Question

hy everyone..
give me the steps to balance a redox reaction...

no spams please... ​

Answer 1

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Assign oxidation numbers and find out what is being oxidized and what is being reduced.

Draw arrows connecting "like" elements on the left and right side of the chemical equation. Note that those elements MUST have different oxidation numbers on the left and right sides of the equation.

Write the number of electrons over your arrows that shows the differences in the oxidation states. You should also note here which 1/2-reaction is the oxidation and which one is the reduction.

Now adjust those electrons accordingly (and formulas) IF one of the formulas has more than one element in it.

Example: Cr2O72- is on the left side and Cr3+ is on the right side. That's an oxidation state for chromium of +6 in the dichromate and +3 for the chromium(III) ion. That is a difference of 3 electrons - you write +3e- over the arrow. WAIT!!! There are 2 Cr's in the dichromate formula and you MUST account for them. So put a coefficient of 2 in front of the chromium(III) ion and CHANGE your total of electrons to +6e- over the arrow. Dichromate reacts TWO chromiums at a time (not one) and therefore 6 electrons at a time.

Now multiply each of your 1/2-reactions by the appropriate scaler so as to make the total number of electrons transferred MATCH in each. Remember to ADJUST the amounts on BOTH sides of the equation. NOTE: If your doing a titration calculation, you can now stop and work the problem. Why?/How? Well because you now know the stoichiometric ratio of the two reactants. However, if the point of the problem is to FINISH balancing the equation then read on...

To balance out the oxygens, add however many H2Os to the side that is short

Now balance the hydrogens with H+s to the side that is short.

Your done! Check out the charge balance now to make sure.

in Basic Conditions

Balance the equation just like you do for acids (no OH- will be in the equation at this point, only H2O and H+ ).

Now ADD as many OH-s as there are H+s to BOTH sides of the equation.

The OH-s and the H+s will combine to give H2O (as many as there were H+s and OH-s). So cross out the OH-s and the H+s and write in place of them that many H2Os.

Now cancel out any H2Os from the two sides of the equation. H2O should only be listed on one side or the other, not both.

Your done! Check out the charge balance now to make sure.

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Answer 2

hii mate...

here we go...

Step 1

Correctly write the formula for the reactants and the products of the chemical reaction.

Step 2

Determine correctly the atoms that undergo oxidation number change in the given reaction by allocating the oxidation number of the individual elements present in the reaction.

Step 3

Calculate the oxidation number on the basis of each atom for the given molecule or ion of the chemical reactions. If the numbers are not equal then multiply it to such a number that overall these numbers become equal. If in a case, two substances are either only oxidized or only reduced then this signifies that something is wrong with the chemical reaction. This signifies that either the formulas of the reactant or the products are incorrect. This can also mean that the allocation of oxidation numbers is incorrect.

Step 4

Keep in mind the involvement of the ions if the reaction occurs in water. Accordingly, add H+ or OH– ions in the appropriate side of the reaction. Overall, the ionic charges of reactant and products will be equal. However, if the reaction takes place in acidic solution then add H+ ions in the chemical equation. Similarly, if the reaction takes place in the basic solution add OH– ions in the chemical equation.

Step 5

It is very important to equate the number of hydrogen atoms on each side of the equation by adding water molecules or H2O molecules. Additionally, it is necessary to check the oxygen atoms present in the equation. It will be a balance reaction if there are equal numbers of oxygen atoms present in both the reactant as well as the product the side.

hope this helped u...