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Write the net ionic equation for the reaction of potassium dichromate (VI), K_(2)Cr_(2)O_(7) with sodium sulphite, Na_(2)SO_(3), in an acid solution to give chromium (III) ion and the sulphate ion. |
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Answer» Solution :Step-1 : The skeletal ionic equation is : `Cr_(2)O_(7(aq))^(2-)+SO_(3(aq))^(2-)toCr_((aq))^(3+)+SO_(4(aq))^(2-)` Step-2 : Assign oxidation numbers for Cr and S. `OVERSET(+6,-2)(Cr_(2)O_(7(aq))^(2-))+overset(+4,-2)(SO_(3(aq))^(2-))tooverset(+3)(Cr_((aq))^(3+))+overset(+6,-2)(SO_(4(aq))^(2-))` This indicates that the dichromate ion is the oxidant and the sulphite ion is the reductant. Step-3 : Calculate the increase and decrease of oxidation number, and make them equal : `overset(+6,-2)(Cr_(2)O_(7(aq))^(2-))+overset(+4,-2)(3SO_(3(aq))^(2-))tooverset(+3)(2Cr_((aq))^(3+))+overset(+6,-2)(3SO_(4(aq))^(2-))` Step-4 : As the reaction occurs in the acidic medium, and further the ionic charges are not equal on both the sides, add `8H^(+)` on the left to make ionic charges equal `Cr_(2)O_(7(aq))^(2-)+3SO_(3(aq))^(2-)+8H^(+)to2Cr_((aq))^(3+)+3SO_(4(aq))^(2-)` Step-5 : Finally, count the hydrogen atoms, and add appropriate number of water molecules (i.e., `4H_(2)O`) on the right to achieve balanced REDOX change. `Cr_(2)O_(7(aq))^(2-)+3SO_(3(aq))^(2-)+8H_((aq))^(+)to2Cr_((aq))^(3+)+3SO_(4(aq))^(2-)+4H_(2)O_((L))` |
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