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Chromium metal in basic medium is oxidised in air to give chromic tetrahydroxide anion. |
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Answer» Solution :a) This IONIC skeleton EQUATION is written as `Cr+O_(2)OVERSET(OH^(-))rarrCr(OH)_(4)^(-)` b) Writing oxidation numbers `overset(0)(Cr)+overset(0)(O_(2))rarr[overset(+3)(Cr)(overset-2)(O)overset(+1)(H))_(4)]^(-)` c) Locating atoms undergoing change in oxidation numbers `overset(0)(Cr)+overset(0)(O_(2))rarr[overset(+3)(Cr)(overset(-2)(O)H)_(4)]^(-)` d) Dividing the reaction into two halves andbalancing in acidic medium. separately Oxidation half - reaction : `Crrarr[Cr(OH)_(4)]^(-)` Step 1 : Balance oxygen atoms `Cr+4H_(2)Orarr[Cr(OH)_(4)]^(-)` Step 2 : Balance hydrogen atoms `Cr+4H_(2)O+4OH^(-)rarr[Cr(OH)_(4)]^(-)+4H_(2)O` Step 3 : Balance the CHARGE `Cr+4H_(2)O+4OH^(-)rarr[Cr(OH)_(4)]^(-)+4H_(2)O+3e^(-)"....(a)"` Reduction half - reaction : `O_(2)rarr[Cr(OH)_(4)]^(-)` Step 1 : Balance chromium atoms `O_(2)+Crrarr [Cr(OH)_(4)]^(-)` Step 2 : Balance oxygen atoms `O_(2)+Cr+2H_(2)Orarr[Cr(OH)_(4)]^(-)` Step 3 : Balance hydrogen atoms `O_(2)+Cr+2H_(2)rarr[Cr(OH)_(4)]^(-)` Step 4: Balance charnge `O_(2)+Cr+2H_(2)O+e^(-)rarr[Cr(OH)_(4)]^(-)"....(b)"` Step 5 : Equalising the electrons and adding the two half reactions `"eq (a)"XX1+"eq (b)"xx"3, we get"` `{:(Cr+4H_(2)O+4OH^(-)rarr),(""[Cr(OH)_(4)]^(-)+4H_(2)O+3e^(-)),(Cr+4H_(2)O+4OH^(-)rarr),(""[Cr(OH)_(4)]^(-)+4H_(2)O+3e^(-)),(3O_(2)+3Cr+6H_(2)O+3e^(-)rarr2[Cr(OH)_(4)]^(-)),(bar(4Cr+3O_(2)+6H_(2)O+4HO^(-)rarr4[Cr(OH)_(4)]^(-))):}` This is the balnced equation. |
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