When a pink aqueous solution of potassium permanganate, faintly acidified with d

Question

When a pink aqueous solution of potassium permanganate, faintly acidified with dilute sulfuric acid was treated with 10% aq. hydrogen peroxide, the reaction took place with the evolution of gas bubbles, and the pink solution was turned colorless. Further chemical analysis revealed that the evolved gas was oxygen, and the resulting solution contains potassium sulfate and manganese (II) sulfate; water was also formed during the same reaction. Please answer the followings: 1) Write down the balanced chemical equation for this reaction. 2) Define the type/types of the reaction; assign the oxidation number of Manganese in potassium permanganate and manganese (II) sulfate. Also, offer an explanation for the color change. 3) Write down the ions present in the solution before & after the reaction. If 100 mL 0.5 M potassium permanganate was mixed with 50 mL 2 M sulfuric acid, what will be the final pH 7 (>7 or

Explanation / Answer

1.)KMnO4+5H2O2+3H2SO42MnSO4+K2SO4+5O2+8H2O

2.) This process is a combination of two reactions. On the one hand, this is an ordinary redox reaction, on the other hand, there is a catalytic disproportionation of hydrogen peroxide in the presence of potassium permanganate.

Oxidation number of Mn in KMnO4

1+ x-8=0

x = 7

Oxidation number of Mn in MnSO4

x-2=0

x= 2

Discoloration of potassium permanganate is due to its reduction to manganese sulfate.

3.)2MnO4 -+6H++5H2O22Mn+2+8H2O + 5O2

pH is less than 7 because when 2M H2SO4 is mixed at the end solution colour change colourless from pink . If it is basic ,the colour of solution at the end will be brown.

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