Redox Concepts: Oxidation Numbers and Balancing Redox Reactions

Welcome! Today, we're diving into a fundamental concept in chemistry: redox reactions. Redox is short for reduction-oxidation. These reactions are crucial because they underpin many processes, from the batteries that power our devices to the respiration that keeps us alive. Understanding redox means understanding how electrons are transferred between chemical species.

What are Oxidation and Reduction?

At its core, a redox reaction involves a change in the oxidation states of atoms.

  • Oxidation: This is the process where a chemical species loses electrons, resulting in an increase in its oxidation number. Think of it as "Leo the lion says 'GUR'" - Lose Electrons Oxidation.
  • Reduction: This is the process where a chemical species gains electrons, resulting in a decrease in its oxidation number. Think of it as "GER" - Gain Electrons Reduction.

Crucially, oxidation and reduction always occur simultaneously. You can't have one without the other. If one substance loses electrons, another substance must gain them. The substance that gets oxidized is called the reducing agent because it causes something else to be reduced. The substance that gets reduced is called the oxidizing agent because it causes something else to be oxidized.

Assigning Oxidation Numbers

To track electron transfer, we use the concept of oxidation numbers (or oxidation states). These are hypothetical charges an atom would have if all its bonds to different atoms were fully ionic. While not always reflecting the real charge, they are a powerful tool for analyzing redox reactions.

Rules for Assigning Oxidation Numbers:

Here are the standard rules, which should be applied in order:

  1. Element in its Free State: The oxidation number of an atom in an element in its free, uncombined state is zero.
    • Examples: $O_2$, $H_2$, $Na$, $Cl_2$, $Fe$, $S_8$ all have an oxidation number of 0 for the respective atoms.
  2. Monatomic Ions: The oxidation number of a monatomic ion is equal to its charge.
    • Examples: $Na^+$ has an oxidation number of +1, $Mg^{2+}$ is +2, $Cl^-$ is -1, $O^{2-}$ is -2.
  3. Oxygen: Oxygen usually has an oxidation number of -2 in its compounds.
    • Exceptions:
      • In peroxides ($ROOR'$), oxygen is -1 (e.g., $H_2O_2$).
      • In superoxides ($ROO^-$), oxygen is -1/2 (e.g., $KO_2$).
      • When bonded to fluorine (which is more electronegative), oxygen can have positive oxidation numbers (e.g., in $OF_2$, oxygen is +2).
  4. Hydrogen: Hydrogen usually has an oxidation number of +1 when bonded to nonmetals.
    • Exception: In metal hydrides (like $NaH$, $CaH_2$), hydrogen is -1 because metals are less electronegative.
  5. Halogens: Halogens (F, Cl, Br, I) usually have an oxidation number of -1.
    • Exception: When bonded to oxygen or a more electronegative halogen, their oxidation number can be positive (e.g., in $HClO_4$, Cl is +7; in $ICl$, I is +1). Fluorine is always -1 in compounds.
  6. Sum of Oxidation Numbers:
    • In a neutral molecule, the sum of the oxidation numbers of all atoms must be zero.
    • In a polyatomic ion, the sum of the oxidation numbers of all atoms must equal the charge of the ion.
Memory Trick: To remember the order of rules, think "Elements are 0, Ions are their charge. Oxygen is -2 (mostly), Hydrogen is +1 (mostly). Halogens are -1 (mostly). Sum it up!"

Applying the Rules: Examples

Let's practice assigning oxidation numbers to specific atoms in various compounds.

Example 1: $KMnO_4$ (Potassium Permanganate)

This is a neutral compound, so the sum of oxidation numbers must be 0.

  1. Potassium (K) is an alkali metal in Group 1, so its oxidation number is +1.
  2. Oxygen (O) usually has an oxidation number of -2. There are 4 oxygen atoms, so their total contribution is $4 \times (-2) = -8$.
  3. Let the oxidation number of Manganese (Mn) be 'x'.
  4. Sum: $(+1) + x + (-8) = 0$
  5. Solving for x: $x - 7 = 0 \implies x = +7$.
  6. So, in $KMnO_4$, Mn has an oxidation number of +7.

Example 2: $SO_4^{2-}$ (Sulfate Ion)

This is a polyatomic ion with a charge of -2. The sum of oxidation numbers must equal -2.

  1. Oxygen (O) usually has an oxidation number of -2. There are 4 oxygen atoms, so their total contribution is $4 \times (-2) = -8$.
  2. Let the oxidation number of Sulfur (S) be 'y'.
  3. Sum: $y + (-8) = -2$
  4. Solving for y: $y = -2 + 8 \implies y = +6$.
  5. So, in $SO_4^{2-}$, S has an oxidation number of +6.

Example 3: $P_4O_{10}$ (Tetraphosphorus Decoxide)

Neutral compound, sum = 0.

  1. Oxygen (O) is -2. There are 10 oxygen atoms, total = $10 \times (-2) = -20$.
  2. Let the oxidation number of Phosphorus (P) be 'z'. There are 4 P atoms, so total = $4z$.
  3. Sum: $4z + (-20) = 0$
  4. Solving for z: $4z = 20 \implies z = +5$.
  5. So, in $P_4O_{10}$, P has an oxidation number of +5.

Example 4: $Na_2S_2O_3$ (Sodium Thiosulfate)

Neutral compound, sum = 0.

  1. Sodium (Na) is +1. There are 2 Na atoms, total = $2 \times (+1) = +2$.
  2. Oxygen (O) is -2. There are 3 oxygen atoms, total = $3 \times (-2) = -6$.
  3. Let the oxidation number of Sulfur (S) be 'w'. There are 2 S atoms, total = $2w$.
  4. Sum: $(+2) + 2w + (-6) = 0$
  5. Solving for w: $2w - 4 = 0 \implies 2w = 4 \implies w = +2$.
  6. This gives an average oxidation state of +2 for sulfur. However, sulfur can have different oxidation states within the same molecule. In thiosulfate, one sulfur is typically +6 (like in sulfate) and the other is -2 (like in sulfide). The average is $(+6 + (-2))/2 = +4/2 = +2$. This highlights that oxidation numbers are often averages.

Identifying Redox Reactions

A reaction is a redox reaction if there is a change in the oxidation numbers of any atoms involved. We compare the oxidation numbers of each element on the reactant side to its oxidation number on the product side.

Example: Reaction of Zinc with Copper Sulfate

Consider the reaction: $Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)$

Let's assign oxidation numbers to each atom:

  • Reactants:
    • $Zn(s)$: Zn is in its free state, so oxidation number = 0.
    • $CuSO_4$: This is a neutral compound. $SO_4^{2-}$ has a charge of -2. So, Cu must be +2 to balance. (Cu: +2, S: +6, O: -2).
  • Products:
    • $ZnSO_4$: This is a neutral compound. $SO_4^{2-}$ has a charge of -2. So, Zn must be +2 to balance. (Zn: +2, S: +6, O: -2).
    • $Cu(s)$: Cu is in its free state, so oxidation number = 0.

Comparing oxidation numbers:

  • Zinc (Zn): Changes from 0 to +2. Its oxidation number increased, so Zn is oxidized.
  • Copper (Cu): Changes from +2 to 0. Its oxidation number decreased, so Cu is reduced.
  • Sulfur (S) and Oxygen (O) remain unchanged (+6 and -2 respectively).

Since there is a change in oxidation numbers (Zn and Cu), this is a redox reaction.

  • $Zn$ is oxidized and acts as the reducing agent.
  • $Cu^{2+}$ (from $CuSO_4$) is reduced and acts as the oxidizing agent.

Balancing Redox Reactions

Balancing redox reactions can be more complex than balancing regular chemical equations because we need to ensure that both the number of atoms of each element and the total charge are balanced on both sides of the equation. The two primary methods are the oxidation number method and the ion-electron (half-reaction) method. We will focus on the ion-electron method, which is generally more robust, especially in aqueous solutions.

The Ion-Electron Method (Half-Reaction Method)

This method breaks the overall redox reaction into two parts: an oxidation half-reaction and a reduction half-reaction.

Steps for Balancing in Acidic Solution:

  1. Identify the species being oxidized and reduced. Assign oxidation numbers to determine this.
  2. Write the unbalanced half-reactions. Separate the overall reaction into two skeletal half-reactions, one for oxidation and one for reduction, including only the species that change oxidation states.
  3. Balance atoms other than O and H in each half-reaction.
  4. Balance oxygen atoms by adding $H_2O$ molecules to the side that needs oxygen.
  5. Balance hydrogen atoms by adding $H^+$ ions to the side that needs hydrogen.
  6. Balance the charges in each half-reaction by adding electrons ($e^-$). The electrons should be added to the more positive side (or less negative side) to make the charges equal on both sides of the half-reaction.
  7. Make the number of electrons equal in both half-reactions. Multiply one or both half-reactions by appropriate integers so that the number of electrons lost in oxidation equals the number of electrons gained in reduction.
  8. Add the balanced half-reactions together. The electrons should cancel out. Combine the remaining species from both half-reactions to form the overall balanced equation.
  9. Check the final equation. Ensure that the number of atoms of each element and the total charge are balanced on both sides.

Example: Balancing $MnO_4^-$ with $C_2O_4^{2-}$ in Acidic Solution

Overall unbalanced reaction: $MnO_4^-(aq) + C_2O_4^{2-}(aq) \rightarrow Mn^{2+}(aq) + CO_2(g)$ (Assume acidic conditions)

  1. Identify oxidation/reduction:
    • $MnO_4^-$: Mn is +7, $Mn^{2+}$: Mn is +2. Mn is reduced (gain of electrons).
    • $C_2O_4^{2-}$: C is +3 (avg, $2C + 4(-2) = -2 \implies 2C = +6 \implies C = +3$), $CO_2$: C is +4. C is oxidized (loss of electrons).
  2. Write unbalanced half-reactions:
    • Reduction: $MnO_4^- \rightarrow Mn^{2+}$
    • Oxidation: $C_2O_4^{2-} \rightarrow CO_2$
  3. Balance atoms other than O, H:
    • Reduction: $MnO_4^- \rightarrow Mn^{2+}$ (Mn is already balanced)
    • Oxidation: $C_2O_4^{2-} \rightarrow 2CO_2$ (Balance C)
  4. Balance O with $H_2O$:
    • Reduction: $MnO_4^- \rightarrow Mn^{2+} + 4H_2O$ (Add 4 $H_2O$ to the right)
    • Oxidation: $C_2O_4^{2-} \rightarrow 2CO_2$ (O is already balanced)
  5. Balance H with $H^+$:
    • Reduction: $MnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O$ (Add 8 $H^+$ to the left)
    • Oxidation: $C_2O_4^{2-} \rightarrow 2CO_2$ (H is balanced)
  6. Balance charge with $e^-$:
    • Reduction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
      • Left side charge: $-1 + 8(+1) + 5(-1) = -1 + 8 - 5 = +2$
      • Right side charge: $+2$
      • Charges are balanced.
    • Oxidation: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$
      • Left side charge: $-2$
      • Right side charge: $2(0) + 2(-1) = -2$
      • Charges are balanced.
  7. Equalize electrons:
    • Multiply reduction half-reaction by 2 (to get 10 $e^-$).
    • Multiply oxidation half-reaction by 5 (to get 10 $e^-$).
    • Reduction (x2): $2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$
    • Oxidation (x5): $5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$
  8. Add half-reactions:
    • Combine the multiplied half-reactions:
    • $(2MnO_4^- + 16H^+ + 10e^-) + (5C_2O_4^{2-}) \rightarrow (2Mn^{2+} + 8H_2O) + (10CO_2 + 10e^-)$
    • Cancel out electrons:
    • $2MnO_4^- + 16H^+ + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 8H_2O + 10CO_2$
  9. Check balance:
    • Atoms:
    • Mn: 2 on left, 2 on right (Balanced)
    • O: $(2 \times 4) + (5 \times 4) = 8 + 20 = 28$ on left. $8 + (10 \times 2) = 8 + 20 = 28$ on right. (Balanced)
    • H: $16 \times 1 = 16$ on left. $8 \times 2 = 16$ on right. (Balanced)
    • C: $5 \times 2 = 10$ on left. $10 \times 1 = 10$ on right. (Balanced)
    • Charge:
    • Left side: $2(-1) + 16(+1) + 5(-2) = -2 + 16 - 10 = +4$
    • Right side: $2(+2) + 8(0) + 10(0) = +4$ (Balanced)

The balanced equation in acidic solution is: $2MnO_4^-(aq) + 16H^+(aq) + 5C_2O_4^{2-}(aq) \rightarrow 2Mn^{2+}(aq) + 8H_2O(l) + 10CO_2(g)$

Steps for Balancing in Basic Solution:

Balance the reaction in acidic solution first, using the steps above. Then, neutralize the $H^+$ ions by adding an equal number of $OH^-$ ions to *both* sides of the equation. Any $H^+$ and $OH^-$ on the same side combine to form $H_2O$. Simplify by canceling out excess $H_2O$ molecules.

Example: Balancing $Cr_2O_7^{2-}$ with $S^{2-}$ in Basic Solution

Overall unbalanced reaction: $Cr_2O_7^{2-}(aq) + S^{2-}(aq) \rightarrow Cr(OH)_3(s) + S(s)$ (Assume basic conditions)

  1. Acidic Balancing (Simplified):
    • Identify: $Cr_2O_7^{2-}$ (Cr +6 to +3, reduction), $S^{2-}$ (S -2 to 0, oxidation)
    • Half-reactions:
    • Reduction: $Cr_2O_7^{2-} \rightarrow Cr(OH)_3$
    • Oxidation: $S^{2-} \rightarrow S$
    • Balancing these leads to (after several steps):
    • Reduction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr(OH)_3 + 4H_2O$
    • Oxidation: $S^{2-} \rightarrow S + 2e^-$
    • Equalizing electrons (multiply oxidation by 3):
    • Reduction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr(OH)_3 + 4H_2O$
    • Oxidation (x3): $3S^{2-} \rightarrow 3S + 6e^-$
    • Add: $Cr_2O_7^{2-} + 14H^+ + 3S^{2-} \rightarrow 2Cr(OH)_3 + 4H_2O + 3S$
  2. Convert to Basic Solution:
    • We have $14H^+$ on the left. Add $14OH^-$ to *both* sides.
    • $(Cr_2O_7^{2-} + 14H^+ + 14OH^- + 3S^{2-}) \rightarrow (2Cr(OH)_3 + 4H_2O + 3S + 14OH^-)$
    • Combine $H^+$ and $OH^-$ to form $H_2O$: $14H^+ + 14OH^- = 14H_2O$.
    • $Cr_2O_7^{2-} + 14H_2O + 3S^{2-} \rightarrow 2Cr(OH)_3 + 4H_2O + 3S + 14OH^-$
  3. Simplify $H_2O$:
    • Cancel $4H_2O$ from the left side with $4H_2O$ from the right side, leaving $10H_2O$ on the left.
    • $Cr_2O_7^{2-} + 10H_2O + 3S^{2-} \rightarrow 2Cr(OH)_3 + 3S + 14OH^-$
  4. Check balance:
    • Atoms:
    • Cr: 2 on left, 2 on right (Balanced)
    • O: $7 + 10 = 17$ on left. $(2 \times 3) + 14 = 6 + 14 = 20$ on right. (Uh oh, mistake somewhere or intermediate product incorrect)
Important Note on Basic Balancing: Balancing in basic solution can sometimes be tricky if the product species are not clearly defined or if there are multiple possible products. The key is to get the acidic balance correct first and then systematically convert. Let's re-evaluate the basic balancing step.

Let's try balancing $Cr_2O_7^{2-}$ with $S^{2-}$ in basic solution using the half-reaction method directly, assuming the products are $CrO_2^-$ and $S$. (Note: $Cr(OH)_3$ is often formed in neutral/basic conditions, but $CrO_2^-$ is also plausible and easier for balancing demonstration here).

Unbalanced: $Cr_2O_7^{2-} + S^{2-} \rightarrow CrO_2^- + S$ (Basic solution)

  1. Identify: $Cr (+6 \rightarrow +3$, reduction), $S (-2 \rightarrow 0$, oxidation)
  2. Half-reactions:
    • Reduction: $Cr_2O_7^{2-} \rightarrow CrO_2^-$
    • Oxidation: $S^{2-} \rightarrow S$
  3. Balance atoms (not O, H):
    • Reduction: $Cr_2O_7^{2-} \rightarrow 2CrO_2^-$
    • Oxidation: $S^{2-} \rightarrow S$
  4. Balance O with $H_2O$ and $OH^-$ in basic solution: Add $H_2O$ to the side deficient in O, and $2OH^-$ to the other side.
    • Reduction: $Cr_2O_7^{2-} + 2H_2O \rightarrow 2CrO_2^- + 4OH^-$
    • Oxidation: $S^{2-} \rightarrow S$
  5. Balance H with $OH^-$: Add $OH^-$ to the side deficient in H.
    • Reduction: $Cr_2O_7^{2-} + 2H_2O + 6OH^- \rightarrow 2CrO_2^- + 4OH^-$
    • Wait, this isn't right. The rule for basic solutions is:
      1. Balance as if in acid (using $H_2O$ for O, $H^+$ for H).
      2. Add $OH^-$ to both sides equal to the number of $H^+$.
      3. Combine $H^+$ and $OH^-$ to form $H_2O$.
      4. Cancel excess $H_2O$.

Let's go back to the acidic balanced equation: $Cr_2O_7^{2-} + 14H^+ + 3S^{2-} \rightarrow 2Cr^{3+} + 4H_2O + 3S$ (Assuming $Cr^{3+}$ as intermediate product for simplicity in acidic)

Now convert to basic:

  1. Add $14OH^-$ to both sides:
  2. $Cr_2O_7^{2-} + 14H^+ + 14OH^- + 3S^{2-} \rightarrow 2Cr^{3+} + 4H_2O + 3S + 14OH^-$
  3. Combine $H^+$ and $OH^-$:
  4. $Cr_2O_7^{2-} + 14H_2O + 3S^{2-} \rightarrow 2Cr^{3+} + 4H_2O + 3S + 14OH^-$
  5. Cancel $4H_2O$:
  6. $Cr_2O_7^{2-} + 10H_2O + 3S^{2-} \rightarrow 2Cr^{3+} + 3S + 14OH^-$
  7. Check charge: Left = $-2 + 0 + 3(-2) = -8$. Right = $2(+3) + 0 + 14(-1) = +6 - 14 = -8$. (Charge balanced)
  8. Check atoms: Cr=2, S=3, H=20, O=7+10=17. Right: Cr=2, S=3, H=14, O=14. (Atoms not balanced - this indicates $Cr^{3+}$ is not the final product in basic solution. It likely forms $Cr(OH)_3$ or similar).
Key Strategy for Basic Solutions: Always balance in acid first, then convert to basic. If the products given in the question are specific (like $Cr(OH)_3$), use those. The conversion step from $Cr^{3+}$ to $Cr(OH)_3$ needs to be handled carefully. For instance, $Cr^{3+} + 3OH^- \rightarrow Cr(OH)_3$. If we have $2Cr^{3+}$ and $14OH^-$ on the right, we'd form $2Cr(OH)_3$ and have $14 - 2 \times 3 = 14 - 6 = 8 OH^-$ remaining.

Let's re-attempt basic balancing with the correct products $Cr(OH)_3$ and $S$.

Unbalanced: $Cr_2O_7^{2-} + S^{2-} \rightarrow Cr(OH)_3 + S$ (Basic solution)

  1. Acidic Balance (as before): $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ $S^{2-} \rightarrow S + 2e^-$ Multiply reduction by 1, oxidation by 3: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ $3S^{2-} \rightarrow 3S + 6e^-$ Add: $Cr_2O_7^{2-} + 14H^+ + 3S^{2-} \rightarrow 2Cr^{3+} + 7H_2O + 3S$
  2. Convert $Cr^{3+}$ to $Cr(OH)_3$: Add $3OH^-$ for each $Cr^{3+}$: $2Cr^{3+} + 6OH^- \rightarrow 2Cr(OH)_3$ So, the reaction becomes: $Cr_2O_7^{2-} + 14H^+ + 6OH^- + 3S^{2-} \rightarrow 2Cr(OH)_3 + 7H_2O + 3S$
  3. Combine $H^+$ and $OH^-$: $14H^+ + 6OH^- \rightarrow 6H_2O + 8H^+$ The equation is now: $Cr_2O_7^{2-} + 8H^+ + 6H_2O + 3S^{2-} \rightarrow 2Cr(OH)_3 + 7H_2O + 3S$
  4. Cancel $H_2O$: Cancel $6H_2O$ from both sides, leaving $1H_2O$ on the right. $Cr_2O_7^{2-} + 8H^+ + 3S^{2-} \rightarrow 2Cr(OH)_3 + H_2O + 3S$
  5. This is still not basic! The $H^+$ must be removed. The correct conversion step is to add $OH^-$ to both sides equal to the number of $H^+$. Back to: $Cr_2O_7^{2-} + 14H^+ + 3S^{2-} \rightarrow 2Cr^{3+} + 7H_2O + 3S$ Add $14OH^-$ to both sides: $Cr_2O_7^{2-} + 14H^+ + 14OH^- + 3S^{2-} \rightarrow 2Cr^{3+} + 7H_2O + 3S + 14OH^-$ Combine $H^+$ and $OH^-$: $Cr_2O_7^{2-} + 14H_2O + 3S^{2-} \rightarrow 2Cr^{3+} + 7H_2O + 3S + 14OH^-$ Cancel $7H_2O$: $Cr_2O_7^{2-} + 7H_2O + 3S^{2-} \rightarrow 2Cr^{3+} + 3S + 14OH^-$ Now, convert $2Cr^{3+}$ to $2Cr(OH)_3$ by adding $6OH^-$: $Cr_2O_7^{2-} + 7H_2O + 3S^{2-} + 6OH^- \rightarrow 2Cr(OH)_3 + 3S + 14OH^-$ Cancel $6OH^-$: $Cr_2O_7^{2-} + 7H_2O + 3S^{2-} \rightarrow 2Cr(OH)_3 + 3S + 8OH^-$
  6. Final Check: Atoms: Cr=2, O=7+7=14, H=14, S=3. Right: Cr=2, O=(2*3)+8=14, H=(2*3)+8=14, S=3. (Balanced) Charge: Left = $-2 + 0 + 3(-2) = -8$. Right = $0 + 0 + 8(-1) = -8$. (Balanced)

The balanced equation in basic solution is: $Cr_2O_7^{2-}(aq) + 7H_2O(l) + 3S^{2-}(aq) \rightarrow 2Cr(OH)_3(s) + 3S(s) + 8OH^-(aq)$

Common Oxidizing and Reducing Agents

Memorizing common oxidizing and reducing agents can be very helpful for predicting reaction outcomes.

Common Oxidizing Agents (Gain electrons, get reduced):

  • Halogens: $F_2, Cl_2, Br_2, I_2$
  • Oxygen: $O_2$
  • Ozone: $O_3$
  • Peroxides: $H_2O_2$ (can be oxidizing or reducing)
  • Acids: Concentrated $HNO_3$, Concentrated $H_2SO_4$
  • Salts of strong oxidizing agents: $KMnO_4$ (permanganate), $K_2Cr_2O_7$ (dichromate), $KIO_3$ (iodate)
  • Metal ions in high oxidation states: $Fe^{3+}, Sn^{4+}$

Common Reducing Agents (Lose electrons, get oxidized):

  • Active metals: $Na, K, Ca, Mg, Al, Zn, Fe$
  • Non-metals in low oxidation states: $H_2, C, S^{2-}, SO_3^{2-}, I^-$
  • Hydrogen Peroxide: $H_2O_2$ (can be oxidizing or reducing)
  • Oxoanions in low oxidation states: $NO_2^-, S_2O_3^{2-}$ (thiosulfate)
Trick for remembering Oxidation States of common elements in Oxoacids/Oxoanions: For elements like S, P, N, Cl in their common oxoanions: * Sulfur: $SO_3^{2-}$ (S=+4), $SO_4^{2-}$ (S=+6) * Phosphorus: $PO_3^{3-}$ (P=+3), $PO_4^{3-}$ (P=+5) * Nitrogen: $NO_2^-$ (N=+3), $NO_3^-$ (N=+5) * Chlorine: $ClO^-$ (Cl=+1), $ClO_2^-$ (Cl=+3), $ClO_3^-$ (Cl=+5), $ClO_4^-$ (Cl=+7) The oxidation states generally increase with the number of oxygen atoms.

Disproportionation Reactions

A special type of redox reaction where the *same* element is simultaneously oxidized and reduced.

Example: Decomposition of Hydrogen Peroxide

$2H_2O_2(aq) \rightarrow 2H_2O(l) + O_2(g)$

  • In $H_2O_2$, Oxygen has an oxidation state of -1.
  • In $H_2O$, Oxygen has an oxidation state of -2 (reduced).
  • In $O_2$, Oxygen has an oxidation state of 0 (oxidized).

The same element (oxygen) is both oxidized and reduced.

Example: Reaction of Chlorine with Water

$Cl_2(g) + H_2O(l) \rightleftharpoons HCl(aq) + HClO(aq)$

  • In $Cl_2$, Cl is 0.
  • In $HCl$, Cl is -1 (reduced).
  • In $HClO$ (hypochlorous acid), Cl is +1 (oxidized).

This reaction is reversible and is more complex in basic solutions.

Comproportionation Reactions

The opposite of disproportionation, where two different species containing the same element in different oxidation states react to form a single product where that element is in an intermediate oxidation state.

Example: Reaction of $I^-$ and $IO_3^-$

$5I^-(aq) + IO_3^-(aq) + 6H^+(aq) \rightarrow 3I_2(s) + 3H_2O(l)$

  • Iodine in $I^-$ is -1.
  • Iodine in $IO_3^-$ is +5.
  • Iodine in $I_2$ is 0 (intermediate oxidation state).

Here, iodine in two different oxidation states (-1 and +5) combine to form iodine in a single, intermediate oxidation state (0).

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