Balancing Redox: Basic — Balance redox reactions in basic solution - Chemistry 2

SECTION 1

Historical Context & Motivation

Chemical reactions that involve the transfer of electrons are among the most important processes in both nature and industry. From rusting iron to the batteries that power your phone, redox reactions (short for reduction-oxidation reactions) drive countless real-world phenomena. Scientists spent centuries developing tools to understand and describe these reactions, eventually creating systematic methods to balance them—even in tricky chemical environments like basic (alkaline) solutions.

Early chemists noticed that some reactions couldn't be balanced simply by adjusting coefficients, because atoms and charges didn't add up at the same time. The challenge became even more complex when reactions occurred in solutions containing excess hydroxide ions (OH⁻). Over time, a step-by-step method known as the half-reaction method was developed to handle these situations reliably.

1770s • Lavoisier & Oxidation

Antoine Lavoisier coined the term oxidation to describe reactions involving oxygen. He laid the groundwork for understanding chemical change in terms of element conservation.

1800 • Volta's Pile & Electrochemistry

Alessandro Volta built the first true battery, showing that chemical reactions could produce electric current. This linked redox chemistry directly to electron transfer.

1830s • Faraday's Laws of Electrolysis

Michael Faraday quantified the relationship between the amount of substance produced at an electrode and the electric charge passed through the solution, making balanced equations essential for predictions.

Early 1900s • Electron Theory of Redox

Scientists redefined oxidation as loss of electrons and reduction as gain of electrons, broadening the concept beyond oxygen-based reactions.

Mid-1900s • Half-Reaction Method Standardized

Chemistry educators and textbooks formalized the half-reaction balancing technique, including separate procedures for acidic and basic solutions. This became the standard approach taught worldwide.

The central question this lesson addresses is: How do you systematically balance a redox equation when the reaction takes place in a basic solution, where OH⁻ ions are abundant? The answer involves splitting the reaction into two halves, balancing each one carefully, and then combining them while accounting for the basic environment.

SECTION 2

Core Principles & Definitions

Before diving into the balancing procedure, you need a solid understanding of several foundational ideas. In a redox reaction, one species loses electrons (it is oxidized) while another species gains those electrons (it is reduced). These two processes always happen together—you can't have one without the other.

Oxidation Number

The oxidation number (or oxidation state) of an atom tracks how many electrons it has effectively gained or lost compared to its neutral state. Changes in oxidation number tell you which atoms are oxidized and which are reduced.

Half-Reactions

A half-reaction shows either the oxidation or the reduction part of a redox reaction in isolation. The oxidation half-reaction shows electrons as products; the reduction half-reaction shows electrons as reactants.

Conservation of Mass & Charge

A balanced equation must conserve both atoms (same number of each element on both sides) and charge (total charge on the left equals total charge on the right).

Basic Solution Environment

In a basic solution, hydroxide ions (OH⁻) are plentiful. The key difference from acidic balancing is that we convert any H⁺ ions into water by adding OH⁻ to both sides of the equation at the end.

Electron Balance

The total number of electrons lost in oxidation must equal the total number gained in reduction. We achieve this by multiplying the half-reactions by appropriate factors before combining them.

✦

KEY TAKEAWAY

Think of balancing redox in basic solution like making a fair trade at school: every electron one atom gives away must be picked up by another atom (conservation of charge), and you have to do the exchange in a hallway full of OH⁻ ions instead of H⁺ ions. The extra step of "neutralizing" H⁺ with OH⁻ is what makes the basic method different from the acidic method.

SECTION 3

Visual Overview: The Half-Reaction Method in Basic Solution

The flowchart below illustrates the complete step-by-step procedure for balancing a redox equation in basic solution. Notice that steps 1 through 5 are identical to balancing in acidic solution—the "basic" twist happens in Step 6, where you add OH⁻ to neutralize any H⁺ ions.

The flowchart shows all nine steps. Steps 1–5 and 7–9 are the same as in acidic solution. Step 6 (highlighted in cyan) is the unique "basic fix" where you add OH⁻ to neutralize H⁺ ions.

As the flowchart shows, you essentially balance the equation as if it were in acidic solution first (using H⁺ and H₂O), and then you "convert" it to basic conditions by adding OH⁻ to both sides. Every H⁺ ion pairs with an OH⁻ to form H₂O. After canceling water molecules that appear on both sides, you're left with a properly balanced equation that only contains OH⁻ and H₂O—no free H⁺ ions.

SECTION 4

The Step-by-Step Procedure

Let's formalize the nine steps you saw in the flowchart. Understanding each rule will make the worked example in the next sections much clearer. Throughout this procedure, remember the two conservation laws: mass balance (same atoms on each side) and charge balance (same total charge on each side).

Steps 1–2: Identify and Separate

Assign oxidation numbers to every atom in the unbalanced equation. Identify which atom is oxidized (its oxidation number increases) and which is reduced (its oxidation number decreases). Then write two separate skeletal half-reactions—one for oxidation and one for reduction—showing only the species that change.

Steps 3–5: Balance as if Acidic

Step 3: Balance all atoms except O and H using coefficients.
Step 4: Balance oxygen atoms by adding H₂O molecules to the side that needs oxygen.
Step 5: Balance hydrogen atoms by adding H⁺ ions to the side that needs hydrogen.

Step 6: The Basic Solution Conversion

NEUTRALIZATION RULE

H⁺ + OH⁻ → H₂O

For every H⁺ ion present, add one OH⁻ to both sides of the half-reaction. On the side where H⁺ was, H⁺ + OH⁻ combine to form H₂O.

Steps 7–9: Combine and Verify

Step 7: Balance electrons by multiplying each half-reaction so that the total electrons lost equals the total electrons gained.
Step 8: Add the two half-reactions together. Electrons should cancel completely.
Step 9: Cancel any species (especially H₂O) that appear on both sides. Verify that atoms and charges balance.

⚠

Pro Tip

A common mistake is adding OH⁻ to only one side in Step 6. You must add the same number of OH⁻ to both the left and the right to keep the equation balanced. Think of it as adding zero in disguise: what you add to one side, you add to the other.

SECTION 5

Acidic vs. Basic: Side-by-Side Comparison

Many students find it helpful to see how balancing in basic solution differs from balancing in acidic solution. The table below compares the two methods step by step. Notice that the only real difference is the addition of Step 6 and the subsequent cancellation of water molecules.

Step	Acidic Solution	Basic Solution
1–2	Assign oxidation numbers; write half-reactions	Same
3	Balance atoms other than O and H	Same
4	Balance O with H₂O	Same
5	Balance H with H⁺	Same (temporarily use H⁺)
6	Done (H⁺ is fine in acid)	Add OH⁻ to both sides to neutralize H⁺
7	Balance electrons	Same
8	Add half-reactions	Same
9	Cancel & verify	Cancel H₂O on both sides & verify

Comparison of acidic vs. basic balancing methods

This diagram traces the conversion of the MnO₄⁻ reduction half-reaction from its acidic form (with H⁺) to its basic form (with OH⁻). Notice how 8H₂O appear on the left after neutralization, 4H₂O are canceled from both sides, leaving 4H₂O on the left and 8OH⁻ on the right.

💡

REMEMBER THIS

The "basic fix" is simple: count the H⁺ ions, add that same number of OH⁻ to both sides, merge H⁺ + OH⁻ into H₂O, then cancel duplicate water molecules. After that, your equation will only contain OH⁻ and H₂O—exactly what you'd expect in a basic solution.

SECTION 6

Worked Example: Balancing Cr(OH)₃ + ClO₃⁻ → CrO₄²⁻ + Cl⁻ in Basic Solution

Let's balance the following reaction in basic solution: Cr(OH)₃ + ClO₃⁻ → CrO₄²⁻ + Cl⁻. Chromium is oxidized and chlorine is reduced. Follow along carefully.

Balancing Cr(OH)₃ + ClO₃⁻ → CrO₄²⁻ + Cl⁻ in Basic Solution

1
Step 1 — Assign Oxidation Numbers

In Cr(OH)₃, chromium has an oxidation number of +3. In CrO₄²⁻, chromium is +6. So Cr goes from +3 to +6 → oxidized (loses 3e⁻). In ClO₃⁻, chlorine is +5. In Cl⁻, chlorine is −1. So Cl goes from +5 to −1 → reduced (gains 6e⁻).
2
Step 2 — Write Skeletal Half-Reactions

Oxidation: Cr(OH)₃ → CrO₄²⁻. Reduction: ClO₃⁻ → Cl⁻.

Two separate half-reactions identified
3
Step 3 — Balance Atoms Other Than O and H

Cr is already balanced (1 Cr on each side in the oxidation half). Cl is already balanced (1 Cl on each side in the reduction half). No coefficients needed yet.
4
Step 4 — Balance O by Adding H₂O

Oxidation: Cr(OH)₃ has 3 O on the left; CrO₄²⁻ has 4 O on the right. Add 1 H₂O to the left → Cr(OH)₃ + H₂O → CrO₄²⁻. Now 4 O on each side. Reduction: ClO₃⁻ has 3 O on the left; Cl⁻ has 0 O on the right. Add 3 H₂O to the right → ClO₃⁻ → Cl⁻ + 3H₂O.

Oxidation: Cr(OH)₃ + H₂O → CrO₄²⁻ | Reduction: ClO₃⁻ → Cl⁻ + 3H₂O
5
Step 5 — Balance H by Adding H⁺

Oxidation: Left side has 3H (from OH) + 2H (from H₂O) = 5H. Right side has 0H. Add 5H⁺ to the right → Cr(OH)₃ + H₂O → CrO₄²⁻ + 5H⁺. Reduction: Left side has 0H. Right side has 6H (from 3H₂O). Add 6H⁺ to the left → ClO₃⁻ + 6H⁺ → Cl⁻ + 3H₂O.

Ox: Cr(OH)₃ + H₂O → CrO₄²⁻ + 5H⁺ | Red: ClO₃⁻ + 6H⁺ → Cl⁻ + 3H₂O
6
Step 5b — Balance Charge with Electrons

Oxidation: Left charge = 0. Right charge = (−2) + 5(+1) = +3. To balance, add 3e⁻ to the right → Cr(OH)₃ + H₂O → CrO₄²⁻ + 5H⁺ + 3e⁻. Reduction: Left charge = (−1) + 6(+1) = +5. Right charge = (−1) + 0 = −1. Difference = 6. Add 6e⁻ to the left → ClO₃⁻ + 6H⁺ + 6e⁻ → Cl⁻ + 3H₂O.

Ox: 3e⁻ lost | Red: 6e⁻ gained
7
Step 6 — Add OH⁻ to Both Sides (Basic Fix)

Oxidation half: 5H⁺ on the right → add 5 OH⁻ to both sides. Left: Cr(OH)₃ + H₂O + 5OH⁻. Right: CrO₄²⁻ + 5H⁺ + 5OH⁻ + 3e⁻ → CrO₄²⁻ + 5H₂O + 3e⁻. Simplify: cancel H₂O (1 on left, 5 on right → net 4H₂O on right). Result: Cr(OH)₃ + 5OH⁻ → CrO₄²⁻ + 4H₂O + 3e⁻. Reduction half: 6H⁺ on the left → add 6 OH⁻ to both sides. Left: ClO₃⁻ + 6H⁺ + 6OH⁻ + 6e⁻ → ClO₃⁻ + 6H₂O + 6e⁻. Right: Cl⁻ + 3H₂O + 6OH⁻. Simplify: cancel 3H₂O from both sides. Result: ClO₃⁻ + 3H₂O + 6e⁻ → Cl⁻ + 6OH⁻.
8
Step 7 — Balance Electrons

The oxidation half produces 3e⁻ and the reduction half consumes 6e⁻. The least common multiple of 3 and 6 is 6. Multiply the oxidation half by 2: 2Cr(OH)₃ + 10OH⁻ → 2CrO₄²⁻ + 8H₂O + 6e⁻. The reduction half stays the same (already 6e⁻).

Both half-reactions now involve 6e⁻
9
Step 8 — Add Half-Reactions

2Cr(OH)₃ + 10OH⁻ + ClO₃⁻ + 3H₂O + 6e⁻ → 2CrO₄²⁻ + 8H₂O + 6e⁻ + Cl⁻ + 6OH⁻. Cancel the 6e⁻ on both sides.
10
Step 9 — Cancel Common Species & Verify

Cancel electrons (6e⁻ both sides). Cancel water: 3H₂O on the left, 8H₂O on the right → net 5H₂O on the right. Cancel OH⁻: 10OH⁻ on the left, 6OH⁻ on the right → net 4OH⁻ on the left. Final balanced equation:

2Cr(OH)₃ + ClO₃⁻ + 4OH⁻ → 2CrO₄²⁻ + Cl⁻ + 5H₂O

💡

Verification Check

Atoms — Cr: 2 = 2 ✓ | Cl: 1 = 1 ✓ | O: 6 + 3 + 4 = 13 left, 8 + 5 = 13 right ✓ | H: 6 + 4 = 10 left, 10 right ✓. Charge — Left: 0 + (−1) + 4(−1) = −5. Right: 2(−2) + (−1) + 0 = −5 ✓. The equation is balanced!

SECTION 7

Common Mistakes & Tips for Success

Even after learning the procedure, students frequently stumble on certain steps. The table below lists the most common errors alongside the correct approach. Studying these pitfalls before attempting practice problems will save you a lot of frustration.

Common Mistake	Why It's Wrong	Correct Approach
Adding OH⁻ to only one side	Breaks charge balance—you've changed the equation	Always add the same number of OH⁻ to both sides
Forgetting to cancel H₂O	Leaves unnecessary water on both sides, inflating coefficients	After combining H⁺ + OH⁻ → H₂O, subtract H₂O common to both sides
Wrong oxidation numbers	Leads to incorrect electron counts, ruining the entire balance	Double-check using the rule: sum of oxidation numbers = charge of species
Leaving H⁺ in the final answer	H⁺ should not appear in a basic solution equation	If you see H⁺ in your final answer, you skipped Step 6
Not verifying at the end	Small arithmetic errors can persist unnoticed	Always count atoms AND check total charge on both sides

Common mistakes when balancing redox in basic solution

✦

KEY TAKEAWAY

Think of the basic solution method like building a LEGO set: first you follow the acidic instructions (Steps 1–5), then you apply the "basic solution upgrade kit" (Step 6, adding OH⁻). If you skip the upgrade, your final model has the wrong pieces showing. If you apply it to only one side, the structure falls apart.

SECTION 8

Connection to Electrochemistry & Advanced Topics

Balancing redox reactions isn't just an exercise for exams—it is the foundation for everything else in electrochemistry. Whether you're calculating cell potentials, predicting the products of electrolysis, or sizing a battery, you need a correctly balanced equation as your starting point. The table below shows how this skill connects to more advanced topics you'll encounter later.

This Lesson	Advanced Topic
Balancing half-reactions	Used to calculate standard cell potential (E°) from reduction potential tables
Counting electrons transferred	The "n" in the Nernst equation and Faraday's law comes directly from balanced half-reactions
OH⁻ in basic solution	Alkaline batteries, fuel cells, and industrial electrolysis all operate in basic media
Mass & charge balance	Stoichiometry of electroplating, corrosion calculations, and energy storage design

How basic-solution balancing connects to advanced electrochemistry

In AP Chemistry and college courses, you'll use the number of electrons transferred (n) from your balanced equation to plug into the Nernst equation (E = E° − (RT/nF) ln Q) and Faraday's law (m = MIt/nF). Without a properly balanced equation, these calculations fall apart. Mastering basic-solution balancing now gives you a strong foundation for every electrochemistry problem you'll face in the future.

SECTION 9

Practice Problems

Test your understanding with these five problems. They increase in difficulty from conceptual recall to critical thinking. Try each one on paper before checking the answer.

Question 1: CONCEPTUAL

In the half-reaction method for basic solution, why do we first balance as if the solution were acidic, and then add OH⁻? Why not just use OH⁻ from the start?

Show answer

Balancing in acidic solution first gives us a clear, systematic way to handle oxygen and hydrogen atoms using only H₂O and H⁺. If we tried to use OH⁻ from the start, we would be adding both O and H atoms simultaneously, which makes it much harder to keep track of atom counts. By converting H⁺ → H₂O with OH⁻ as a separate step, we avoid confusion and ensure that the equation is balanced at every stage. The acidic-first approach is essentially a bookkeeping strategy that guarantees accuracy.

Question 2: BASIC CALCULATION

Balance the following half-reaction in basic solution: MnO₄⁻ → MnO₂.

Show answer

Start with MnO₄⁻ → MnO₂. Mn goes from +7 to +4 (gains 3e⁻). Balance O: add 2H₂O to the right → MnO₄⁻ → MnO₂ + 2H₂O. Balance H: add 4H⁺ to the left → MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O. Balance charge: left = −1 + 4 = +3, right = 0. Add 3e⁻ to the left → MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O. Convert to basic: add 4OH⁻ to both sides → MnO₄⁻ + 4H₂O + 3e⁻ → MnO₂ + 2H₂O + 4OH⁻. Cancel 2H₂O: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻.

Question 3: INTERMEDIATE

Balance the full redox equation in basic solution: Al + MnO₄⁻ → Al(OH)₄⁻ + MnO₂.

Show answer

Oxidation: Al → Al(OH)₄⁻. Al goes from 0 to +3 (loses 3e⁻). Balance O: add 4H₂O to the left. Balance H: 8H on left, 4H on right → add 4H⁺ to right. Charge: left = 0, right = −1 + 4 = +3 → add 3e⁻ to right. Acidic form: Al + 4H₂O → Al(OH)₄⁻ + 4H⁺ + 3e⁻. Basic fix: add 4OH⁻ both sides → Al + 4H₂O + 4OH⁻ → Al(OH)₄⁻ + 4H₂O + 3e⁻. Cancel 4H₂O → Al + 4OH⁻ → Al(OH)₄⁻ + 3e⁻. Reduction (from Problem 2): MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. Both involve 3e⁻, so add directly: Al + MnO₄⁻ + 2H₂O → Al(OH)₄⁻ + MnO₂. The 4OH⁻ from each half cancel. Verify: Al 1=1 ✓, Mn 1=1 ✓, O 4+2=6 and 4+2=6 ✓, H 4=4 ✓, charge −1 = −1 ✓.

Question 4: APPLIED

In an alkaline battery, zinc is oxidized and manganese dioxide is reduced. The unbalanced reaction is: Zn + MnO₂ → ZnO + Mn₂O₃. Balance this in basic solution. (Hint: Mn goes from +4 to +3.)

Show answer

Oxidation: Zn → ZnO. Zn goes 0 → +2 (loses 2e⁻). Balance O: add H₂O to left → Zn + H₂O → ZnO. Balance H: add 2H⁺ to right → Zn + H₂O → ZnO + 2H⁺ + 2e⁻. Basic fix (add 2OH⁻ both sides): Zn + H₂O + 2OH⁻ → ZnO + 2H₂O + 2e⁻. Simplify: Zn + 2OH⁻ → ZnO + H₂O + 2e⁻. Reduction: 2MnO₂ → Mn₂O₃. Mn goes +4 → +3 (each gains 1e⁻, total 2e⁻). Balance O: 4O left, 3O right → add H₂O to right → 2MnO₂ → Mn₂O₃ + H₂O. Already need 2H on right → add 2H⁺ to left → 2MnO₂ + 2H⁺ + 2e⁻ → Mn₂O₃ + H₂O. Basic fix: add 2OH⁻ → 2MnO₂ + 2H₂O + 2e⁻ → Mn₂O₃ + H₂O + 2OH⁻. Simplify: 2MnO₂ + H₂O + 2e⁻ → Mn₂O₃ + 2OH⁻. Both have 2e⁻. Add: Zn + 2MnO₂ → ZnO + Mn₂O₃. (OH⁻ and H₂O cancel completely.) Verify: Zn 1=1 ✓, Mn 2=2 ✓, O 4=4 ✓, charge 0=0 ✓.

Question 5: CRITICAL THINKING

Balance in basic solution: Bi₂O₃ + ClO⁻ → BiO₃⁻ + Cl⁻. Then determine how many moles of electrons are transferred per mole of Bi₂O₃ oxidized.

Show answer

Oxidation: Bi₂O₃ → 2BiO₃⁻. Bi goes +3 → +5, each Bi loses 2e⁻, total 4e⁻. Balance O: 3O left, 6O right → add 3H₂O to left → Bi₂O₃ + 3H₂O → 2BiO₃⁻. Balance H: 6H left → add 6H⁺ right → Bi₂O₃ + 3H₂O → 2BiO₃⁻ + 6H⁺ + 4e⁻. Basic fix (6OH⁻ both sides): Bi₂O₃ + 3H₂O + 6OH⁻ → 2BiO₃⁻ + 6H₂O + 4e⁻. Simplify: Bi₂O₃ + 6OH⁻ → 2BiO₃⁻ + 3H₂O + 4e⁻. Reduction: ClO⁻ → Cl⁻. Cl goes +1 → −1 (gains 2e⁻). Balance O: add H₂O right → ClO⁻ → Cl⁻ + H₂O. Wait, O is on the left. Add H₂O to right: ClO⁻ → Cl⁻, add H₂O right for 0 extra O? Actually, 1O left 0O right → add 1H₂O to right → but that adds O. Instead: add H₂O to right side: we need 1O on right → ClO⁻ → Cl⁻; 1O left, 0O right; add 1H₂O right? No. Correct: ClO⁻ has 1O. Cl⁻ has 0O. So the right side needs 1O → no, the LEFT has excess O. Add 1H₂O to the RIGHT is wrong. Standard approach: balance O by adding H₂O to the O-deficient side (right). But that introduces O. We need to remove O from the left. Add H₂O to the right to carry the O away? Actually: there's 1 O on left, 0 on right. Add 1 H₂O to the right: ClO⁻ → Cl⁻ + H₂O (no, H₂O has O). Wait — that's wrong, we want less O on the left. The rule is: add H₂O to the side deficient in O. Right side has 0 O, left has 1 → right is deficient, but adding H₂O to the right doesn't help; it creates O there. Correctly: to balance, the O must end up somewhere. Think again: left 1O, right 0O → we need 1 O on the right → add 1 H₂O (which contains 1 O) to the right? That gives right 1O. But H₂O also adds 2H to the right. Actually this is correct! ClO⁻ → Cl⁻ + H₂O is wrong because that's 1O = 1O but we must track it properly. ClO⁻ has 1 O, and Cl⁻ + H₂O has 1 O. Nope: that IS balanced for O. No wait, there's no additional O to create. The O from ClO⁻ goes into the H₂O. But we haven't actually balanced O; we need to use H₂O solely to balance the count. So: left: 1O. Right (with H₂O added): 1O. But I added H₂O, so right now has Cl⁻ + H₂O. The O from ClO⁻ essentially goes into the water. No, that's the correct approach: the O in ClO⁻ needs a place to go. Since Cl⁻ has no O, add H₂O to the product side to absorb the oxygen. That's the standard method. So: ClO⁻ → Cl⁻ + H₂O (nope, now check O: 1=1 ✓). Okay, but that seems like we pulled the H₂O from nowhere. But the H atoms will be supplied by H⁺. So Step 4 already done? No — I think I confused myself. Let me redo: ClO⁻ → Cl⁻. Left: 1O. Right: 0O. Deficient side = right. Add 1H₂O to right? No, that puts 1O on the right: 1O left, 1O right ✓. But it also adds 2H to the right. Hmm, but the O count doesn't work because we're adding a whole H₂O molecule. Okay this is fine: Step 4 gives ClO⁻ → Cl⁻ + H₂O (wrong — this doesn't seem right either because the one O on the left IS going into the Cl⁻ + H₂O product side). Hmm, I think the issue is that the left side already has 1 O in ClO⁻. If I add nothing, there's 1O left, 0O right. If I want O balanced, I can't just add O to the product side; I need to conserve atoms. The technique is: for each extra O on the side with more O, add 1 H₂O to the other side. So: left has 1 more O → add 1 H₂O to the right. Now: ClO⁻ → Cl⁻ + H₂O? But check: O: left 1, right 1 ✓ (the O in H₂O). However we've now got 2H on the right that we must balance. Step 5: add 2H⁺ to the left → ClO⁻ + 2H⁺ → Cl⁻ + H₂O. Charge: left (−1 + 2 = +1), right (−1 + 0 = −1). Add 2e⁻ to left → ClO⁻ + 2H⁺ + 2e⁻ → Cl⁻ + H₂O. Check charge: left −1+2−2 = −1, right −1 ✓. Basic fix: 2OH⁻ → ClO⁻ + 2H₂O + 2e⁻ → Cl⁻ + H₂O + 2OH⁻. Cancel 1H₂O: ClO⁻ + H₂O + 2e⁻ → Cl⁻ + 2OH⁻. Now, to match electrons: oxidation has 4e⁻, reduction has 2e⁻. Multiply reduction by 2: 2ClO⁻ + 2H₂O + 4e⁻ → 2Cl⁻ + 4OH⁻. Combine: Bi₂O₃ + 6OH⁻ + 2ClO⁻ + 2H₂O → 2BiO₃⁻ + 3H₂O + 2Cl⁻ + 4OH⁻. Simplify OH⁻: 6−4 = 2OH⁻ left. Simplify H₂O: 2 left, 3 right → net 1H₂O right. Final: Bi₂O₃ + 2ClO⁻ + 2OH⁻ → 2BiO₃⁻ + 2Cl⁻ + H₂O. Verification: Bi 2=2 ✓, Cl 2=2 ✓, O 3+2+2=7 left, 6+1=7 right ✓, H 2=2 ✓. Charge: −2−2 = −4 left, −2−2 = −4 right ✓. Per mole of Bi₂O₃, 4 moles of electrons are transferred (2 Bi atoms each lose 2 electrons).

SUMMARY

Summary

Balancing redox reactions in basic solution follows the same half-reaction method used in acidic solution, with one critical extra step. First, you assign oxidation numbers to identify which atoms are oxidized (lose electrons) and which are reduced (gain electrons). You write separate half-reactions, balance all atoms using H₂O and H⁺ (as if in acid), and then balance charge with electrons.

The key distinction is Step 6: the basic fix, where you add OH⁻ to both sides of each half-reaction to neutralize every H⁺ ion, converting H⁺ + OH⁻ into H₂O. After canceling duplicate water molecules, you equalize electrons by multiplying the half-reactions, add them together, and verify that both mass and charge are conserved. This skill is foundational for electrochemistry topics including cell potentials, the Nernst equation, and Faraday's law.