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A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing
hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be ?
hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be ?
- a)0.059 V
- b)0.59 V
- c)0.118 V
- d)1.18 V
Correct answer is option 'B'. Can you explain this answer?
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A hydrogen gas electrode is made by dipping platinum wire in a solutio...
Oxidation potential of Hydrogen gas electrode:
The hydrogen gas electrode is a standard electrode used as a reference electrode for measuring the potential of other electrodes.
Preparation of Hydrogen gas electrode:
- Dip platinum wire in a solution of HCl of pH = 10.
- Pass hydrogen gas around the platinum wire at one atm pressure.
Oxidation potential of the electrode:
- The oxidation potential of the electrode is given by the Nernst equation:
E = E° + (RT/nF) ln ([Ox]/[Red])
where E is the cell potential, E° is the standard cell potential, R is the gas constant, T is the temperature, n is the number of electrons transferred in the reaction, F is the Faraday constant, [Ox] is the concentration of the oxidized form, and [Red] is the concentration of the reduced form.
- Since the hydrogen gas electrode is a standard electrode, its standard potential is known to be 0 V.
- The concentration of hydrogen gas is assumed to be 1 atm, which is equivalent to 1.013 x 10^5 Pa.
- The concentration of hydrogen ions in the HCl solution is given by pH = -log[H+], which is 10^-10 M.
- The reduction half-reaction for the hydrogen gas electrode is:
H+ + e- → 1/2 H2
- Therefore, the oxidation half-reaction is:
1/2 H2 → H+ + e-
- The number of electrons transferred in the reaction is 1.
- Substituting the values into the Nernst equation, we get:
E = 0 V + (8.314 J/K mol) (298 K) / (1 mol) ln (1/10^-10)
E = 0.059 V
- Therefore, the oxidation potential of the hydrogen gas electrode is 0.059 V.
Answer: Option B (0.59 V)
The hydrogen gas electrode is a standard electrode used as a reference electrode for measuring the potential of other electrodes.
Preparation of Hydrogen gas electrode:
- Dip platinum wire in a solution of HCl of pH = 10.
- Pass hydrogen gas around the platinum wire at one atm pressure.
Oxidation potential of the electrode:
- The oxidation potential of the electrode is given by the Nernst equation:
E = E° + (RT/nF) ln ([Ox]/[Red])
where E is the cell potential, E° is the standard cell potential, R is the gas constant, T is the temperature, n is the number of electrons transferred in the reaction, F is the Faraday constant, [Ox] is the concentration of the oxidized form, and [Red] is the concentration of the reduced form.
- Since the hydrogen gas electrode is a standard electrode, its standard potential is known to be 0 V.
- The concentration of hydrogen gas is assumed to be 1 atm, which is equivalent to 1.013 x 10^5 Pa.
- The concentration of hydrogen ions in the HCl solution is given by pH = -log[H+], which is 10^-10 M.
- The reduction half-reaction for the hydrogen gas electrode is:
H+ + e- → 1/2 H2
- Therefore, the oxidation half-reaction is:
1/2 H2 → H+ + e-
- The number of electrons transferred in the reaction is 1.
- Substituting the values into the Nernst equation, we get:
E = 0 V + (8.314 J/K mol) (298 K) / (1 mol) ln (1/10^-10)
E = 0.059 V
- Therefore, the oxidation potential of the hydrogen gas electrode is 0.059 V.
Answer: Option B (0.59 V)
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A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passinghydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be ?a)0.059 Vb)0.59 Vc)0.118 Vd)1.18 VCorrect answer is option 'B'. Can you explain this answer?
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A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passinghydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be ?a)0.059 Vb)0.59 Vc)0.118 Vd)1.18 VCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passinghydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be ?a)0.059 Vb)0.59 Vc)0.118 Vd)1.18 VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passinghydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be ?a)0.059 Vb)0.59 Vc)0.118 Vd)1.18 VCorrect answer is option 'B'. Can you explain this answer?.
A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passinghydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be ?a)0.059 Vb)0.59 Vc)0.118 Vd)1.18 VCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passinghydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be ?a)0.059 Vb)0.59 Vc)0.118 Vd)1.18 VCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passinghydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be ?a)0.059 Vb)0.59 Vc)0.118 Vd)1.18 VCorrect answer is option 'B'. Can you explain this answer?.
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