\n", n / lost, n / gained, n))

# ---- 4. E°cell, ΔG° and K at 25 °C ---------------------------------------------
E_cell <- 1.51 - 0.771                             # E°(MnO4-/Mn2+) - E°(Fe3+/Fe2+)
dG <- -n * 96485.33212 * E_cell / 1000             # kJ/mol
logK <- n * E_cell / 0.05916
cat(sprintf("E°cell = %.3f V, ΔG° = %.1f kJ/mol, log K = %.1f\n", E_cell, dG, logK))

# ---- 5. Checks -------------------------------------------------------------
stopifnot(all(coef == c(1, 5, 8, 1, 5, 4)),
          all(A %*% coef == 0),                    # atoms and charge balance
          round(E_cell, 3) == 0.739, round(dG, 1) == -356.5)
cat("Verdict: Verified (all checks pass)\n")

# ---- 6. Figure: atoms and charge on each side --------------------------------
lhs <- colSums(sp[sp$side == 1, c("Mn", "Fe", "O", "H", "charge")] * coef[sp$side == 1])
rhs <- colSums(sp[sp$side == -1, c("Mn", "Fe", "O", "H", "charge")] * coef[sp$side == -1])
png("redox_balance.png", width = 2100, height = 1350, res = 300)
par(mar = c(4, 4.5, 3, 1))
barplot(rbind(lhs, rhs), beside = TRUE, col = c("#2563eb", "#16a34a"), legend.text = c("reactants", "products"),
        main = "Balanced redox equation: every pair matches", ylab = "count")
invisible(dev.off())
cat("Saved redox_balance.png\n")

Output: Balanced: MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O, Oxidation x 5, reduction x 1, n = 5 electrons, E°cell = 0.739 V, ΔG° = -356.5 kJ/mol, log K = 62.5, Verdict: Verified.

10.2 R traps

• The matrix method alone cannot fix reactions with two independent balances (MnO₄⁻ with H₂O₂); add the electron ratio as an extra row, or use the half-reaction method.
• svd returns a unit vector with decimals; divide by the smallest non-zero entry and multiply up until every value is whole.
• Multiply each row by the side (+1 or −1) so products count as negative.

10.3 What each part does

10.4 The figure

The script saves redox_balance.png at 2100 × 1350 pixels and 300 dpi: atoms of each element and the total charge on the reactant and product sides, which must match pair by pair.

This script uses SymPy’s exact nullspace, so the coefficients are whole numbers with no rounding: one row per element plus a charge row, H⁺ and H₂O added for acidic solution. It then counts the electrons, calculates E°cell, ΔG° and K, checks the answer and saves a 300 dpi chart. It is the same matrix check the redox reaction balancer runs in your browser.

Install once: pip install sympy matplotlib

11.1 The complete script

Python, saves redox_balance.png

# ============================================================
# Redox reaction balancer in Python (half-reaction method check):
# balance atoms + charge exactly, count electrons, E°cell, ΔG° and K
# StatsUnlock - https://statsunlock.com/redox-reaction-balancer/
# pip install sympy matplotlib
# ============================================================
from math import lcm
from sympy import Matrix, ilcm, igcd
import matplotlib
matplotlib.use("Agg")
import matplotlib.pyplot as plt

# ---- 1. Species: element counts and charge (MnO4- + Fe2+ in acid) --------
#  side +1 = reactant, -1 = product; H+ and H2O are added for acidic solution
species = [
    ("MnO4-", {"Mn": 1, "O": 4}, -1, +1),
    ("Fe2+",  {"Fe": 1},          2, +1),
    ("H+",    {"H": 1},           1, +1),
    ("Mn2+",  {"Mn": 1},          2, -1),
    ("Fe3+",  {"Fe": 1},          3, -1),
    ("H2O",   {"H": 2, "O": 1},   0, -1),
]

# ---- 2. One row per element plus one for charge; solve A x = 0 exactly ----
elements = sorted({e for _, c, _, _ in species for e in c})
rows = [[c.get(e, 0) * s for _, c, _, s in species] for e in elements]
rows.append([q * s for _, _, q, s in species])          # charge balance
basis = Matrix(rows).nullspace()
assert len(basis) == 1, "more than one balance: give the half-reactions"
v = basis[0]
v = v * ilcm(*[x.q for x in v])                            # clear fractions
g = abs(igcd(*[int(x) for x in v]))
coef = [int(x) // g for x in v]
if coef[0] < 0:
    coef = [-x for x in coef]

def side(sign):
    return " + ".join(f"{'' if k == 1 else k}{n}" for (n, _, _, s), k in zip(species, coef) if s == sign)
print("Balanced:", side(+1), "->", side(-1))

# ---- 3. Electrons: Fe +2 -> +3 loses 1 e-, Mn +7 -> +2 gains 5 e- --------
lost, gained = 1, 5
n = lcm(lost, gained)                                       # electrons transferred
print(f"Oxidation x {n // lost}, reduction x {n // gained}, n = {n} e-")

# ---- 4. Standard potentials (V), E°cell, ΔG° and K at 25 °C --------------
E_cathode, E_anode = 1.51, 0.771                            # MnO4-/Mn2+ and Fe3+/Fe2+
E_cell = E_cathode - E_anode
F = 96485.33212                                             # C/mol (CODATA 2018)
dG = -n * F * E_cell / 1000                                 # kJ/mol
logK = n * E_cell / 0.05916
print(f"E°cell = {E_cell:.3f} V, ΔG° = {dG:.1f} kJ/mol, K = 10^{logK:.1f}")

# ---- 5. Checks ------------------------------------------------------------
assert coef == [1, 5, 8, 1, 5, 4]
assert round(E_cell, 3) == 0.739 and round(dG, 1) == -356.5
print("Verdict: Verified (all checks pass)")

# ---- 6. Figure: atoms and charge on each side -----------------------------
labels = elements + ["charge"]
left = [sum(k * c.get(e, 0) for (_, c, _, s), k in zip(species, coef) if s == 1) for e in elements]
right = [sum(k * c.get(e, 0) for (_, c, _, s), k in zip(species, coef) if s == -1) for e in elements]
left.append(sum(k * q for (_, _, q, s), k in zip(species, coef) if s == 1))
right.append(sum(k * q for (_, _, q, s), k in zip(species, coef) if s == -1))
x = range(len(labels))
fig, ax = plt.subplots(figsize=(7, 4.5))
ax.bar([i - 0.2 for i in x], left, 0.4, label="reactants", color="#2563eb")
ax.bar([i + 0.2 for i in x], right, 0.4, label="products", color="#16a34a")
ax.set_xticks(list(x)); ax.set_xticklabels(labels)
ax.set_title("Balanced redox equation: every pair matches"); ax.legend()
fig.tight_layout(); fig.savefig("redox_balance.png", dpi=300)
print("Saved redox_balance.png")

Output: Balanced: MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O, Oxidation x 5, reduction x 1, n = 5 e-, E°cell = 0.739 V, ΔG° = -356.5 kJ/mol, K = 10^62.5, Verdict: Verified.

11.2 Python traps

• len(basis) != 1 means the species can react in more than one way; give the half-reactions or the electron ratio.
• NumPy’s floating-point null space gives 0.0909 instead of 1/11; SymPy keeps exact fractions.
• math.lcm needs Python 3.9 or later.

11.3 Packages

pip install chempy balances ordinary equations with balance_stoichiometry; for redox in acidic or basic solution, add H⁺ or OH⁻ and H₂O to the species yourself, as this script does.

11.4 What each part does

11.5 The figure

The script saves redox_balance.png at 300 dpi: reactant and product totals for every element and for charge.

12.1 Standard reduction potentials and balanced half-reactions

Typical tabulated values at 25 °C, 1 mol/L and 1 bar, from strongest oxidising agent to strongest reducing agent (Bratsch, 1989; tables differ by a few hundredths of a volt). Each half-reaction was balanced by the same engine as the redox reaction balancer; these are the couples it uses for E°cell.

12.2 Balanced redox reaction examples in acidic and basic solution

Every row was balanced by the redox reaction balancer and passed the atom, charge and electron checks.

12.3 Balancing redox reactions worksheet with answers

Twenty-five questions in three groups: half-reactions, redox reactions in acidic solution, and redox reactions in basic or neutral solution and by the oxidation number method. Every answer was checked by the balancer. Cover the last column and try them first.

Eight balanced redox equations run through the balancer, from a titration in acid to a disproportionation in base. Each figure is drawn from the computed numbers: the electrons lost in the oxidation half (orange) and gained in the reduction half (blue) before and after they are made equal.

This is the written method examiners expect, on the same problem as Sections 7, 9, 10 and 11: balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic solution.

14.1 Steps to balance a redox reaction (half-reaction method)

1. Assign oxidation numbers and find what changes: Mn +7 → +2 (reduced), Fe +2 → +3 (oxidised).
2. Separate the redox reaction into its component half-reactions: Fe²⁺ → Fe³⁺ and MnO₄⁻ → Mn²⁺.
3. Balance the atoms other than O and H in each half (here they already balance).
4. Balance O by adding H₂O: MnO₄⁻ → Mn²⁺ + 4H₂O.
5. Balance H by adding H⁺: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O.
6. Balance charge by adding electrons to the more positive side: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and Fe²⁺ → Fe³⁺ + e⁻.
7. Make the electrons equal: multiply the iron half by 5.
8. Add the halves and cancel anything on both sides, including the 5 electrons.
9. Basic solution only: add OH⁻ to both sides for every H⁺, combine into water and cancel.
10. Check atoms, charge (+17 = +17) and that the coefficients are the smallest whole numbers.

14.2 How to balance redox reactions in acidic solution

Dichromate oxidising iron(II) is the classic “complete and balance the following redox reaction in acidic solution” question. Chromium goes from +6 to +3, and there are two Cr atoms, so the reduction half takes 6 electrons:

14.3 How to balance redox reactions in basic solution

Balance as if in acid, then neutralise every H⁺. For permanganate and iodide giving MnO₂ and I₂: the halves are 2I⁻ → I₂ + 2e⁻ (×3) and MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O (×2). The acidic sum is 2MnO₄⁻ + 6I⁻ + 8H⁺ → 2MnO₂ + 3I₂ + 4H₂O. Add 8OH⁻ to both sides; the 8H⁺ + 8OH⁻ become 8H₂O, and 4 cancel:

A shortcut that gives the same answer: in basic solution, for each O atom a side needs, add 2OH⁻ to that side and one H₂O to the other; for each H atom a side needs, add one H₂O to that side and one OH⁻ to the other. Either way, the final equation contains OH⁻ and never H⁺.

14.4 Balancing redox reactions by the oxidation number method

The oxidation number method works directly on molecular equations with spectator ions, the usual Class 11 example being potassium permanganate with hydrochloric acid:

1. Oxidation numbers: Mn +7 in KMnO₄ → +2 in MnCl₂ (gains 5 e⁻); Cl −1 in HCl → 0 in Cl₂ (2 e⁻ lost per Cl₂).
2. Cross-multiply: LCM(5, 2) = 10, so 2 KMnO₄ and 5 Cl₂.
3. Balance the rest by inspection: 2 K → 2KCl; 2 Mn → 2MnCl₂; Cl on the right = 2 + 4 + 10 = 16, so 16 HCl; 16 H → 8H₂O, which also balances the 8 O.

Only 10 of the 16 HCl are oxidised; the other 6 supply chloride to KCl and MnCl₂. That double role is why the oxidation number method handles molecular equations better than the half-reaction method.

14.5 Balancing redox reactions in neutral solution

In neutral solution, water is the reactant, and H⁺ or OH⁻ can only be formed. Permanganate oxidising iodide to iodate gives 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻: the solution becomes slightly basic as the reaction runs.

14.6 Disproportionation

For Cl₂ → Cl⁻ + ClO₃⁻ in base, write Cl₂ in both halves: Cl₂ + 2e⁻ → 2Cl⁻ (×5) and Cl₂ + 12OH⁻ → 2ClO₃⁻ + 6H₂O + 10e⁻. Adding gives 6Cl₂ + 12OH⁻ → 10Cl⁻ + 2ClO₃⁻ + 6H₂O; divide by 2: 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O.

14.7 Balancing complex redox reactions with 3 reactants

When two elements in one formula are oxidised, count both. In As₂S₃ + NO₃⁻ → H₃AsO₄ + SO₄²⁻ + NO, As goes +3 → +5 (2 atoms, 4 e⁻) and S goes −2 → +6 (3 atoms, 24 e⁻): 28 e⁻ per As₂S₃. Nitrogen gains 3, so LCM = 84: 3 As₂S₃ and 28 NO₃⁻, giving 3As₂S₃ + 28NO₃⁻ + 10H⁺ + 4H₂O → 6H₃AsO₄ + 9SO₄²⁻ + 28NO.

14.8 Do it in the lab: permanganate titration of iron(II)

Example data. Using the wrong ratio (1 : 1 instead of 1 : 5) would give 0.0195 mol/L, five times too low: the balanced redox equation is the whole calculation.

14.9 Balancing redox reactions practice problems with answers

Ten practice problems, easy to hard. Open each one to see the half-reactions and the answer.

14.10 Balancing redox reactions rules and mistakes that lose marks

Research on electrochemistry and redox learning finds that most lost marks come from the conventions, not the arithmetic (Garnett & Treagust, 1992; Rahayu et al., 2011).

• Leaving electrons in the final equation, or putting them on the wrong side of a half-reaction.
• Balancing atoms but not charge.
• Writing H⁺ in an answer for basic solution.
• Multiplying only one species when scaling a half-reaction.
• Counting electrons per atom instead of per formula (Cr₂O₇²⁻ takes 6, not 3).
• Changing a subscript (writing Fe₂ to balance iron) instead of a coefficient.
• Calling the element (Mn) the oxidising agent instead of the species (MnO₄⁻).

Four real examples

Decision tree

1. Does any oxidation number change? No: it is not redox; balance it by inspection. Yes: go on.
2. Is it written with ions (MnO₄⁻, Fe²⁺)? Yes: the half-reaction method, in the medium the question gives.
3. Is it a molecular equation with spectator ions (KMnO₄ + HCl)? Yes: the oxidation-number method.
4. Only one change (Cr₂O₇²⁻ → Cr³⁺)? Use “Balance one half-reaction”.
5. Already balanced? Use “Check my balanced equation”.

Limitations. A redox reaction balancer balances the reaction you give it; it does not predict products. Oxidation numbers are a bookkeeping device, not real charges, and for some species (azide, superoxide, mixed-valence oxides, organic carbons) they are averages (Karen, 2015; Woolf, 1988). The electron count is still correct because it uses totals per formula. Standard potentials describe thermodynamics only: a reaction with a positive E°cell can still be too slow to see.

What the answer is

A balanced redox equation conserves atoms, charge and electrons. For permanganate and iron(II) in acid it is MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, with 5 electrons transferred.

What it means chemically

The coefficients are the mole ratios for titrations and electrochemistry, n links the equation to ΔG° and the Nernst equation, and the two half-reactions are the electrode reactions of a cell.

Watch out for

Balance charge as well as atoms, keep electrons out of the final equation, use OH⁻ (never H⁺) in basic solution, and count electrons per formula, not per atom.

Next steps

Practise with the worksheet in Section 12.3 and the ten problems in Section 14.9, check your own answers with Check mode, and use the balancer for your next titration calculation.

Balancing is linear algebra, and comparing titration results is statistics; these tools cover both.

Peer-reviewed sources for the methods and data used in this redox reaction balancer: the IUPAC definition of oxidation state, standard electrode potentials, the Faraday constant, algebraic and inspection methods of balancing, and research on how students learn redox.

1. Österlund, L.-L., Berg, A., & Ekborg, M. (2010). Redox models in chemistry textbooks for the upper secondary school: Friend or foe? Chemistry Education Research and Practice, 11(3), 182–192. https://doi.org/10.1039/C005467B
2. Alberty, R. A. (1991). Chemical equations are actually matrix equations. Journal of Chemical Education, 68(12), 984. https://doi.org/10.1021/ed068p984
3. Blakley, G. R. (1982). Chemical equation balancing: A general method which is quick, simple, and has unexpected applications. Journal of Chemical Education, 59(9), 728–734. https://doi.org/10.1021/ed059p728
4. Brandriet, A. R., & Bretz, S. L. (2014). The development of the Redox Concept Inventory as a measure of students’ symbolic and particulate redox understandings and confidence. Journal of Chemical Education, 91(8), 1132–1144. https://doi.org/10.1021/ed500051n
5. Bratsch, S. G. (1989). Standard electrode potentials and temperature coefficients in water at 298.15 K. Journal of Physical and Chemical Reference Data, 18(1), 1–21. https://doi.org/10.1063/1.555839
6. Calzaferri, G. (1999). Oxidation numbers. Journal of Chemical Education, 76(3), 362. https://doi.org/10.1021/ed076p362
7. Garnett, P. J., & Treagust, D. F. (1992). Conceptual difficulties experienced by senior high school students of electrochemistry: Electric circuits and oxidation-reduction equations. Journal of Research in Science Teaching, 29(2), 121–142. https://doi.org/10.1002/tea.3660290204
8. Herndon, W. C. (1997). On balancing chemical equations: Past and present. Journal of Chemical Education, 74(11), 1359–1362. https://doi.org/10.1021/ed074p1359
9. Jensen, W. B. (2007). The origin of the oxidation-state concept. Journal of Chemical Education, 84(9), 1418. https://doi.org/10.1021/ed084p1418
10. Karen, P. (2015). Oxidation state, a long-standing issue! Angewandte Chemie International Edition, 54(16), 4716–4726. https://doi.org/10.1002/anie.201407561
11. Karen, P., McArdle, P., & Takats, J. (2014). Toward a comprehensive definition of oxidation state (IUPAC Technical Report). Pure and Applied Chemistry, 86(6), 1017–1081. https://doi.org/10.1515/pac-2013-0505
12. Karen, P., McArdle, P., & Takats, J. (2016). Comprehensive definition of oxidation state (IUPAC Recommendations 2016). Pure and Applied Chemistry, 88(8), 831–839. https://doi.org/10.1515/pac-2015-1204
13. Kolb, D. (1978). The chemical equation. Part I: Simple reactions. Journal of Chemical Education, 55(3), 184–189. https://doi.org/10.1021/ed055p184
14. Olson, J. A. (1997). An analysis of the algebraic method for balancing chemical reactions. Journal of Chemical Education, 74(5), 538–542. https://doi.org/10.1021/ed074p538
15. Rahayu, S., Treagust, D. F., Chandrasegaran, A. L., Kita, M., & Ibnu, S. (2011). Assessment of electrochemical concepts: A comparative study involving senior high-school students in Indonesia and Japan. Research in Science & Technological Education, 29(2), 169–188. https://doi.org/10.1080/02635143.2010.536949
16. Smith, W. R., & Missen, R. W. (1997). Using Mathematica and Maple to obtain chemical equations. Journal of Chemical Education, 74(11), 1369–1371. https://doi.org/10.1021/ed074p1369
17. Tiesinga, E., Mohr, P. J., Newell, D. B., & Taylor, B. N. (2021). CODATA recommended values of the fundamental physical constants: 2018. Reviews of Modern Physics, 93(2), 025010. https://doi.org/10.1103/RevModPhys.93.025010
18. Tóth, Z. (1997). Balancing chemical equations by inspection. Journal of Chemical Education, 74(11), 1363–1364. https://doi.org/10.1021/ed074p1363
19. Woolf, A. A. (1988). Oxidation numbers and their limitations. Journal of Chemical Education, 65(1), 45. https://doi.org/10.1021/ed065p45
20. Yarroch, W. L. (1985). Student understanding of chemical equation balancing. Journal of Research in Science Teaching, 22(5), 449–459. https://doi.org/10.1002/tea.3660220507