InterviewSolution
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InterviewSolution
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Evaluate equivalent weight of reductant or oxidant given on left hand side of each reaction: (a) `As_(2)O_(3)+5H_(2)Orarr2AsO_(4)^(3-)+10H^(+)+4e` (b) `MnO_(4)^(-)+8H^(+)+5erarrMn^(2+)+4H_(2)O` (c ) `Cr_(2)O_(7)^(2-)+14H^(+)+6erarr2Cr^(3+)+7H_(2)O` (d) `C_(2)O_(4)^(2-)rarr2CO_(2)+2e` (e ) `FeC_(2)O_(4)rarrFe^(3+)+2CO_(2)+3e` (f) `2CuSO_(4)+2erarrCu_(2)^(1+)+SO_(4)^(2-)` |
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Answer» `E_("red"//"oxi")={:(ul("Mol.weight of reductant or oxidant")),("Number of electrons gained or lost by"),("one molecule of reductant or oxidant"),("or valence factor"):}` (a) `E_(As_(2)O_(3)) =(M_(As_(2)O_(3)))/(4)` `(As_(2)^(3+) rarr 2As^(5+) + 4e)` (b) `E_(MnO_($)^(-)) = (M_(MnO_(4)^(-)))/(5)` `(Mn^(7+) + 5e rarr Mn^(2+))` ( c) `E_(Cr_(2)O_(7)^(2-)) = (M_(Cr_(2)O_(7)^(2-)))/(6)` `(Cr_(2)^(6+) rarr 2Cr^(3+) + 6e)` (d) `E_(C_(2)O_(4)^(2-)) = (M_(C_(2)O_(4)^(2-)))/(2)` `(C_(2)^(3+) rarr 2C^(4+) + 2e)` (e) `E_(FeC_(2)O_(4)) = (M_(FeC_(2)O_(4)))/(3)` `(Fe^(2+)+C_(2)^(3+) rarr Fe^(3+) + 2C^(4+) + 3e)` (f) `E_(CuSO_(4)) = (M_CuSO_(4))/(1)` `(2Cu^(2+) + 2e rarr Cu_(2)^(+))` |
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