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Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :
- a)2.24 L
- b)4.48 L
- c)6.72 L
- d)8.96 L
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Oxidation of succinate ion produces ethylene and carbon dioxide gases....
To solve this question, we need to calculate the total volume of gases produced during the oxidation of succinate ion.
**Given:**
- 0.2 Faraday of electricity is passed through the solution.
- Succinate ion is oxidized to produce ethylene and carbon dioxide gases.
- We need to calculate the total volume of gases at both the cathode and anode at STP (1 atm and 273 K).
**Solution:**
1. Calculate the moles of succinate ion:
- The succinate ion has a chemical formula of C4H4O4^2-.
- The molar mass of succinate ion is 4 * (12.01) + 4 * (1.008) + 4 * (16.00) = 118.09 g/mol.
- Let's assume the mass of succinate ion in the solution is 'm' grams.
- Therefore, the number of moles of succinate ion is m / 118.09.
2. Calculate the moles of electrons passed:
- The Faraday constant is 96485 C/mol.
- Given that 0.2 Faraday of electricity is passed, the number of moles of electrons passed is 0.2 * 96485.
3. Calculate the moles of ethylene and carbon dioxide produced:
- The balanced equation for the oxidation of succinate ion is:
C4H4O4^2- -> C2H4 + 2CO2 + 2e^-
- From the equation, we can see that 2 moles of succinate ion produce 1 mole of ethylene and 2 moles of carbon dioxide.
- Therefore, the number of moles of ethylene produced is (1/2) * (moles of succinate ion).
- And the number of moles of carbon dioxide produced is 2 * (moles of succinate ion).
4. Calculate the total volume of gases:
- The ideal gas equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.
- At STP, the pressure is 1 atm, the temperature is 273 K, and the gas constant is 0.0821 L.atm/(mol.K).
- The total volume of gases produced is (moles of ethylene + moles of carbon dioxide) * RT / P.
5. Substitute the values and calculate:
- The total volume of gases produced is ((1/2) * (moles of succinate ion) + 2 * (moles of succinate ion)) * 0.0821 * 273 / 1.
- Simplifying the expression, we get (5/2) * (moles of succinate ion) * 0.0821 * 273.
6. Substitute the moles of succinate ion:
- The moles of succinate ion is m / 118.09.
- Therefore, the total volume of gases produced is (5/2) * (m / 118.09) * 0.0821 * 273.
7. Simplify the expression:
- The total volume of gases produced is (0.922 * m) / 118.09.
8
**Given:**
- 0.2 Faraday of electricity is passed through the solution.
- Succinate ion is oxidized to produce ethylene and carbon dioxide gases.
- We need to calculate the total volume of gases at both the cathode and anode at STP (1 atm and 273 K).
**Solution:**
1. Calculate the moles of succinate ion:
- The succinate ion has a chemical formula of C4H4O4^2-.
- The molar mass of succinate ion is 4 * (12.01) + 4 * (1.008) + 4 * (16.00) = 118.09 g/mol.
- Let's assume the mass of succinate ion in the solution is 'm' grams.
- Therefore, the number of moles of succinate ion is m / 118.09.
2. Calculate the moles of electrons passed:
- The Faraday constant is 96485 C/mol.
- Given that 0.2 Faraday of electricity is passed, the number of moles of electrons passed is 0.2 * 96485.
3. Calculate the moles of ethylene and carbon dioxide produced:
- The balanced equation for the oxidation of succinate ion is:
C4H4O4^2- -> C2H4 + 2CO2 + 2e^-
- From the equation, we can see that 2 moles of succinate ion produce 1 mole of ethylene and 2 moles of carbon dioxide.
- Therefore, the number of moles of ethylene produced is (1/2) * (moles of succinate ion).
- And the number of moles of carbon dioxide produced is 2 * (moles of succinate ion).
4. Calculate the total volume of gases:
- The ideal gas equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.
- At STP, the pressure is 1 atm, the temperature is 273 K, and the gas constant is 0.0821 L.atm/(mol.K).
- The total volume of gases produced is (moles of ethylene + moles of carbon dioxide) * RT / P.
5. Substitute the values and calculate:
- The total volume of gases produced is ((1/2) * (moles of succinate ion) + 2 * (moles of succinate ion)) * 0.0821 * 273 / 1.
- Simplifying the expression, we get (5/2) * (moles of succinate ion) * 0.0821 * 273.
6. Substitute the moles of succinate ion:
- The moles of succinate ion is m / 118.09.
- Therefore, the total volume of gases produced is (5/2) * (m / 118.09) * 0.0821 * 273.
7. Simplify the expression:
- The total volume of gases produced is (0.922 * m) / 118.09.
8
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Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :a)2.24 Lb)4.48 Lc)6.72 Ld)8.96 LCorrect answer is option 'D'. Can you explain this answer?
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Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :a)2.24 Lb)4.48 Lc)6.72 Ld)8.96 LCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :a)2.24 Lb)4.48 Lc)6.72 Ld)8.96 LCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :a)2.24 Lb)4.48 Lc)6.72 Ld)8.96 LCorrect answer is option 'D'. Can you explain this answer?.
Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :a)2.24 Lb)4.48 Lc)6.72 Ld)8.96 LCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :a)2.24 Lb)4.48 Lc)6.72 Ld)8.96 LCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :a)2.24 Lb)4.48 Lc)6.72 Ld)8.96 LCorrect answer is option 'D'. Can you explain this answer?.
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Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :a)2.24 Lb)4.48 Lc)6.72 Ld)8.96 LCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :a)2.24 Lb)4.48 Lc)6.72 Ld)8.96 LCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing 0.2 Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP (1 atm and 273 K) is :a)2.24 Lb)4.48 Lc)6.72 Ld)8.96 LCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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