# Balancing the equation

I've got a question:

$$\ce{Cr2O7^2- -> Cr^3+}$$

This reaction takes place in a acidic medium - how many $\ce{H+}$ are required to balance the equation?

How do I balance this reaction.

Cr had a oxidation state of $+6$ at first then it gained three electrons and now has an oxidation state of $+3$. What does that have to do with $\ce{H+}$?

Additionally, how would I balance this reaction if it were to take place in a basic medium.

• $\ce{14H+ + Cr2O7^2- + 6e- <=> 7H2O + 2Cr^3+}\\\ce{7H2O + Cr2O7^2- + 6e- <=> 2Cr^3+ + 14OH-}$ – DHMO Oct 15 '16 at 7:34
• This is a basic stoichiometry question. Reaction equations must be balanced due to the law of conservation of mass and charge – Jan Oct 15 '16 at 12:56

First balance the number of atoms both side (except H and O). $$\ce{Cr2O7^2- -> 2 Cr^3+}$$

Since the oxidation number of Cr on the left is $+ 6$ while on the right, it is $+ 3$, since 2 atoms of Cr are involved so reduction has occur. Add 6 electrons on the left hand side thus, $$\ce{Cr2O7^2- + 6e- -> 2 Cr^3+}$$ To balance the charge we have to add $\ce{H+}$ on either side, since charge on the left is $-8=-2+ -6$ and on the right it is $+ 6$, so add $\ce{14H+}$ to the left hand side. Thus, $$\ce{Cr2O7^2- + 6e- + 14 H^+ -> 2 Cr^3+}$$ To balance the O atoms add $\ce{H2O}$ on the other side, $$\ce{Cr2O7^2- + 6e- + 14 H^+ -> 2 Cr^3+ + 7 H2O}.$$ The H atoms will autometically get balanced. The next question is what will happen in basic medium, proceed till $\ce{Cr2O7^2- + 6 e -> 2 Cr^3+}$ as earlier. In acidic medium we add hydrogen to balance the charge, do it here too but also add same number of $\ce{OH-}$ to both sides. As in the example, $$\ce{Cr2O7^2- + 6e- -> 2 Cr^3+},$$ it will become, $$\ce{Cr2O7^2- + 6e- + 14 H^+ + 14 OH- -> 2Cr^3+ + 14 OH-}$$ $$\ce{Cr2O7^2- + 6e- + 14 H2O -> 2Cr^3+ + 14 OH-}$$

Add $\ce{H2O}$ on either side to balance the O atoms, $$\ce{Cr2O7^2- + 6e- + 14 H2O -> 2Cr^3+ + 14 OH- + 7H2O}$$

$$\ce{Cr2O7^2- + 6e- + 7 H2O -> 2Cr^3+ + 14 OH-}$$

• You don’t have to first add $\ce{H+}$ and then additionally add $\ce{OH-}$ — you can just add $\ce{OH-}$ to balance the charge, just remember to do it on the side with excessive positive charge. Balancing with water should then happen on the other side. – Jan Oct 15 '16 at 13:01
• Nice answer. $+1$ – kovid vishesh Jan 9 '17 at 16:31