Redox Reactions and Oxidation Numbers

4.3 Identifying Redox Reactions

Skill #1 – Oxidation Numbers for Elements & Monatomic Ions

  • Simple redox reactions are easy to identify by observing which substance becomes more positive (oxidized) and which becomes less positive (reduced).
  • More complex reactions require assigning "Oxidation Numbers" (States) to identify the two half-reactions.
Oxidation Number (O.N.)
  • Definition: A number indicating how many electrons a substance has lost or gained compared to its isolated state.
1st Set of Rules: Isolated Elements & Monatomic Ions
A. Isolated Elements
  • O.N. = 0
  • Examples: Ca(s), Hg(l), F2(g), H2(g), O2(g) all have O.N. = 0
  • Note: Oxygen (O) and Hydrogen (H) are special cases (explained later).
B. Monatomic Ions (and Multivalent Ions) in Solution, Isolated
  • O.N. = ionic charge as given in the question
  • Examples:
    • Cu2+^{2+} (aq) = +2 (positive sign goes first)
    • S2^{2-} (aq) = -2 (negative sign goes first)
    • Mn2+^{2+} (aq) = +2
C. Monatomic Ions (not Multivalent) as Part of a Compound
  • O.N. = ionic charge as given on the Periodic Table
  • Examples:
    • NaCl: Na = +1, Cl = -1
    • ZnI2_2: Zn = +2, I = -1
    • Ag<em>2<em>2CO</em>3</em>3: Ag = +1, CO3_3??? (explained later)
Examples: Writing Oxidation Numbers
  • Write the oxidation number (state) under each term in the chemical reaction.
  • Ex 1: 3Mg(s) + 2Cr3+^{3+} (aq) → 3Mg2+^{2+} (aq) + 2Cr (s)
  • Ex 2: 8Cl<em>2<em>2 (g) + 8K</em>2</em>2S (aq) → S8_8 (s) + 16 KCl (aq)
Answers to Examples
  • Ex 1: 3Mg(s) + 2Cr3+^{3+} (aq) → 3Mg2+^{2+} (aq) + 2Cr (s)
    • 0 +3 +2 0
  • Ex 2: 8Cl<em>2<em>2 (g) + 8K</em>2</em>2S (aq) → S8_8 (s) + 16 KCl (aq)
    • 0 +1,-2 0 +1,-1

Skill #2 – 2nd Set of Rules: Complete Compounds, Polyatomic Ions

D. Complete Compounds (not Ions)
  • O.N. = 0
  • Examples:
    • NaCl = 0
    • Li<em>2<em>2CO</em>3</em>3 = 0
    • C<em>6<em>6H</em>12</em>{12}O6_6 = 0
E. Polyatomic Ions
  • O.N. = the regular ionic charge (either isolated or as part of an ionic compound)
  • Examples:
    • SO4_4 2^{2-} = -2 (overall, for the whole ion)
    • NH4_4 1+^{1+} = +1
    • NO<em>3<em>3 in Cu(NO</em>3</em>3)2_{2} = -1
F. Special Case: Oxygen Atoms within Compounds
  • Each atom of oxygen, O.N. = -2
  • Examples:
    • Within CrO4_4 2^{2-}, each O = -2
    • Within H2_2O, O = -2
  • Exception: O within O2_2 (g) has O.N. = 0 (Rule A: Isolated elements)
G. Special Case: Hydrogen Atoms within Compounds
  • Each atom of hydrogen, O.N. = +1
  • Examples:
    • Within KOH, H = +1
    • Within H<em>2<em>2PO</em>3</em>3$^- , each H = +1
  • Exception: H_2 (g) has O.N. = 0 (Rule A: Isolated elements)
H. The Tricky One: Elements (other than O & H) in Polyatomic Ions
  • No direct rule! You have to solve for it.
  • It varies depending on the compound.
  • Procedure:
    1. Take the known O.N. for the compound (or ion).
    2. Assign O.N. for O, H, and monatomic ions using previous rules.
    3. Solve for the O.N. of the remaining element.
  • Example: AgNO_3 … solve for N
    • Mark in O.N. for each element: +1 x -2
    • Set up equation & solve: 1(+1) + 1(x) + 3(-2) = 0
    • 1 + x - 6 = 0
    • x = +5 for N (in THIS compound!)
Examples: Calculating O.N. in Polyatomic Ions
  • Ex2: Calculate O.N. for N in NH_3
    • 1(x) + 3(1) = 0
    • X = -3 in THIS compound
  • Ex3: S in the sulfate ion SO_4^{2-}
    • 1(x) + 4(-2) = -2
    • X = +6 for S in THIS ion

Skill #3: Using O.N. to Determine if a Reaction is Redox or Not

  • Redox reactions involve the transfer of electrons from one element to another.
  • If two elements undergo changes in O.N., then the reaction is a redox reaction.
  • If none of the elements experience a change in O.N., then the reaction is not a redox reaction.
  • Procedure: Mark the O.N. for all elements before & after the reaction and compare.
The "Usual Four"
  • The element whose O.N. increases is being oxidized (and is the reducing agent).
  • The element whose O.N. decreases is being reduced (and is the oxidizing agent).
Examples: Identifying Redox Reactions and the "Usual Four"
  • Ex1: Cu (s) + 2 AgNO3(aq)2Ag(s)+Cu(NO(aq) → 2 Ag (s) + Cu(NO3))_{2} (aq)
    • 0 +1+5-2 0 +2 +5 -2
    • Cu goes from O.N. = 0 to +2… Increase = oxidized
    • Ag goes from O.N. = +1 to 0… Decrease = reduced.
    • Changes in O.N. indicate that the reaction IS redox.
    • Substance oxidized = Cu(s)
    • Substance reduced = Ag^+ (aq)
    • Oxidizing agent = Ag^+(aq)[orAgNO(aq) [or AgNO_3]
    • Reducing agent = Cu (s)
    • NO_3^{1-}(aq) is a spectator ion (does nothing).
  • Ex2: NaCl(aq) + AgNO3(aq)NaNO(aq) → NaNO3(aq) + AgCl(s)
    • +1-1 +1+5-2 +1+5-2 +1-1
    • Is it redox? No.
    • Why not? Because none of the oxidation numbers change. Nothing is oxidized (and so nothing is reduced).
  • Ex3: C4HH8(g)+5O(g) + 5O2(aq)3CO(aq) → 3CO2(g)+4H(g) + 4H_2O(g)
    • -2+1 0 +4-2 +1-2
    • Is it redox? Yes
    • Why? Because the O.N. of C & O have changed!
    • Substance oxidized = C
    • Substance reduced = O
    • Oxidizing agent = O (or O_2, I suppose..)
    • Reducing agent = C (or C4HH8, I suppose…)

Skill #4: Spontaneity of Redox Reactions

Example 1
  • If a strip of Zn (s) is placed in a 0.1 mol/L solution of Cu(NO3)){2}, after 5 minutes orange-brown deposits have formed on the surface of the Zn (s) and the blue-colour intensity of the solution has faded.
    • A) Was the redox reaction spontaneous or non-spontaneous?
    • B) What was the oxidation half-reaction?
    • C) What was the reduction half-reaction?
    • D) What was the oxidizing agent?
Answers
  • A) Spontaneous
  • B) Zn (s) → Zn^{2+}(aq)+2e(aq) + 2e^-
  • C) Cu^{2+}(aq)+2e(aq) + 2e^- → Cu (s)
  • D) Cu^{2+}(aq)becauseitsthesubstancebeingreduced.Thereactionwasspontaneous,meaningCu(aq) because it's the substance being reduced. The reaction was spontaneous, meaning Cu^{2+} (aq) is a relatively strong oxidizing agent.
Example 2
  • Another strip of Zn (s) is now placed in a 0.1 mol/L solution of Al(NO3)){3}. After 5 minutes nothing has happened!
    • A) Was the redox reaction spontaneous or non-spontaneous?
    • B) What would have been the oxidizing agent?
    • C) Between Ex1 & Ex2, which was the stronger oxidizing agent?
Answers
  • A) Non-spontaneous
  • B) Al^{3+} (aq)
  • C) Cu^{2+}(aq)isastrongeroxidizingagent(wasabletooxidizeZn)thanAl(aq) is a stronger oxidizing agent (was able to oxidize Zn) than Al^{3+}$$ (aq) (was not able to oxidize Zn)
    • This leads us to the S.R.P. data sheet, which lists oxidizing agents in order of increasing oxidizing strength.