Q. When a pink aqueous solution of potassium permanganate, faintly acidified wit
ID: 1067040 • Letter: Q
Question
Q. 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 or <7) of the solution?
4) After the completion of reaction, when evolution of gas-bubbles ceased, what will happen if an aqueous solution of sodium carbonate was added to the reaction mixture? Please write down the net ionic equation.
5) If initially 5.65 g of potassium permanganate was taken for the reaction, calculate the total mass of manganese (II) sulfate present in the solution after the completion of the reaction. What is the Manganese ion concentration of the solution (assume that the total volume of the final reaction mixture is 500 mL)?
Useful hints: hydrogen peroxide is a strong oxidizing agent and considered to be a molecular compound
Explanation / Answer
Overall reaction is 2KMnO4(aq) + 5H2O2(aq) + 3H2SO4(aq) --> K2SO4(aq) + 2MnSO4(aq) + 5O2(g) + 8H2O(l) (1)
8H+ + MnO4- + 5e- --> Mn2+ + 4H2O ( Reduction of MnO4- ) (1) and H2O2 --> O2 + 2H+ + 2e- ( Oxidation of oxygen) (2)
Eq.1*2 +Eq.2* 5 givesThe pink color is due to the presence of the permanganate ion
6H+ + 2MnO4- + 5H2O2 --> 2Mn2+ + 5O2 + 8H2O
when Na2CO3 is added, the following reactino takes place
MnSO4 (aq) + Na2CO3 (aq) --> Na2SO4 (aq) + MnCO3 (s)
Molar mass of KMnO4= 158, moles of KMnO4 inn 5.65 gm= 5.65/158 =0.03579 moles
From Reaction -1, 2 moles of KMnO4 produces 2 moles of Manganese sulfate
0.03579 moles gives 0.035759 moles of MnSO4.
Concentration= moles/L= 0.035759/0.5 L =0.071519 M
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.