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{:("Column I","Column II"),((A)underset(3.5"mole")(Sn^(+2))+underset(1.2"mole")(MnO_4^(-)("acidic")),(p)"Amount of oxidant available decides the number of electrons transfer"),((B)underset(8.4"mole")(H_2C_2O_4)+underset(3.6"mole")(MnO_4^(-)("acidic")),(q)"Amount of reductant available decides the number of electrons transfer"),((C )underset(7.2"mole")(S_2O_3^(-2))+underset(3.6"mole")(l_2),(r)"Number of electrons involved per mole of oxidant gt Number of electrons involved per mole of reductant"),((D )underset(9.2"mole")(Fe^(+2))+underset(1.6"mole")(Cr_2O_7^(-2)"(acidic)"),(s)"Number of electrons involved per mole of oxidant lt Number of electrons involved per mole of reductant"):} |
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Answer» Eq of `MnO_4^(-)` =Moles xv.f=1.2x5=6 Since `MnO_4^(-)`(O.A) is the LR, so the amount of oxidant available DECIDES the number of ELECTRON transfer. Also, electron involved per mole of OA `(5)gt` electron involved per mole of RA(2). (B)Eq of `H_2C_2O_4`=Moles x v.f. =8.4x2=16.8 Eq of `MnO_4^(-)` =Moles x v.f=3.6x5=18 Since `H_2C_2O_4`(RA) is the LR, so the amount of reductantavailable decides the number of electron transfer. Also, electron involved per mole of `OA (5)gt` electron involved per mole of RA(2). (C )Eq of `S_2O_3^(2-)`=Moles x v.f. =7.2x1=7.2 Eq of `I_2` =Moles x v.f=3.6x2=7.2 Since `S_2O_3^(2-)`(RA)and `I_2` (OA) both completely get consumed, so both the amount of REDUCTANT and oxidantdecides the number of electron transfer. Also, electron involved per mole of `OA (2)gt` electron involved per mole of RA(1). (D)Eq of `Fe^(2+)`=Moles x v.f. =9.2x1=9.2 Eq of `Cr_2O_7^(2-)` =Moles x v.f=1.6x6=9.6 Since `Fe^(2+)`(RA) is the LR, so the amount of reductantavailable decides the number of electron transfer. Also, electron involved per mole of `OA (6)gt` electron involved per mole of RA(1). |
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