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# Sir please help me balance this equation-Cr(oh)3+IO3^-1-------->I-+CrO4^2-(in basic medium) by ion electron method.Sir it is written in my book to first balance all the atoms other that o and h by seeing increase and decrease in oxidation number. After that balance o by adding h20 to excess side and adding double of that to other side. But in this case the ans is the other way round. I am getting 2CR(oh)3+io3-   >>>>I-+2crO4 2-After that I feel that there is one extra oxygen in lhs but in the ans there are 5h20 on the other side. Pls help me.

6 years ago
General Steps
Step 1:Try to balance the atoms in the equation by inspection, that is, by the standard technique for balancing non-redox equations. (Many equations for redox reactions can be easily balanced by inspection.) If you successfully balance the atoms, go to Step 2. If you are unable to balance the atoms, go to Step 3.

Step 2:Check to be sure that the net charge is the same on both sides of the equation. If it is, you can assume that the equation is correctly balanced. If the charge is not balanced, go to Step 3.

Step 3:If you have trouble balancing the atoms and the charge by inspection, determine the oxidation numbers for the atoms in the formula, and use them to decide whether the reaction is a redox reaction. If it is not redox, return to Step 1 and try again. If it is redox, go to Step 4.
Step 4:Determine the net increase in oxidation number for the element that is oxidized and the net decrease in oxidation number for the element that is reduced.

Step 5:Determine a ratio of oxidized to reduced atoms that would yield a net increase in oxidation number equal to the net decrease in oxidation number (a ratio that makes the number of electrons lost equal to the number of electrons gained).

Step 6:Add coefficients to the formulas so as to obtain the correct ratio of the atoms whose oxidation numbers are changing. (These coefficients are usually placed in front of the formulas on the reactant side of the arrow.)

Step 7:Balance the rest of the equation by inspection.