Electrochemistry worksheet covering oxidation numbers, redox reactions, and identifying oxidizing and reducing agents.

Let’s go step by step through each question. I’ll explain clearly so you can follow along and understand how to find oxidation numbers, identify redox reactions, and spot oxidizing/reducing agents.

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Question 1: Find oxidation number of the bold atom

We use these rules:
- Element alone (like S₈) → oxidation number = 0
- For ions, total charge = sum of oxidation numbers
- Oxygen is usually -2 (except in peroxides like H₂O₂ or with fluorine)
- Hydrogen is +1 when bonded to nonmetals, -1 when bonded to metals
- Alkali metals (Group 1) are always +1
- Halogens (F, Cl, Br, I) are usually -1 unless with oxygen or more electronegative element

Let’s solve each:

a) S₈
Sulfur is an element by itself → oxidation number = 0

b) NO₂⁻
Bold is N. Let x = oxidation number of N.
Oxygen = -2 each → 2 × (-2) = -4
Total charge = -1
So: x + (-4) = -1 → x = +3

c) CrO₄²⁻
Bold is Cr. Let x = Cr
Oxygen = -2 × 4 = -8
Total charge = -2
x + (-8) = -2 → x = +6

d) Cr₂O₇²⁻
Bold is Cr. Two Cr atoms → let each be x
Oxygen = -2 × 7 = -14
Total charge = -2
2x + (-14) = -2 → 2x = 12 → x = +6

e) KMnO₄
Bold is Mn. K = +1, O = -2 × 4 = -8
Total compound neutral → +1 + x + (-8) = 0 → x = +7

f) Na₃⁻
Wait — this looks odd. Sodium doesn’t form Na₃⁻ ion. Probably typo? Maybe it’s N³⁻? But it says “Na₃⁻” — if we take it literally:
Three Na atoms, total charge -1. Each Na normally +1, but here they’re in a weird ion.
Assume all Na same oxidation state: 3x = -1 → x = -1/3
But that’s unusual. Likely meant N³⁻ (nitride ion). If it’s N³⁻, then N = -3
I think it’s a typo — probably N³⁻, so answer = -3

*(Note: In real chemistry, Na₃⁻ isn’t standard. We’ll assume it’s N³⁻ for sense.)*

g) C₂H₆
Bold is C. H = +1 each → 6 × (+1) = +6
Compound neutral → 2x + 6 = 0 → 2x = -6 → x = -3

h) KHSO₃
Bold is S. K = +1, H = +1, O = -2 × 3 = -6
Total neutral: +1 +1 + x + (-6) = 0 → x -4 = 0 → x = +4

i) Al(OH)₄⁻
Bold is Al. OH group: O = -2, H = +1 → each OH = -1
Four OH groups → 4 × (-1) = -4
Total charge = -1
So: x + (-4) = -1 → x = +3

j) S₂F₁₀
Bold is S. F = -1 each → 10 × (-1) = -10
Neutral molecule → 2x + (-10) = 0 → 2x = 10 → x = +5

k) N₂O₃
Bold is N. O = -2 × 3 = -6
Neutral → 2x + (-6) = 0 → 2x = 6 → x = +3

l) Na₂MnO₄
Bold is Mn. Na = +1 × 2 = +2, O = -2 × 4 = -8
Neutral: +2 + x + (-8) = 0 → x -6 = 0 → x = +6

m) N₂H₅⁺
Bold is N. H = +1 × 5 = +5
Total charge = +1
2x + 5 = 1 → 2x = -4 → x = -2

n) Li₂O₂
Bold is O. This is lithium peroxide. Peroxide rule: O = -1 each
Check: Li = +1 × 2 = +2, O₂ = 2 × (-1) = -2 → total 0 ✔️

o) HCO₃⁻
Bold is C. H = +1, O = -2 × 3 = -6
Total charge = -1
+1 + x + (-6) = -1 → x -5 = -1 → x = +4

p) K₂UO₄
Bold is U. K = +1 × 2 = +2, O = -2 × 4 = -8
Neutral: +2 + x + (-8) = 0 → x -6 = 0 → x = +6

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Question 2: Assign oxidation numbers to bold species and say which undergoes oxidation

Oxidation = increase in oxidation number (loss of electrons)

a) ClO₂ + C → ClO₂⁻ + CO²⁻

Bold: Cl in ClO₂, C in C, Cl in ClO₂⁻, C in CO₃²⁻

Cl in ClO₂: O = -2 × 2 = -4 → Cl = +4
C in C (element) = 0
Cl in ClO₂⁻: O = -4, total charge -1 → Cl = +3
C in CO₃²⁻: O = -6, total -2 → C = +4

Changes:
Cl: +4 → +3 → reduced
C: 0 → +4 → oxidized ← this one undergoes oxidation

b) Sn²⁺ + Cl⁻ + BrO₃⁻ → SnCl₆²⁻ + Br⁻

Bold: Sn in Sn²⁺, Br in BrO₃⁻, Sn in SnCl₆²⁻, Br in Br⁻

Sn²⁺ = +2
Br in BrO₃⁻: O = -6, total -1 → Br = +5
Sn in SnCl₆²⁻: Cl = -1 × 6 = -6, total -2 → Sn = +4
Br in Br⁻ = -1

Changes:
Sn: +2 → +4 → oxidized ← oxidation
Br: +5 → -1 → reduced

c) MnO₄⁻ + C₂O₄²⁻ → MnO₂ + CO₂

Bold: Mn in MnO₄⁻, C in C₂O₄²⁻, Mn in MnO₂, C in CO₂

Mn in MnO₄⁻: O = -8, total -1 → Mn = +7
C in C₂O₄²⁻: O = -8, total -2 → 2x -8 = -2 → 2x=6 → x=+3
Mn in MnO₂: O = -4 → Mn = +4
C in CO₂: O = -4 → C = +4

Changes:
Mn: +7 → +4 → reduced
C: +3 → +4 → oxidized ← oxidation

d) NO₃⁻ + H₂Te → NO + TeO₄²⁻

Bold: N in NO₃⁻, Te in H₂Te, N in NO, Te in TeO₄²⁻

N in NO₃⁻: O = -6, total -1 → N = +5
Te in H₂Te: H = +1 × 2 = +2 → Te = -2
N in NO: O = -2 → N = +2
Te in TeO₄²⁻: O = -8, total -2 → Te = +6

Changes:
N: +5 → +2 → reduced
Te: -2 → +6 → oxidized ← oxidation

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Question 3: Which are redox reactions?

Redox = at least one element changes oxidation number.

a) I₂ + 5HOBr + H₂O → 2IO₃⁻ + 5Br⁻ + 7H⁺

I₂: 0 → IO₃⁻: I = +5 (since O=-6, total -1 → I=+5) → oxidized
Br in HOBr: H=+1, O=-2 → Br=+1 → Br⁻ = -1 → reduced
→ Redox ✔

b) 4Ag⁺ + Cr₂O₇²⁻ + H₂O → 2Ag₂CrO₄ + 2H⁺

Ag⁺ = +1 → Ag in Ag₂CrO₄: still +1 (chromate ion CrO₄²⁻, so Ag must be +1)
Cr in Cr₂O₇²⁻: 2x -14 = -2 → x=+6
Cr in CrO₄²⁻: x -8 = -2 → x=+6 → no change
No oxidation number change → NOT redox ✘

c) KHCO₃ + HI → KI + CO₂ + H₂O

K: +1 everywhere
H: +1 everywhere
C: in HCO₃⁻: H=+1, O=-6, total -1 → C=+4; in CO₂: C=+4 → no change
I: in HI = -1, in KI = -1 → no change
→ NOT redox ✘

d) 2H₂O → 2H₂ + O₂

H in H₂O: +1 → H₂: 0 → reduced
O in H₂O: -2 → O₂: 0 → oxidized
→ Redox ✔

e) H₂SO₄ + BaCl₂ → BaSO₄ + 2HCl

All ions just swap partners. No change in oxidation states.
S: +6 in both, Ba: +2, Cl: -1, H: +1, O: -2
→ NOT redox ✘

f) Fe + H₂SO₄ → FeSO₄ + H₂

Fe: 0 → +2 → oxidized
H: +1 → 0 → reduced
→ Redox ✔

✔ So redox reactions: a, d, f

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Question 4: Identify oxidizing and reducing agents

Oxidizing agent gets reduced (causes oxidation in other)
Reducing agent gets oxidized (causes reduction in other)

a) 2Na(s) + O₂(g) → Na₂O(s)

Na: 0 → +1 → oxidized → reducing agent
O: 0 → -2 → reduced → oxidizing agent

→ Oxidizing agent: O₂
→ Reducing agent: Na

b) Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s)

Fe: 0 → +2 → oxidized → reducing agent
Cu²⁺: +2 → 0 → reduced → oxidizing agent

→ Oxidizing agent: Cu²⁺
→ Reducing agent: Fe

c) Sn⁴⁺(aq) + Fe²⁺(aq) → Sn²⁺(aq) + Fe³⁺(aq)

Sn⁴⁺: +4 → +2 → reduced → oxidizing agent
Fe²⁺: +2 → +3 → oxidized → reducing agent

→ Oxidizing agent: Sn⁴⁺
→ Reducing agent: Fe²⁺

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Final Answer:

1. Oxidation Numbers:
a) 0
b) +3
c) +6
d) +6
e) +7
f) -3 (assuming N³⁻)
g) -3
h) +4
i) +3
j) +5
k) +3
l) +6
m) -2
n) -1
o) +4
p) +6

2. Species undergoing oxidation:
a) C
b) Sn²⁺
c) C in C₂O₄²⁻
d) Te in H₂Te

3. Redox reactions: a, d, f

4. Agents:
a) Oxidizing: O₂; Reducing: Na
b) Oxidizing: Cu²⁺; Reducing: Fe
c) Oxidizing: Sn⁴⁺; Reducing: Fe²⁺