Question

Assume the following reaction is carried out in acidic medium.
$6e^{-}+C{r}_2{O_7}^{2-}⟶2{Cr}^{3+}+7H_{2}O$

How many $H^{+}$ do we need to add to the LHS to balance the H atoms?

• A. 10
• B. 6
• C. 7
• D. 14

Answer 1

The correct option is D

14

Count number of Hydrogen on right hand side 7 $\times$2 = 14
Hence, add 14 $H^{+}$ ions to the side which is deficient.
$C{r}_2{O_7}^{2-}+6e^{-}+14H^{+}⟶2{Cr}^{3+}+7H_{2}O$