Redox and Equivalent ConceptHard
Question
Calculate equivalent weight of KHC2O4 (potassium acid oxalate) when :-
(a) It is neutralized by KOH
(b) It is oxidised by MnO4- in basic medium
(a) It is neutralized by KOH
(b) It is oxidised by MnO4- in basic medium
Options
A.128, 128
B.64, 128
C.64, 64
D.128, 64
Solution
(a) KHC2O4 + KOH → K2C2O4 + H2O
∴ eq.ωt = M/1 = 128
(b) KHC2O4 + MnO4- → MnO2 + 2CO2
↑ ↑ ↑ ↑
+6 +7 +4 +8
∴ eq. ωt of KHC2O4 = M/2 = 64
∴ eq.ωt = M/1 = 128
(b) KHC2O4 + MnO4- → MnO2 + 2CO2
↑ ↑ ↑ ↑
+6 +7 +4 +8
∴ eq. ωt of KHC2O4 = M/2 = 64
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