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Balance the following equations by the oxidation number method |
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Answer» `Fe^(2+) +H^(+) +Cr_(2)O_(7)^(2)rarrCr^(3)+Fe^(3+)+H_(2)O`  Step 2 find out the total increase and decrease in O.N since there is only fe atom on either side of Eq (a) therefore total increase in O.N is 1 further since there are two cr atoms in `Cr_(2)O_(7)^(2-)` on L.H.S and only in `Cr^(3+)` oin L.H.S of Eq (a) therefore total decrease in O.N=`2xx3=6` step 3 balance increase / decrease in O.N sincethe total increase in O.N is 1 and decrease is 6 therefore multiply `Fe^(2+)` on the L.H.S of Eq (a) by 6 we have `6FE^(2+)+h^(+)+r_(2)O_(7)^(2-)rarrCr^(3+)+6Fe^(3+)+H_(2)O` step 4 balance all atoms other than O and H To balance Cr on either side of Eq (b) multiply `Cr^(3+)` by 2 on R.H.S of Eq (b) we have `6Fe^(2+)+H^(+)Cr_(2)O_(7)^(2+)rarr 2Cr^(3+)+6Fe^(3+)+H_(2)O` Step 5 balance O and H atoms by hit and trial method to balance H multiply `H^(+)` pmL.H.S of Eq (d) by 14we have `6Fe^(2+)+14H^(+)+Cr_(2)O_(7)^(2-) rarr 6 Fe^(3+)+2 Cr^(3+)+7H_(2)O` step 1 find out the element which undergo a change in O.N  step 2 find out the total increase/ decrease in o.n since there are tow I atoms of the L.H.S of eq since there is only one N atom on either side of Eq (a) therefore total decrease in O.N =`1xx1=1` step 3 balance increase /decrease in O.N since total increase in O.N is 10 and decrease is only 1 therefore multiply `NO_(3)^(-)` on L.H.S of Eq (a) 10 we have step 4 balance all atoms otehr than O and H sincethere are tow I atoms on L.H.S and only 1 On R.H.S of Eq (b) therefore multiple `IO_(3)^(-)` by 2 combining these two steps we have step 5 balance O and H atoms by hit and trial method to balance O atoms on either side of eq (c ) add 4 `H_(2)O` to the R.H.S and tobalance H atoms add 8 `H^(+)` the L.H.S of Eq we have (iii) SETP 1 find out the element which undergo a change in O.N  step 2 findout the total increase / decrease in O.N since there are two I atoms on L.H.S of Eq (a) and only on R.H.S therefore the total decrease in O.N =`2xx1=2` further since there are two s atoms on L.H.S and four S atoms on the R.H.S of eq (IV) step 1 find outthe element which undergo a change in O.N  step 2 find out the total increase / decrease in o.n since there is only ione mn atoms on either side ofEq (a) there fore total decrease in o.n =`1xx2=-2` further sincethere are two c atoms on the L.H.S and only one on the r.H.s of eq (a) there foretotal increase in O.N `=2xx1=2` step 3 balance increase / decrease in O.N the totalincrease or decrease in O.N is already balanced step 5 balance all atoms other than o and h atoms since there are six o atoms on L.H.S and onlyfour on the R.H.S of Eq (b) add 2 `H_(2)O` to the R.H.s of R.H.S eq (c ) we have thus eq (d) represent the CORRECT balanced equation |
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