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Write the net ionic equation for the rection of potassium dichromate (VI) K_(2)Cr_(2)O_(7) with sodium suphite Na_(2)SO_(3) in acid solution to give chromium (III) ion and suphate ion |
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Answer» Solution :Step 1 WRITE the skeleton equation for the GIVEN reaction `Cr_(2)O_(7)^(2-)(aq)+SO_(3)^(2-)(aq)rarrCr^(3+)(aq)+SO_(4)^(2-)(aq)` Step 2 find out the element which undergo a change in oxidiatoin number (O.N)  here O.N of Cr decrease from +6 `Cr_(2)O_(7)^(2-)` to +3 in `Cr^(3+)` while that of s increases from +4 in `SO_(3)^(2+) to +6 in SO_(4)^(2-)` step 3 find out the total increase and decrease in O.N SINCE thereare two Cr atoms on L.H.S and only one on R.H.S therefore multiply `Cr^(3+)` onR.H.S of eq (i) by 2 and thus the total decreasein O.HN of Cr is `2xx3=6` step 4 balance increase / decrease in O.N since the toal increasein O.N is 2 and decrease is 6 therefore multiply `SO_(3)^(2-)` on L.H.S and `SO_(4)^(2-)` on R.H.S of Eq (ii) 3 combining steps 2 and 3 we have step 5 balance all atoms than H and O not needed since both cr and s atoms are already balanced step 6 balance o atoms by adding `H_(2)O` moleculed since there are seven O atoms in `Cr_(2)O_(7)^(2-)` and nine in `3So_(3)^(2-)` on L.H.S and only 12 on the R.H.S of eq (iii) therefore add 4 molecules of `H_(2)O` to R.H.S of eq (iii) we have `Cr_(2)O_(7)^(2-) (aq)+3SO_(3)^(2-)(aq)rarr2cr^(3+)(aq)+3SO_(4)^(2-)(aq)+4H_(2)O(l)` step 7 balance H toms by adding `H^(+)` IONS since the reaction occurs in the acidic medium since there are 8H on R.H.s and none o the L.H.S therefore add `8H^(+)` to the L.H.S of Eq (iv) we have `Cr_(2)O_(7)^(2-)+3SO_(3)^(2-)(aq)+8H^(+)(aq)rarr2Cr^(3+)r(aq)+3SO_(4)^(2-)(aq)+4H_(2)O(l)` thus equ (v) REPRESENT the correct balanced equation |
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