11th Class Chemistry Chapter 10 Electro Chemistry Short Questions Answer
11th Class Chemistry Chapter 10 Electro Chemistry Short Question Answer
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The oxidation number is the apparent charge positive or negative which an atom has in a molecule
K +Mn +40 = 0
+1 + Mn +4 (-2) = 0
Mn – 7 = 0
Mn = +7
Mn + 20 = 0
Mn + 2(-2) = 0
Mn – 4 = 0
Mn = +4
Zn(s)/Zn+2(aq) (1M) ‖ Cu+2(aq) (1M) │Cu(s)
Zn+2(aq) + 2e-1 →Zn(s)
Cu(s) → Cu+2(aq)+2e-1
The electrodes above She in electrochemical series have negative reduction potential and positive oxidation potential. They have greater tendency to deliver the electrons to SHE.
Zn → Zn+2 + 2e– (oxidation at anode)
2H+ + 2e– → H2 (reduction at cathode)
The electrodes below SHE in electrochemical series have negative oxidation potential. They accept the electrons from SHE.
Cu+2 + 2e– → Cu0 (reduction at anode)
H2 → 2H+ + 2e– (oxidation at cathode)
Zn2+ + 2e– → Zn0 , is – 0.76 V. When this reaction is reversed, then it is an oxidation half reaction.
Zn → Zn+2 + 2e– E0 = +0.76V
The value of oxidation potential will be the same but sign is reversed. i.e., + 0. 76V.
At anode Pb(s) → Pb+2(aq) + 2e– (oxidation)
At cathode PbO2(s) + 4H+(aq) + 2e– → Pb+(aq) + 2H2O(2) (reduction)
2PbSO4(s) + 2H2O(1) → Pb(s) + PbO2(s) + 4H+(aq)+ 2SO42-(aq)
Hence lead accumulator is a chargeable battery.
At anode Zn(s) + 2H2O(aq) + 2e– → Zn(oh)(s) + 2e– (oxidation)
At cathode 2MnO4 + 2H2O(aq) + 2e– → 2MnO(OH)(s) + 2OH–(aq) (reduction)
At anode Zn(s) + 2OH+(aq) →Zn (OH)2(s) + 2e– (oxidation)
At cathode Ag2O + H2O(aq) + 2e– → 2Ag(s) + 2OH–(aq) (reduction)
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