Chemical reaction formula of the oxidation-reduction reaction
1) Reaction of potassium iodide KI and the hydrogen peroxide H2O2 Add oxygenated water H2O2 to a potassium iodide KI water solution. (the acidity sulfate) Change of the material Potassium iodide KI Hydrogen peroxide H2O2 I- changed in ( I2 ) H2O2 changed in (( H2O ) Change of the oxidation state (-1 ) → (0 ) (-1 ) → (-2 ) Work by this reaction Oxidizer, reducing agent Reducing agent Oxidizer Expression including e- (half equation) 2I- →( I2 )+( 2e- ) H2O2+2H++(2e- )→(H2O )
【2】About the atom of the underline part of the next chemical reaction formulas, you show oxidation state, and answer it in an oxidizer, a reducing agent. (例1) 2CuO+C→2Cu+CO2 +2 0 0 +4 oxidizer=(The material which was reduced)〔 CuO 〕
reducing agent=(The material which was oxidized)〔 C 〕 (2) 2Na+2H2O→2NaOH+H2 0 +1 +1 0 Oxidizer 〔 H2O 〕
reducing agent〔 Na 〕 (3) Mg+2HCl→MgCl2+H2 0 +1 +2 0