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In one reaction,. Cr2o7^2- 14H 6e- = 2Cr^3 7H2o. In this reaction how to find that how much electron are involved ? Explain if anybody wants to.?
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In one reaction,. Cr2o7^2- 14H 6e- = 2Cr^3 7H2o. In this reactio...
Calculate total charge on both the sides of reaction...here in reactant side u have 12 +ve charge and in product syd 6+ve charge so in order to neutralise u have to add 6electron in reactant syd
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In one reaction,. Cr2o7^2- 14H 6e- = 2Cr^3 7H2o. In this reactio...
Explanation:
This reaction involves the transfer of electrons from the reactant to the product. The reactant Cr2O7^2- gains electrons to form Cr^3, which means that 6 electrons are involved in this reaction.
The balanced chemical equation for this reaction is:
Cr2O7^2- + 14H+ + 6e- → 2Cr^3+ + 7H2O
Breaking down the reaction:
- Cr2O7^2- gains 6 electrons to form 2 Cr^3+ ions
- 14 H+ ions are added to balance the charge
- 7 H2O molecules are formed as a product
Conclusion:
In this reaction, 6 electrons are involved in the reduction of Cr2O7^2- to Cr^3+. The transfer of electrons is a fundamental concept in chemistry, and it is essential to understand the number of electrons involved in a reaction to balance the chemical equation and calculate stoichiometry.
This reaction involves the transfer of electrons from the reactant to the product. The reactant Cr2O7^2- gains electrons to form Cr^3, which means that 6 electrons are involved in this reaction.
The balanced chemical equation for this reaction is:
Cr2O7^2- + 14H+ + 6e- → 2Cr^3+ + 7H2O
Breaking down the reaction:
- Cr2O7^2- gains 6 electrons to form 2 Cr^3+ ions
- 14 H+ ions are added to balance the charge
- 7 H2O molecules are formed as a product
Conclusion:
In this reaction, 6 electrons are involved in the reduction of Cr2O7^2- to Cr^3+. The transfer of electrons is a fundamental concept in chemistry, and it is essential to understand the number of electrons involved in a reaction to balance the chemical equation and calculate stoichiometry.
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In one reaction,. Cr2o7^2- 14H 6e- = 2Cr^3 7H2o. In this reaction how to find that how much electron are involved ? Explain if anybody wants to.?
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In one reaction,. Cr2o7^2- 14H 6e- = 2Cr^3 7H2o. In this reaction how to find that how much electron are involved ? Explain if anybody wants to.? 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 In one reaction,. Cr2o7^2- 14H 6e- = 2Cr^3 7H2o. In this reaction how to find that how much electron are involved ? Explain if anybody wants to.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In one reaction,. Cr2o7^2- 14H 6e- = 2Cr^3 7H2o. In this reaction how to find that how much electron are involved ? Explain if anybody wants to.?.
In one reaction,. Cr2o7^2- 14H 6e- = 2Cr^3 7H2o. In this reaction how to find that how much electron are involved ? Explain if anybody wants to.? 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 In one reaction,. Cr2o7^2- 14H 6e- = 2Cr^3 7H2o. In this reaction how to find that how much electron are involved ? Explain if anybody wants to.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In one reaction,. Cr2o7^2- 14H 6e- = 2Cr^3 7H2o. In this reaction how to find that how much electron are involved ? Explain if anybody wants to.?.
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