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+ (aq) = +2 (positive sign goes first)
- S2− (aq) = -2 (negative sign goes first)
- Mn2+ (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: Zn = +2, I = -1
- Ag<em>2CO</em>3: Ag = +1, CO3??? (explained later)
Examples: Writing Oxidation Numbers
- Write the oxidation number (state) under each term in the chemical reaction.
- Ex 1: 3Mg(s) + 2Cr3+ (aq) → 3Mg2+ (aq) + 2Cr (s)
- Ex 2: 8Cl<em>2 (g) + 8K</em>2S (aq) → S8 (s) + 16 KCl (aq)
Answers to Examples
- Ex 1: 3Mg(s) + 2Cr3+ (aq) → 3Mg2+ (aq) + 2Cr (s)
- Ex 2: 8Cl<em>2 (g) + 8K</em>2S (aq) → S8 (s) + 16 KCl (aq)
Skill #2 – 2nd Set of Rules: Complete Compounds, Polyatomic Ions
D. Complete Compounds (not Ions)
- O.N. = 0
- Examples:
- NaCl = 0
- Li<em>2CO</em>3 = 0
- C<em>6H</em>12O6 = 0
E. Polyatomic Ions
- O.N. = the regular ionic charge (either isolated or as part of an ionic compound)
- Examples:
- SO4 2− = -2 (overall, for the whole ion)
- NH4 1+ = +1
- NO<em>3 in Cu(NO</em>3)2 = -1
F. Special Case: Oxygen Atoms within Compounds
- Each atom of oxygen, O.N. = -2
- Examples:
- Within CrO4 2−, each O = -2
- Within H2O, O = -2
- Exception: O within O2 (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>2PO</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:
- Take the known O.N. for the compound (or ion).
- Assign O.N. for O, H, and monatomic ions using previous rules.
- 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(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_3]
- Reducing agent = Cu (s)
- NO_3^{1-}(aq) is a spectator ion (does nothing).
- Ex2: NaCl(aq) + AgNO3(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: C4H8(g)+5O2(aq)→3CO2(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 C4H8, 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^-
- C) Cu^{2+}(aq)+2e^- → Cu (s)
- D) Cu^{2+}(aq)becauseit′sthesubstancebeingreduced.Thereactionwasspontaneous,meaningCu^{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^{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.