Question

identify the oxidizing and reducing agents in the skeletal (unbalanced) reaction. Then, balance the reaction, including the phase (solid, liquid, etc.) of each species. The reaction takes place in basic aqueous solution. Cl 2 O 7 ( g ) + H 2 O 2 ( aq ) ⟶ ClO − 2 ( aq ) + O 2 ( g )

Answer 1

Answer:

$Cl_2O_7(g)+4H_2O_2+2OH^-\rightarrow 2(ClO_2)^-+4O_2+5H_2O$

Explanation:

Hello,

In this case, the reaction taking place with its oxidation states is:

$Cl_2^{+7} O_7^{-2} ( g ) + H_2^+ O_2^- ( aq ) \rightarrow (ClO_2)^ - ( aq ) + O_2^0 ( g )$

Thus, the half reactions are:

$Cl_2^{+7} O_7^{-2} ( g ) +8e^-+3H_2O \rightarrow 2(Cl^{+3}O^{-2}_2)^ - ( aq ) +6OH^-\\H_2^+O_2^-+2OH^- ( aq ) \rightarrow O_2^0+2e^-+2H_2O ( g )$

Which balanced result:

$2Cl_2^{+7} O_7^{-2} ( g ) +16e^-+6H_2O \rightarrow 4(Cl^{+3}O^{-2}_2)^ - ( aq ) +12OH^-\\8H_2^+O_2^-+16OH^- ( aq ) \rightarrow 8O_2^0+16e^-+16H_2O ( g )\\\\2Cl_2^{+7} O_7^{-2} ( g ) +16e^-+6H_2O +8H_2^+O_2^-+16OH^- ( aq )\rightarrow 4(Cl^{+3}O^{-2}_2)^ - ( aq ) +12OH^-+8O_2^0+16e^-+16H_2O ( g )$

Which is reduced to:

$Cl_2O_7(g)+4H_2O_2+2OH^-\rightarrow 2(ClO_2)^-+4O_2+5H_2O$

Best regards.