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Mno4_ ions are reduced in acidic conditions to Mn2+ ions where they are reduced in neutral conditions to Mno2.the oxidation of 25ml of a solution x containing fe2.+.ions Required in acidic conditions 20 ml of of a solution y contain Mno4 Ions what value of a solution Mno4ions . What value of solution y containingMno4ions.what value if solution y would be requires to oxidise 25ml of solution x containing fe2+ ionsin neutral condition?
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Mno4_ ions are reduced in acidic conditions to Mn2+ ions where they ar...
Introduction:
In this scenario, we have a solution x containing Fe2+ ions that need to be oxidized. The oxidation can be carried out in two different conditions - acidic and neutral. The volume of solution x is given as 25 ml, and we need to determine the volume of solution y containing MnO4- ions required for the oxidation in both conditions.
Acidic Conditions:
In acidic conditions, MnO4- ions are reduced to Mn2+ ions. The balanced equation for this reaction is:
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
To balance the equation, we need 5 moles of electrons for each mole of MnO4-.
Given that the volume of solution x is 25 ml, we can calculate the number of moles of Fe2+ ions present:
n(Fe2+) = (25 ml) / (1000 ml/mol) * C(Fe2+)
To balance the equation, the number of moles of MnO4- required is equal to the number of moles of Fe2+:
n(MnO4-) = 5 * n(Fe2+)
Now, let's calculate the volume of solution y containing MnO4- ions required:
V(MnO4-) = (n(MnO4-) / C(MnO4-)) * 1000 ml
Neutral Conditions:
In neutral conditions, MnO4- ions are reduced to MnO2. The balanced equation for this reaction is:
MnO4- + 4H2O + 3e- → MnO2 + 8OH-
To balance the equation, we need 3 moles of electrons for each mole of MnO4-.
The number of moles of MnO4- required is the same as in the acidic conditions:
n(MnO4-) = 5 * n(Fe2+)
Now, let's calculate the volume of solution y containing MnO4- ions required in neutral conditions:
V(MnO4-) = (n(MnO4-) / C(MnO4-)) * 1000 ml
Conclusion:
In conclusion, we have determined the volume of solution y containing MnO4- ions required to oxidize solution x containing Fe2+ ions in both acidic and neutral conditions. The volume of solution y is calculated based on the stoichiometry of the balanced equations for each condition. Remember to use the appropriate molar concentrations (C) of the respective ions to calculate the volumes accurately.
In this scenario, we have a solution x containing Fe2+ ions that need to be oxidized. The oxidation can be carried out in two different conditions - acidic and neutral. The volume of solution x is given as 25 ml, and we need to determine the volume of solution y containing MnO4- ions required for the oxidation in both conditions.
Acidic Conditions:
In acidic conditions, MnO4- ions are reduced to Mn2+ ions. The balanced equation for this reaction is:
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
To balance the equation, we need 5 moles of electrons for each mole of MnO4-.
Given that the volume of solution x is 25 ml, we can calculate the number of moles of Fe2+ ions present:
n(Fe2+) = (25 ml) / (1000 ml/mol) * C(Fe2+)
To balance the equation, the number of moles of MnO4- required is equal to the number of moles of Fe2+:
n(MnO4-) = 5 * n(Fe2+)
Now, let's calculate the volume of solution y containing MnO4- ions required:
V(MnO4-) = (n(MnO4-) / C(MnO4-)) * 1000 ml
Neutral Conditions:
In neutral conditions, MnO4- ions are reduced to MnO2. The balanced equation for this reaction is:
MnO4- + 4H2O + 3e- → MnO2 + 8OH-
To balance the equation, we need 3 moles of electrons for each mole of MnO4-.
The number of moles of MnO4- required is the same as in the acidic conditions:
n(MnO4-) = 5 * n(Fe2+)
Now, let's calculate the volume of solution y containing MnO4- ions required in neutral conditions:
V(MnO4-) = (n(MnO4-) / C(MnO4-)) * 1000 ml
Conclusion:
In conclusion, we have determined the volume of solution y containing MnO4- ions required to oxidize solution x containing Fe2+ ions in both acidic and neutral conditions. The volume of solution y is calculated based on the stoichiometry of the balanced equations for each condition. Remember to use the appropriate molar concentrations (C) of the respective ions to calculate the volumes accurately.
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Mno4_ ions are reduced in acidic conditions to Mn2+ ions where they are reduced in neutral conditions to Mno2.the oxidation of 25ml of a solution x containing fe2.+.ions Required in acidic conditions 20 ml of of a solution y contain Mno4 Ions what value of a solution Mno4ions . What value of solution y containingMno4ions.what value if solution y would be requires to oxidise 25ml of solution x containing fe2+ ionsin neutral condition?
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Mno4_ ions are reduced in acidic conditions to Mn2+ ions where they are reduced in neutral conditions to Mno2.the oxidation of 25ml of a solution x containing fe2.+.ions Required in acidic conditions 20 ml of of a solution y contain Mno4 Ions what value of a solution Mno4ions . What value of solution y containingMno4ions.what value if solution y would be requires to oxidise 25ml of solution x containing fe2+ ionsin neutral condition? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Mno4_ ions are reduced in acidic conditions to Mn2+ ions where they are reduced in neutral conditions to Mno2.the oxidation of 25ml of a solution x containing fe2.+.ions Required in acidic conditions 20 ml of of a solution y contain Mno4 Ions what value of a solution Mno4ions . What value of solution y containingMno4ions.what value if solution y would be requires to oxidise 25ml of solution x containing fe2+ ionsin neutral condition? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Mno4_ ions are reduced in acidic conditions to Mn2+ ions where they are reduced in neutral conditions to Mno2.the oxidation of 25ml of a solution x containing fe2.+.ions Required in acidic conditions 20 ml of of a solution y contain Mno4 Ions what value of a solution Mno4ions . What value of solution y containingMno4ions.what value if solution y would be requires to oxidise 25ml of solution x containing fe2+ ionsin neutral condition?.
Mno4_ ions are reduced in acidic conditions to Mn2+ ions where they are reduced in neutral conditions to Mno2.the oxidation of 25ml of a solution x containing fe2.+.ions Required in acidic conditions 20 ml of of a solution y contain Mno4 Ions what value of a solution Mno4ions . What value of solution y containingMno4ions.what value if solution y would be requires to oxidise 25ml of solution x containing fe2+ ionsin neutral condition? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Mno4_ ions are reduced in acidic conditions to Mn2+ ions where they are reduced in neutral conditions to Mno2.the oxidation of 25ml of a solution x containing fe2.+.ions Required in acidic conditions 20 ml of of a solution y contain Mno4 Ions what value of a solution Mno4ions . What value of solution y containingMno4ions.what value if solution y would be requires to oxidise 25ml of solution x containing fe2+ ionsin neutral condition? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Mno4_ ions are reduced in acidic conditions to Mn2+ ions where they are reduced in neutral conditions to Mno2.the oxidation of 25ml of a solution x containing fe2.+.ions Required in acidic conditions 20 ml of of a solution y contain Mno4 Ions what value of a solution Mno4ions . What value of solution y containingMno4ions.what value if solution y would be requires to oxidise 25ml of solution x containing fe2+ ionsin neutral condition?.
Solutions for Mno4_ ions are reduced in acidic conditions to Mn2+ ions where they are reduced in neutral conditions to Mno2.the oxidation of 25ml of a solution x containing fe2.+.ions Required in acidic conditions 20 ml of of a solution y contain Mno4 Ions what value of a solution Mno4ions . What value of solution y containingMno4ions.what value if solution y would be requires to oxidise 25ml of solution x containing fe2+ ionsin neutral condition? in English & in Hindi are available as part of our courses for Class 11.
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