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Chemical oxidizing of water to produce O2 gas is an energy demanding reaction, done routinely by plants using the process called photosynthesis. By how many eV will it be uphill if the water oxidation reaction be carried out at pH = 0 versus at pH = 7.0:
- a)0.41 eV
- b)–1.6 eV
- c)–0.41 eV
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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Chemical oxidizing of water to produce O2 gas is an energy demanding r...
Explanation:
The oxidation of water to produce O2 gas is an energy demanding reaction. The energy required to carry out this reaction can be calculated by considering the difference in the standard reduction potentials of the two half-reactions involved in the process.
The half-reactions involved in the process are:
2H2O(l) → O2(g) + 4H+(aq) + 4e- (oxidation half-reaction)
2H+(aq) + 2e- → H2(g) (reduction half-reaction)
The standard reduction potential for the reduction half-reaction is 0.00 V, while the standard reduction potential for the oxidation half-reaction is +1.23 V. The standard potential difference (ΔE°) between the two half-reactions is therefore:
ΔE° = E°(oxidation) - E°(reduction) = +1.23 V - 0.00 V = +1.23 V
This means that the oxidation of water to produce O2 gas is an uphill reaction, requiring an input of energy to proceed. The energy required can be calculated using the equation:
ΔG° = -nFE°
where ΔG° is the standard free energy change, n is the number of electrons transferred, F is the Faraday constant (96,485 C/mol), and E° is the standard reduction potential.
For the water oxidation reaction, n = 4 (since 4 electrons are transferred in the oxidation half-reaction) and ΔG° = +nFE° = +(4)(96,485 C/mol)(+1.23 V) = +472,311 J/mol = +0.47 kJ/mol.
Now, the pH of the solution also affects the standard potential of the half-reactions involved in the process. The standard reduction potential is pH-dependent and can be calculated using the Nernst equation:
E = E° - (RT/nF) ln([H+]/[H+])
where E is the reduction potential at the given pH, R is the gas constant, T is the temperature, n is the number of electrons transferred, F is the Faraday constant, [H+] is the hydrogen ion concentration, and [H+] is the standard hydrogen ion concentration.
At pH = 7.0, the hydrogen ion concentration is [H+] = 10^-7 M, and the standard reduction potential can be calculated as:
E = E° - (RT/nF) ln([H+]/[H+]) = 0.00 V - (0.0257 V/n) ln(10^-7/1) = 0.00 V - 0.0257 V = -0.026 V
At pH = 0, the hydrogen ion concentration is [H+] = 1 M, and the standard reduction potential can be calculated as:
E = E° - (RT/nF) ln([H+]/[H+]) = 0.00 V - (0.0257 V/n) ln(1/1) = 0.00 V - 0.0257 V = -0.026 V
The change in the standard reduction potential between pH = 7.0 and pH = 0 is therefore:
ΔE° = E°(pH=0) - E°(pH=7.0) =
The oxidation of water to produce O2 gas is an energy demanding reaction. The energy required to carry out this reaction can be calculated by considering the difference in the standard reduction potentials of the two half-reactions involved in the process.
The half-reactions involved in the process are:
2H2O(l) → O2(g) + 4H+(aq) + 4e- (oxidation half-reaction)
2H+(aq) + 2e- → H2(g) (reduction half-reaction)
The standard reduction potential for the reduction half-reaction is 0.00 V, while the standard reduction potential for the oxidation half-reaction is +1.23 V. The standard potential difference (ΔE°) between the two half-reactions is therefore:
ΔE° = E°(oxidation) - E°(reduction) = +1.23 V - 0.00 V = +1.23 V
This means that the oxidation of water to produce O2 gas is an uphill reaction, requiring an input of energy to proceed. The energy required can be calculated using the equation:
ΔG° = -nFE°
where ΔG° is the standard free energy change, n is the number of electrons transferred, F is the Faraday constant (96,485 C/mol), and E° is the standard reduction potential.
For the water oxidation reaction, n = 4 (since 4 electrons are transferred in the oxidation half-reaction) and ΔG° = +nFE° = +(4)(96,485 C/mol)(+1.23 V) = +472,311 J/mol = +0.47 kJ/mol.
Now, the pH of the solution also affects the standard potential of the half-reactions involved in the process. The standard reduction potential is pH-dependent and can be calculated using the Nernst equation:
E = E° - (RT/nF) ln([H+]/[H+])
where E is the reduction potential at the given pH, R is the gas constant, T is the temperature, n is the number of electrons transferred, F is the Faraday constant, [H+] is the hydrogen ion concentration, and [H+] is the standard hydrogen ion concentration.
At pH = 7.0, the hydrogen ion concentration is [H+] = 10^-7 M, and the standard reduction potential can be calculated as:
E = E° - (RT/nF) ln([H+]/[H+]) = 0.00 V - (0.0257 V/n) ln(10^-7/1) = 0.00 V - 0.0257 V = -0.026 V
At pH = 0, the hydrogen ion concentration is [H+] = 1 M, and the standard reduction potential can be calculated as:
E = E° - (RT/nF) ln([H+]/[H+]) = 0.00 V - (0.0257 V/n) ln(1/1) = 0.00 V - 0.0257 V = -0.026 V
The change in the standard reduction potential between pH = 7.0 and pH = 0 is therefore:
ΔE° = E°(pH=0) - E°(pH=7.0) =
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Chemical oxidizing of water to produce O2 gas is an energy demanding reaction, done routinely by plants using the process called photosynthesis. By how many eV will it be uphill if the water oxidation reaction be carried out at pH = 0 versus at pH = 7.0:a)0.41 eVb)–1.6 eVc)–0.41 eVd)None of theseCorrect answer is option 'A'. Can you explain this answer?
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