# Question 8340d

Jan 15, 2015

Both examples are redox reactions.

Let's take the general form of the first one (I'm not going to write the ions for $B a {\left(O H\right)}_{2}$, I'll just leave it like this):

$B a {\left(O H\right)}_{2 \left(a q\right)} + {H}_{2} {O}_{2 \left(a q\right)} + 2 C l {O}_{2 \left(a q\right)} \to B a {\left(C l {O}_{2}\right)}_{2 \left(s\right)} + 2 {H}_{2} {O}_{\left(l\right)} + {O}_{2 \left(g\right)}$

If you assign oxidation numbers to all the atoms involved in the reaction, you'll get

$B {a}^{+ 2} {\left({O}^{- 2} {H}^{+ 1}\right)}_{2} + {H}_{2}^{+ 1} {O}_{2}^{- 1} + 2 C {l}^{+ 4} {O}_{2}^{- 2} \to B {a}^{+ 2} {\left(C {l}^{+ 3} {O}_{2}^{- 2}\right)}_{2} + 2 {H}_{2}^{+ 1} {O}^{- 2} + {O}_{2}^{0}$

So, $\text{Cl}$'s oxidation number changes from +4 in ${\text{ClO}}_{2}$, to +3 in "Ba(ClO"_2)_2#, while $\text{O}$'s oxidation nubmer changes from -1 in ${\text{H"_2"O}}_{2}$, to 0 in ${\text{O}}_{2}$.

${\text{Cl"^(+4) + 1"e"^(-) -> "Cl}}^{+ 3}$
${\text{O"^(-1) -> "O"^(0) + 1"e}}^{-}$

For the second reaction, you'll have

${I}_{2}^{+ 5} {O}_{5}^{- 2} + 5 {C}^{+ 2} {O}^{- 2} \to {I}_{2}^{0} + 5 {C}^{+ 4} {O}_{2}^{- 2}$

Iodine's oxidation number changes from +5 in ${\text{I"_2"O}}_{5}$, to 0 in ${\text{I}}_{2}$, while carbon's oxidation number changes from +2 in $\text{CO}$, to +4 in ${\text{CO}}_{2}$.

${\text{C"^(+2) -> "C"^(+4) + 2"e}}^{-}$
${\text{I"^(+5) + 5"e"^(-) -> "I}}^{0}$