HS Chemistry - Redox Reactions

Balancing Redox Reactions

Overview of The Page

This page will cover:

• How are redox reactions balanced?
• What are oxidation numbers?

So far, we've had to balance equations so that each side has an equal number of reactants. However, we also have to balance charges in chemical reactions, particularly when balancing redox reactions.

In the following reaction betwen Silver $Ag$ and Scandium $Sc$:

Ag⁺ + Sc → Ag + Sc³⁺

In this reaction, Ag is gaining electrons, and Sc is losing electrons. Ag is being reduced, and Sc is being oxidized. This is a redox reaction.

The number of atoms on both sides are equal - 1 Ag atom and 1 Sc atom - but the charges on both sides are different. The equation isn't balanced - we can't just lose two electrons.

We could just plug in coefficients until it does balance out. In this case, it might be pretty easy to see what coefficients should be plugged in. But that might not be the case in more complex redox reactions.

Fortunately, there's a better method.

Let's split up the reduction half-equation and oxidation half-equation of this redox reaction:

Reduction half-equation: Ag⁺ → Ag

Oxidation half-equation: Sc → Sc³⁺

These equations are unbalanced, so let's balance them.

Reduction half-equation: Ag⁺ + e^- → Ag

Oxidation half-equation: Sc → Sc³⁺ + 3e^-

Now these half-equations are balanced. If we combined them, we might expect to get a redox reaction:

Ag⁺ + e^- + Sc → Ag + Sc³⁺ + 3e^-

Now the mass and the charges are balanced on both sides. Both sides have one Ag atom, one Sc atom, and a net charge of 0.

But the electrons on both sides aren't balanced. There's one on the reactant side, but three on the product side. We need three on the reactant side to balance this out.

Let's go back to our half-equations.

Reduction half-equation: Ag⁺ + e^- → Ag

Oxidation half-equation: Sc → Sc³⁺ + 3e^-

We need three electrons on the product side to balance our redox reaction. Only the reduction half-equation has electrons on the product side, and it only has one. We can multiply the reactants and products of the reduction half-equation by three to get three electrons on the product side.

Reduction half-equation: 3$Ag^\+^ + e^\-^$ → 3$Ag$

Expanding this gives us:

Reduction half-equation: 3Ag⁺ + 3e^- → 3Ag

And now we have three electrons on the product side as well. When we multiplied the reactants and products of the reduction half-equation by three, we didn't do the same for the oxidation reaction. This is because the reduction reaction is separate from the oxidation reaction, and therefore we can treat them separately.

Now when we combine them together we get:

3Ag⁺ + 3 e^- + Sc → 3Ag + Sc³⁺ + 3e^-

The electrons cancel each other out, and the resulting, balanced redox reaction is:

3Ag⁺ + Sc → 3Ag + Sc³⁺

Oxidation Numbers

So far, we've looked at this very conceptually. While it makes sense, this method can easily get tedious for more complex equations. There should be an easier method.

And there is an easier method: using oxidation numbers.

The oxidation number is how many electrons an atom can use in bonding - that is, how many electrons can that atom lose, gain, or share in bonding?

Oxidation numbers have signs. If the electron is lost, there's a + sign, and if it's gained, there's a - sign. If the electrons are being shared, we act as if the more electronegative atom completely steals the electron$s$ away from the less electronegative atom, and then calculate the oxidation number.

Oxidation numbers aren't always the same for an atom. For example, in NH₃, Nitrogen has an oxidation number of -3, and Hydrogen has an oxidation number of +1, but in NF₃, Nitrogen has an oxidation number of +3, and Fluorine has an oxidation number of -1. Oxidation numbers change depending on the circumstance.

That doesn't mean they're difficult to use - in fact, this is what makes them so handy. There's just a few rules that need to be kept in mind when using oxidation numbers:

• In any compound, the oxidation numbers of all the elements will add up to the net charge of the compound.

• Group 1 & 2 elements will always have oxidation numbers of 1 & 2 respectively.

• Elements bonded with each other $like O~2~, 3Fe, etc.$ have an overall charge and oxidation number of 0.

• Monatomic ions $like Br^\-^, K^\+^, Sc^3\+^$ have the same oxidation number as their charge.

• Some elements will always have the same oxidation number:

  • Fluorine will always have an oxidation number of -1.
  • Hydrogen will always have an oxidation number of either +1 or -1.
    • If Hydrogen is bonded to another metal in ionic bonding $NaH, CaH~2~, etc.$ then its oxidation number is -1. Otherwise, it is +1.
  • Oxygen will almost always have an oxidation number of -2.
    • The exceptions are when it is a preoxide $\-1$ or in F₂O $\+2$.

Using these rules, we can not only calculate oxidation numbers for other elements in a compound, but we can also easily find net ionic equations and redox reactions in a complex ionic equation.

If an element's oxidation number is unknown, it can be calculated through:

(net charge of molecule) - (combined charges of everything else in the molecule)

A change in an element's oxidation number $a change in its oxidation state$ over the course of a reaction shows that that element is being reduced/oxidized in the reaction. The amount that the oxidation number changes by is the amount by which it is reduced/oxidized. Thus, by calculating elements' oxidation numbers on both sides of a reaction, we can easily tell which elements are being reduced/oxidized in a reaction, and by how much.

There is one more thing to note. In a redox reaction, one atom gains electrons from another atom, which loses electrons. The atom which gaines electrons is called the oxidizing agent $as it caused the other atom to be oxidized$ and the atom which loses electrons is called the reducing agent $by losing electrons, it caused the other atom to be reduced$.

The atom which undergoes reduction is called the oxidizing agent, and the atom which undergoes oxidation is called the reducing agent.

Practice

What is the oxidation number of P in POF₃? Enter your answer as a number, with a sign to indicate the oxidation number (e.g. -2, +1, etc.):

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What is the oxidation number of N in NO₃^-? Enter your answer as a number, with a sign to indicate the oxidation number (e.g. -2, +1, etc.):

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What is the oxidation number of P in PO₄^3-? Enter your answer as a number, with a sign to indicate the oxidation number (e.g. -2, +1, etc.):

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What is the oxidation number of Hg in Hg₃(PO₄)₂? Enter your answer as a number, with a sign to indicate the oxidation number (e.g. -2, +1, etc.):

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In the equation CaF₂ + Mg → MgF₂ + Ca:

• Which element is being reduced? Ca F Mg
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• Which element is being oxidized? Ca F Mg
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In the equation PH₅ → PH₃ + H₂:

• Which element is being reduced? P H
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• Which element is being oxidized? P H
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In the equation PCl₅ + 2NaF → PCl₃ + 2NaCl + F₂:

• Which element is being reduced? P Cl Na F
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• Which element is being oxidized? P Cl Na F
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Balance the following redox reaction: Zn + V⁵⁺ → Zn²⁺ + V²⁺ 2Zn + 3V⁵⁺ → 2Zn²⁺ + 3V²⁺ 3Zn + 2V⁵⁺ → 3Zn²⁺ + 2V²⁺ 5Zn + 4V⁵⁺ → 5Zn²⁺ + 4V²⁺

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