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Hydrogen peroxide solution (20 mL) reacts quantitatively with a solution of KMnO_(4) (20 mL) acidified with dilute H_(2)SO_(4). The same volume of KMnO_(4) solution is just decolurised by 10 mL of MnSO_(4) in neutral medium simultaneously forming a dark brown precipitate of hydrated MnO_(2) . The brown precipitate is dissolved in 10 mL of 0.2 M sodium oxalate under boiling conditions in the presence of dilute H_(2)SO_(4) . Write the balanced equations involved in the reactios and calculate the molarity of H_(2)O_(2). |
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Answer» Solution :Step 1. To write balanced equations for the reactions involved. (i) In acidic MEDIUM, `MnO_(4)^(-) ` oxidises `H_(2)O_(2)` to `O_(2)` `{:(MnO_(4)^(-)+8H^(+)+ 5e^(-) to Mn^(2+) + 4H_(2)O"]"xx2),(""H_(2)O_(2) to O_(2) + 2H^(+) + 2e^(-) "]"XX5):}/(2MnO_(4)^(-) + 5H_(2)O_(2) + 6H^(+) to 2Mn^(2+) + 5O_(2)+ 8H_(2)O)""...(i)` (ii) In neutral medium , `MnO_(4)^(-)` oxidises `Mn^(2+) ` to `MnO_(2)` . `{:(MnO_(4)^(-)+ 2H_(2)O + 3e^(-) to MnO_(2)+ 4OH^(-)"]"xx2),(""Mn^(2+)+ 4OH^(-) to MnO_(2) + 2H_(2)O + 2e^(-)"]"xx3):}/(2MnO_(4)^(-)+3Mn^(2+) + 4OH^(-) to 5 MnO_(2) + 2H_(2)O)""...(ii)` (iii) In acidic medium. `MnO_(2)` oxidises sodium oxalate to `CO_(2)` `{:(MnO_(2)+ 4H^(+) + 2e^(-) to Mn^(2+) + 2H_(2)O),(""C_(2)O_(4)^(2-) to 2CO_(2) + 2e^(-)):}/(MnO_(2) + C_(2)O_(4)^(2-) + 4H^(+) to Mn^(2+) + 2CO_(2) + 2H_(2)O) ` or `5MnO_(2) + 5C_(2)O_(4)^(2-) + 20 H^(+) to 5Mn^(2+) + 10 CO_(2) + 10 H_(2)O""...(iii)` From the above three balanced equations, it follows that `5C_(2)O_(4)^(2-) -=2MnO_(4)^(-) -=5H_(2)O_(2)""...(iv)` Step 2. To determine the number of moles of `C_(2)O_(4)^(2-)` present in 10 mL of 0.2 M sodium oxalate. No. of moles of sodium oxalate =Molarity `xx` volume in litres `=0.2 xx(10)/(1000)` `=2xx10^(-3)` mole Step 3. To calculate the molarity of `H_(2)O_(2)` From eq. (iv), it follows that `5C_(2)O_(4)^(2-)-= 5H_(2)O_(2)` or `2xx10^(-3)` mole of `C_(2)O_(4)^(2-) -= 2xx10^(-3)` mole of `H_(2)O_(2)` Now `2xx10^(-3) ` mole of `H_(2)O_(2)` is present in 10 mL of `H_(2)O_(2)` `therefore ` Molarity of `H_(2)O_(2)=(2xx10^(-3))/(10)xx1000=0.2` M |
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