Question #8340d

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Question #8340d

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Answer 1 · 2015-01-15T18:19:06

Both examples are redox reactions.

Let's take the general form of the first one (I'm not going to write the ions for $B a {(O H)}_{2}$ $Ba(OH)_2$ , I'll just leave it like this):

$B a {(O H)}_{2 (a q)} + H_{2} O_{2 (a q)} + 2 C l O_{2 (a q)} \to B a {(C l O_{2})}_{2 (s)} + 2 H_{2} O_{(l)} + O_{2 (g)}$ $Ba(OH)_(2(aq)) + H_2O_(2(aq)) + 2ClO_(2(aq)) -> Ba(ClO_2)_(2(s)) + 2H_2O_((l)) + O_(2(g))$

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

$B a^{+ 2} {(O^{- 2} H^{+ 1})}_{2}^{- 2} + H_{2}^{+ 1} O_{2}^{- 1} + 2 C l^{+ 4} O_{2}^{- 2} \to B a^{+ 2} {(C l^{+ 3} O_{2}^{- 2})}_{2}^{+ 3} + 2 H_{2}^{+ 1} O^{- 2} + O_{2}^{0}$ $Ba^(+2)(O^(-2)H^(+1))_2 + H_2^(+1)O_2^(-1) + 2Cl^(+4)O_2^(-2) ->Ba^(+2)(Cl^(+3)O_2^(-2))_2 + 2H_2^(+1)O^(-2) + O_2^(0)$

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

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

For the second reaction, you'll have

$I_{2}^{+ 5} O_{5}^{- 2} + 5 C^{+ 2} O^{- 2} \to I_{2}^{0} + 5 C^{+ 4} O_{2}^{- 2}$ $I_2^(+5)O_5^(-2) + 5C^(+2)O^(-2) -> I_2^(0) + 5C^(+4)O_2^(-2)$

Iodine's oxidation number changes from +5 in $I_{2} O_{5}$ $"I"_2"O"_5$ , to 0 in $I_{2}$ $"I"_2$ , while carbon's oxidation number changes from +2 in $CO$ $"CO"$ , to +4 in ${CO}_{2}$ $"CO"_2$ .

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

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Question #8340d

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