Oxidation Reduction Reactions: Test Tube No. Solution A Chlorine bleach and 2 dr
ID: 588350 • Letter: O
Question
Oxidation Reduction Reactions:
Test Tube No. Solution A Chlorine bleach and 2 drops of 6 M HCI (the active ingredient in chlorine bleach is CIO) Chlorine bleach and 2 drops of 6 M HCI Addition of Solution B Drops of 0.1 M Fe(NHSO Add 1 drop of starch solution followed by 0.01 M KMnO4 and 2 drops of 6 M H2SO4 0.01 M KMnO4 and 2 drops of 6 M H2SO4 Deionized water and 2 drops of 6 M H SO Deionized water and 2 drops of 6 M H SO drops of 0.1 M KI Drops of 0.1 M Fe(NHASO, (Figure 27.3) Drops of 1 M K2C20 Drops of 0.1 M Fe(NH2(SO2 Add 1 drop of starch solution followed by 4 0.1 M H202 and 2 drops of 6 M H2SO4 0.1 M H202 and 2 drops of 6 M H2SO4 drops of 0.1 M KI Drops of 0.1 M Fe(NH2SO Add 1 drop of starch solution followed by drops of 0.1 M KIExplanation / Answer
In case of 5 and 6 there is no suitable reactant(oxidizing agent) to react with Fe(NH4)2SO4 and HI. hence no reaction will occur and thus there is no formation of products.
In case of 7 and 8 the oxidizing agent H2O2 reacts with Fe(NH4)2SO4 and KI and forms the corresponding oxidized and reduced products
The reactions may be given as
H2O2 reacts with Fe(NH4)2SO4 in presence of acidic medium (H2SO4) to form Fe2(SO4)3 and H2O
The two half reactions may be written as
2Fe^2+ ----> 2Fe^3+ + 2e- (oxidation)
H2O2 +2H+ + 2e- ----> 2H2O (reduction)
The overall reaction may be written as
H2O2 + 2Fe^2+ + 2H+ -----> 2Fe^3+ + 2H2O
In this reaction, Iron is oxidized and peroxide is reduced
Similarly, H2O2 reacts with KI to form H2O and I2
The reaction may be written as
2I- ---> I2 + 2e- (oxidation)
H2O2 + 2H+ + 2e- ----> 2H2O (reduction)
Overall reaction may be written as
H2O2 + 2KI + H2SO4 ---> K2SO4 + 2H2O + I2
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