xCr(OH)3+yIO−3+zOH−→xCrO2−4+yI−+5H2O
n-factor for Cr(OH)3 [Cr3+ to Cr6+]
=|6−3|×1=3
n-factor for IO−3=[+5 to −1]
=|−1−5|=6
Balancing the number of electrons
2Cr(OH)3+IO−3→2CrO2−4+I−
Balancing charge by OH− on the left hand side
2Cr(OH)3+IO−3+4OH−→2CrO2−4+I−
Balancing H by adding water
2Cr(OH)3+IO−3+4OH−→2CrO2−4+I−+5H2O
x+y+z=2+1+4=7