Chemical Equation Balancer

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balance redox equation oxidation numbers

Balancing by oxidation numbers: electrons and agents

When one element gains electrons and another loses them, the coefficients come from conservation of electrons, not from counting atoms in isolation. Here is the oxidation-number method in four classic examples, all checked by the tool.

The oxidation-number method in four steps

First, assign an oxidation number to each element: group 1 metals are +1, group 2 are +2, hydrogen is +1 (except in metal hydrides, -1), oxygen is -2 (except in peroxides, -1) and chlorine is -1 when bonded to less electronegative elements. The rest comes from requiring the sum of oxidation numbers to equal the charge of the species. In KMnO4, K = +1 and O = -2 x 4 = -8, so Mn is +7. Second, see what changes: in KMnO4 + HCl, Mn goes from +7 to +2 (gains 5 electrons) and Cl goes from -1 to 0 (each atom loses 1). The reactant containing the element that is reduced is the oxidizing agent, here KMnO4; the one containing the element that is oxidized is the reducing agent, here HCl.
Third, equalize the electrons. If each Mn gains 5 and each Cl2 formed loses 2 (two Cl atoms), the least common multiple is 10: two Mn (2 x 5 = 10) for every five Cl2 (5 x 2 = 10). Fourth, complete the rest by conserving atoms: 2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O. Of the 16 HCl, only 10 become Cl2; the other 6 stay as chloride in KCl and MnCl2, which is why the acid also acts as the medium. The tool computes the same coefficients from the matrix kernel and, in the Redox tab, shows the oxidation numbers, the 10 electrons transferred and the two agents, so you can check your path by hand.
The method has traps the tool helps to spot. An element can be in two environments at once: in Zn + dilute HNO3, part of the nitrogen is reduced to NH4+ (oxidation number -3) while part stays as nitrate (+5), and the result 4Zn + 10HNO3 → 4Zn(NO3)2 + NH4NO3 + 3H2O only closes with both. In compounds with two elements of unknown oxidation number, such as Fe3O4 (average +8/3), the general rules are not enough without the structure, and the tool marks those assignments with an asterisk instead of presenting them as certain. Disproportionation reactions, where the same substance is both oxidized and reduced, are detected and flagged.

Redox equations worked out

Potassium permanganate and hydrochloric acid

Input: KMnO4 + HCl -> KCl + MnCl2 + Cl2 + H2O
Expected output: 2KMnO4 + 16HCl -> 2KCl + 2MnCl2 + 5Cl2 + 8H2O

Mn +7 → +2 and Cl -1 → 0; 10 electrons transferred.

Potassium dichromate and hydrochloric acid

Input: K2Cr2O7 + HCl -> KCl + CrCl3 + Cl2 + H2O
Expected output: K2Cr2O7 + 14HCl -> 2KCl + 2CrCl3 + 3Cl2 + 7H2O

Each Cr goes from +6 to +3, so the 2 Cr gain 6 electrons, which 3 Cl2 supply.

Copper and dilute nitric acid

Input: Cu + HNO3 -> Cu(NO3)2 + NO + H2O
Expected output: 3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O

Cu 0 → +2 (loses 2) and N +5 → +2 (gains 3): 3 Cu for 2 NO.

Zinc and very dilute nitric acid

Input: Zn + HNO3 -> Zn(NO3)2 + NH4NO3 + H2O
Expected output: 4Zn + 10HNO3 -> 4Zn(NO3)2 + NH4NO3 + 3H2O

N +5 → -3 in NH4+ gains 8 electrons; 4 Zn lose 8.

Diarsenic trisulfide with nitric acid

Input: As2S3 + HNO3 + H2O -> H3AsO4 + H2SO4 + NO
Expected output: 3As2S3 + 28HNO3 + 4H2O -> 6H3AsO4 + 9H2SO4 + 28NO

Two elements oxidize (As and S) and one is reduced (N): 84 electrons per 3 As2S3.

Full tool FAQ

How do you balance a chemical equation?

Count the atoms of each element on both sides and adjust the coefficients (never the subscripts) until the counts match. By hand this is done by trial; the tool solves the exact linear system. For Fe + O2 → Fe2O3 the answer is 4Fe + 3O2 → 2Fe2O3, with 4 Fe atoms and 6 O atoms on each side.

Why does an equation need to be balanced?

How does balancing by trial and error work?

How do you balance equations with ions?

How do you balance a redox reaction?

What does "more than one independent solution" mean?

Why does the tool say a substance is on the wrong side or takes no part?

Can I type the equation with coefficients?

How do you find the limiting reagent and the yield?

Does the tool predict the products of a reaction?

Frequently asked questions

How do I know whether a reaction is redox?

If any element changes oxidation number between reactants and products. In H2SO4 + 2NaOH → Na2SO4 + 2H2O nothing changes (Na +1, S +6, O -2, H +1), so it is acid-base. In Zn + 2HCl → ZnCl2 + H2 Zn goes from 0 to +2 and H from +1 to 0, so it is redox.

What is the difference between the oxidizing agent and the substance that is oxidized?

The oxidizing agent is the substance that causes the other to be oxidized and so is the one that is reduced. In KMnO4 + HCl, KMnO4 is the oxidizing agent (its Mn is reduced) and HCl is the reducing agent (its Cl is oxidized). Many people swap the two on exams.

Why does the tool mark some oxidation numbers with an asterisk?

Because the textbook rules cannot decide on their own when a substance has two elements of unknown oxidation number, such as CuSO4 (Cu and S). In those cases it applies a tie-break rule, for example S +6 in sulfates, and says the value was assumed.

Next steps

Atom count, before and after the coefficients
Element	Before (all 1)	After	Check
C	3	1	3	3	✓
H	8	2	8	8	✓
O	2	3	10	10	✓

Atom count, before and after the coefficients

Element

Before (all 1)

After

Check

React.

Prod.

React.

Prod.

✓

The oxidation-number method in four steps

First, assign an oxidation number to each element: group 1 metals are +1, group 2 are +2, hydrogen is +1 (except in metal hydrides, -1), oxygen is -2 (except in peroxides, -1) and chlorine is -1 when bonded to less electronegative elements. The rest comes from requiring the sum of oxidation numbers to equal the charge of the species. In KMnO4, K = +1 and O = -2 x 4 = -8, so Mn is +7. Second, see what changes: in KMnO4 + HCl, Mn goes from +7 to +2 (gains 5 electrons) and Cl goes from -1 to 0 (each atom loses 1). The reactant containing the element that is reduced is the oxidizing agent, here KMnO4; the one containing the element that is oxidized is the reducing agent, here HCl.

Third, equalize the electrons. If each Mn gains 5 and each Cl2 formed loses 2 (two Cl atoms), the least common multiple is 10: two Mn (2 x 5 = 10) for every five Cl2 (5 x 2 = 10). Fourth, complete the rest by conserving atoms: 2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O. Of the 16 HCl, only 10 become Cl2; the other 6 stay as chloride in KCl and MnCl2, which is why the acid also acts as the medium. The tool computes the same coefficients from the matrix kernel and, in the Redox tab, shows the oxidation numbers, the 10 electrons transferred and the two agents, so you can check your path by hand.

The method has traps the tool helps to spot. An element can be in two environments at once: in Zn + dilute HNO3, part of the nitrogen is reduced to NH4+ (oxidation number -3) while part stays as nitrate (+5), and the result 4Zn + 10HNO3 → 4Zn(NO3)2 + NH4NO3 + 3H2O only closes with both. In compounds with two elements of unknown oxidation number, such as Fe3O4 (average +8/3), the general rules are not enough without the structure, and the tool marks those assignments with an asterisk instead of presenting them as certain. Disproportionation reactions, where the same substance is both oxidized and reduced, are detected and flagged.

Redox equations worked out

Potassium permanganate and hydrochloric acid

Input: KMnO4 + HCl -> KCl + MnCl2 + Cl2 + H2O
Expected output: 2KMnO4 + 16HCl -> 2KCl + 2MnCl2 + 5Cl2 + 8H2O

Mn +7 → +2 and Cl -1 → 0; 10 electrons transferred.

Potassium dichromate and hydrochloric acid

Input: K2Cr2O7 + HCl -> KCl + CrCl3 + Cl2 + H2O
Expected output: K2Cr2O7 + 14HCl -> 2KCl + 2CrCl3 + 3Cl2 + 7H2O

Each Cr goes from +6 to +3, so the 2 Cr gain 6 electrons, which 3 Cl2 supply.

Copper and dilute nitric acid

Input: Cu + HNO3 -> Cu(NO3)2 + NO + H2O
Expected output: 3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O

Cu 0 → +2 (loses 2) and N +5 → +2 (gains 3): 3 Cu for 2 NO.

Zinc and very dilute nitric acid

Input: Zn + HNO3 -> Zn(NO3)2 + NH4NO3 + H2O
Expected output: 4Zn + 10HNO3 -> 4Zn(NO3)2 + NH4NO3 + 3H2O

N +5 → -3 in NH4+ gains 8 electrons; 4 Zn lose 8.

Diarsenic trisulfide with nitric acid

Input: As2S3 + HNO3 + H2O -> H3AsO4 + H2SO4 + NO
Expected output: 3As2S3 + 28HNO3 + 4H2O -> 6H3AsO4 + 9H2SO4 + 28NO

Two elements oxidize (As and S) and one is reduced (N): 84 electrons per 3 As2S3.

Full tool FAQ

How do you balance a chemical equation?

Why does an equation need to be balanced?

How does balancing by trial and error work?

How do you balance equations with ions?

How do you balance a redox reaction?

What does "more than one independent solution" mean?

Why does the tool say a substance is on the wrong side or takes no part?

Can I type the equation with coefficients?

How do you find the limiting reagent and the yield?

Does the tool predict the products of a reaction?

Frequently asked questions