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Find the oxidation number of
a. S in SO42- ion
b. S in HSO3- ion
c. Pt in (PtCl6)2-
a. Let the oxidation state of S in SO42 = x
We know that ox. no. of O = -2
So, x + 4(-2) = -2
or x =+6
b. Let the oxidation state of S in HSO3- = x
& ox. no. of H = +1
So, +1 +x+3(-2) = -1
or x = +4
c. Let the oxidation state of Pt in (PtCl6)2- = x
We know that ox. no. of Cl = -1
So, x+ 6(-1) = -2
or x = -2 + 6 = +4
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What is the oxidation number of Mn is KMnO4 and of S in Na2S2O3?
Let the Ox.no. of Mn is KMnO4 be x.
We know that
Ox.no. of K = +1
Ox.no. of O = –2
So Ox.no. K + Os.No. Mn+4 (Ox.no.O) = 0
or +1+ x + (4–2) = 0
or +1+ x – 8 = 0
or x = +8 – 1 = +7
Hence, Ox.no. of Mn in KMnO4 is +7
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What is the oxidation number of Cr in K2Cr3O7 ?
Solution:
Let the oxidation number of Fe be x.
Ox.no. of K =+1
Ox.no. of (CN)– = –1
So 4(Ox.no. K) + Os.No. Fe+6(Ox.no. CN–) = 0
or 4(+1) + x + (6–1) = 0
or +4 + x – 6 = 0
or x = +6 – 4 = +2
The oxidation number of iron in K4Fe(CN)6 is +2.
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Which compound amongst the following has the highest oxidation number for Mn?
KMnO4, K2MnO4, MnO2 and Mn2O3
KMnO4, +1+x –8 = 0 ot x = + 7
Ox.no. of Mn =- +7
K2MnO4,
+2 +x –8 = 0
x = +6
MnO2
x – 4 = 0
Mn2O3
2x – 6 = 0
x = + 3
Thus, the highest oxidation number for the Mn is in KMnO4.
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Can oxidation number of any element in a compound ever be zero? Justify the answer.
Sometimes, oxidation numbers have such values which at first sight appear strange. For example, the oxidation number of carbon in cane sugar (C12H22O11), glucose (C6H12O6), dichloromethane, etc., is zero.
Cane Sugar (C12H22O11) Glucose (C6H12O6)
12×x+22×1+11(–2)=0 6×x+12×1+6(–2)=0
12x+22 – 22 = 0 6x+12–12=0
So x = 0 So x = 0
Dichloromethane (CH2Cl2)
x = 2 × 1 + 2(–1) = 0
x + 2 – 2 =0
So x = 0
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Balance the equation Cr(OH)3 + IO3 - → I - + CrO42-
Oxidation : Cr(OH)3 → CrO42- Reduction : IO3- → I-
Balancing oxidation reaction : Balancing atoms other than H and O
Cr(OH)3 → CrO42-
Balancing O; 2OH- + Cr(OH)3 → CrO42- + H2O
Balancing H; 3OH- + 2OH- + Cr(OH)3 → CrO42- + H2O + 3H2O
Balancing charge; 5OH- + Cr(OH)3 → CrO42- + 4H2O + 3e- (1)
Balancing reduction reaction :
Balancing O; IO3- + 3H2O → 6OH- + I-
Balance the charges 6e- + IO3- + 3H2O ® 6OH- + I- (2)
Multiply (1) by 2 and adding to (2)
2Cr(OH)3 + IO3- + 4OH- → 2CrO42- + 5H2O + I-.
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Balance the equation
Cr2O72- + C2O42- + H+ ® Cr+3 + CO2 + H2O
Reduction : Cr2O72- →Cr+3 Oxidation : C2O42- → CO2
Balancing atoms other than O and H : Cr2O72- → 2Cr+3
Reaction is taking place in acidic medium.
Balancing O
Cr2O72- → 2Cr+3 + 7H2O
Balancing H
14H+ + Cr2O72- → 2Cr+3 + 7H2O
Balancing charges, we get reduction half-reaction.
6e- + 14H+ + Cr2O72- → 2Cr+3 + 7H2O(1)
Balancing oxidation reaction :
Balancing C C2O42- →2CO2
Balancing O C2O42- → 2CO2
Balancing charges, we get oxidation half-reaction
C2O42- → 2CO2 + 2e- (2)
Multiplying (2) by 3 and adding to (1) (To cancel out the electrons)
Cr2O72- + 14H+ + 3C2O42- → 2Cr+3 + 7H2O + 6CO2
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One mole of N2H4 loses 10 mole electrons to form a new compound Y. Assuming that all the N2 appears in new compound, what is the oxidation state of Nitrogen in Y? (There is no change in the oxidation state of H)
N2H4 → (Y) + 10e-
( Y contains all N atoms)
i.e. (2N)x + 10e-
2x - (-4) = 10
x = +3
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You can also have a look at balancing redox reactions
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