Saved Bookmarks
| 1. |
Try all possible approaches to justify that the following reactions re redox reaction (a)CuO(s)+H_(2)rarrCu(s)+H_(2)O(g) (b)Fe_(2)O_(3)(s)+3CO(g)rarr2Fe(s)+3CO_(2)(g) (c )4BCI_(3)(g)+3 LiAIH(s)rarr2B_(2)H_(6)(g)+3LiCI(s)+3AICI(s) (d)2K(s)+F_(2)(g)rarr2K^(+)F^(-)(s) (e )4NH_(33)(g)+5O_(2)(g)rarr4NO(g)+6H_(2)O(g) |
|
Answer» Solution :(a) `CuO(s)+H_(2)(g)rarrCu(s)+H_(2)O(g)` here O is removed form CuO therefore it is reduced to Cu while o is added to `H_(2)`form `H_(2)` therefore it is oxidised furhter O.N of Cu DECREASES form +2 in CuO to 0 in Cu that of H increases from 0 in `H_(2)` to +1 `H_(2)O` therefore CuO reduceed to Cu but `H_(2)` is oixdised to `H_(2)O` thus this is REDOX reactoin (b) `Fe_(2)O_(3)(s)+3cO(g)rarr2fe(s)+3 CO_(2)(g)` here O.N of Fe decreases form +3 is `Fe_(2)O_(3)` to 0 in Fe while that of c increases form +2 in CO to + in `CO_(2)` further oxygen is aremoved from `Fe_(2)O_(3)`and added to CI therefore `Fe_(3)O_(3)` is reduced while CO is oxidised thus this is a redox reaction ltbnrgt (c ) `4B CI_(3)(g)+Li AI H_(4)(s)rarr 2B_(2)H_(6)(g)+3 Li CI (s) + 3AICI_(3)` here O.N of B decreases form + in `CrCI_(3)` to -3 Li `CI(s) +3 AICI_(3)(s)` here O.N of b decrease form +3 in `BrCI_(3)` to -3 in `B_(2)H_(6)` while that of H increases from -1 Li `AIH_(4)`to +1 in `B_(2)H_(6)` (d) `2K(s) +F_(2)rarr2F^(+)F^(-)(s)` here each k atom has lost one letron to form `K(+)` while `F_(2)` has gained two electrons to form two ions therefore k is oixdised while `F_(2)` is reduced thus it is a redox rection (e ) `4HN_(3)(g)+5O_(2)(g)rarr4NO(g)+6H_(2)O(g)` here O.N of N increases from -3 is `NH_(3)` ot +32 is NO while that O decreaess from 0 is `O_(2)` to -2 in NO or `H^(@)O` therefore `NH_(3)` is oxidised while `O_(2)` is reduced further H been removed has been oxidised while `O_(2)` is reduced thus this isa redox reaction |
|