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Balance the following equation by ion electron method in the acidic medium. Cr_(2)O_(7)^(2-)+C_(2)O_(4)^(2-)+H^(+)toCr^(3+)+CO_(2)+H_(2)O |
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Answer» SOLUTION :Step 1. The given equation is `Cr_(2)O_(7)^(2-)+C_(2)O_(4)^(2-)+H^(+)toCr^(3+)+CO_(2)+H_(2)O` Step 2. Writing the oxidation numbers of all atoms, we have `overset(+6-2)(Cr_(2)O_(7)^(2-))+overset(+3-2)(C_(2)O_(4)^(2-))+overset(+1)(H^(+))tooverset(+3)(CR^(3+))+overset(+4-2)(CO_(2))+overset(+1-2)(H_(2)O)` Step 3. The half REACTION corresponding to reduction process is `C_(2)O_(7)^(2-)toCr^(3+)` while the half reaction corresponding to oxidation is `C_(2)O_(4)^(2-)toCO_(2)` Step 4. (a) Balancing of reduction half reaction : (i) Balancing Cr atoms, we get `C_(2)O_(7)^(2-)to2Cr^(3+)` (ii) Since the reaction proceeds in acidic medium,oxygen atoms can be balanced by the addition of seven `H_(2)O` molecules at the right hand side. `Cr_(2)O_(7)^(2-)to2Cr^(3+)+7H_(2)O` Hydrogen atoms can be balanced by the addition of `14H^(+)` on left. `Cr_(2)O_(7)^(2-)+14H^(+)to2Cr^(3+)+7H_(2)O` (III) At this stage, we have +14 - 2 = +12 charge onleft and `+3xx2=+6` charge on the right. The charge on the two sides can be balanced by the addition of six electrons on the left hand side. This gives the balanced half equation for the reduction process. `Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)to2Cr^(3+)+7H_(2)O` (b) Balancing of oxidation half reaction : (i) Balancing C atoms, we get `C_(2)O_(4)^(2-)to2CO_(2)` (ii) Oxygen atoms are already balanced in it. (iii) The equation contains no hydrogen atom. (iv) The charge can be balanced by ADDING two electrons on the right hand side. This gives the balanced half equation for the oxidation process. `C_(2)O_(4)^(2-)to2CO_(2)+2e^(-)` Step 5. The oxidation half reaction contains 6 electrons while the reduction half reaction has 2 electrons. These electrons can be cancelled by multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction as shown. `{:(Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)to2Cr^(3+)+7H_(2)O),(""[C_(2)O_(4)^(2-)to2CO_(2)+2e^(-)]xx3),(bar(Cr_(2)O_(7)^(2-)+3C_(2)O_(4)^(2-)+14H^(+)to2Cr^(3+)+6CO_(2)+7H_(2)O)):}` This is the final balanced equation in acidic medium. Note that in this equation, all atoms and charge are completely balanced. |
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