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Justify that the following reactions are redoxreactions : (a) CuO_((s))+H_(2(g))toCu_((s))+H_(2)O_((g)) (b) Fe_(2)O_(3(s))+3CO_((g))to2Fe_((s))+3CO_(2(g)) ( c) 4BCl_(3(g))+3LiAlH_(4(s))to2B_(2)H_(6(g))+3LiCl_((s))+3AlCl_(3(s)) (d) 2K_((s))+F_(2(g))to2K^(+)F_((s))^(-) (e) 4NH_(3(g))+5O_(2(g))to4NO_((g))+6H_(2)O_((g)) |
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Answer» Solution :(a) `overset(+2)(Cu)overset(-2)(O_((s)))+overset(0)(H_(2(g)))tooverset(0)(Cu_((s)))+overset(+1)(H_(2))overset(-2)(O_((g)))` Here, O is being removed from CuO. Therefore Cu gets reduced. Where O is being added to `H_(2)`, and from `H_(2)O`. Therefore it gets oxidized. Therefore, oxidation number of Cu decreases from +2 to 0. therefore it gets reduced. Where oxidation number of H increases from 0 to +1. Therefore it gets oxidized. Therefore given reaction is a redox reaction. (B) `overset(+3)(Fe_(2))overset(-2)(O_(3(s)))+overset(+2)(3CO_((g)))tooverset(0)(2Fe_((s)))+overset(+4)(3CO_(2(g)))` : Oxidation number of FE gets reduced from +3 to 0. Whereas oxidation number of C increases from +2 to +4. O is being removed from `Fe_(2)O_(3)` and added into CO. `Fe_(2)O_(3)` gets reduced and CO gets oxidized. Therefore given reaction is a redox reaction. ( c) `overset(+3)(4B)overset(-1)(Cl_(3(g)))+overset(+1)(3Li)overset(+3)(Al)overset(-1)(H_(4(s)))tooverset(-3)(2B_(2))overset(+1)(H_(6(g)))+overset(+1)(3Li)overset(-1)(Cl_((s)))+overset(+3)(3Al)overset(-1)(Cl_(3(s)))` Here, oxidation number of B reduces from +3 to -3. therefore their is reduction. Same an oxidation number of hydrogen increases from-1 to +1. Therefore their is oxidation. H is added to `BCl_(3)` and hydrogen is removed from `LiAlH_(4)`. Therefore given reaction is redox reaction. (d) `overset(0)(2K_((s)))+overset(0)(F_(2(g)))to2K^(+)F_((s))^(-)` : Here, oxidation number of K increases from 0 to +1 and oxidation number of F reduces from 0 to -1. Therefore K gets reduced and F gets oxidized. Therefore given reaction is redox reaction. (e) `overset(-3)(4N)overset(+1)(H_(3(g)))+overset(0)(5O_(2(g)))tooverset(+2)(4N)overset(-2)(O_((g)))+overset(+1)(6H_(2))overset(-2)(O_((g)))` : Here, oxidation number of N is INCREASED from -3 to +2. Oxidation number of O is reduced from 0 to -2. Therefore their is reduction. Here, H get removed from `NH_(3)`, and again added therefore `NH_(3)` gets oxidized and `O_(2)`, gets reduced. Therefore given reaction is redox reaction. |
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