PbO2/PbSO4 = 1.69V
PbSO4/Pb = -0.36V
So I've figured out that PbO2 couple is reduction and PbSO4 is oxidation? Is that right? How do you write the half equations ?
thanks in advance
In electrochemical cell, when discharge happens here, how do you write the half equations?
2017-02-20 4:39 pm
回答 (1)
2017-02-20 5:28 pm
The reaction is feasible when E°(cell) > 0
Hence, E°(cell) = E°(PbO₂/PbSO₄) - E°(PbSO₄/Pb) = 1.69 V - (-0.36 V) = 2.05 V > 0
E°(cell) = E°(reduction) - E°(oxidation)
Hence, PbO₂ is reduced to PbSO₄ at the cathode, while Pb is oxidized to PbSO₄ at the anode.
Half equation for the cathodic (reduction) reaction :
PbO₂(s) + 4H⁺(aq) + SO₄²⁻(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)
Half equation for the anodic (oxidation) reaction :
Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
Hence, E°(cell) = E°(PbO₂/PbSO₄) - E°(PbSO₄/Pb) = 1.69 V - (-0.36 V) = 2.05 V > 0
E°(cell) = E°(reduction) - E°(oxidation)
Hence, PbO₂ is reduced to PbSO₄ at the cathode, while Pb is oxidized to PbSO₄ at the anode.
Half equation for the cathodic (reduction) reaction :
PbO₂(s) + 4H⁺(aq) + SO₄²⁻(aq) + 2e⁻ → PbSO₄(s) + 2H₂O(l)
Half equation for the anodic (oxidation) reaction :
Pb(s) + SO₄²⁻(aq) → PbSO₄(s) + 2e⁻
收錄日期: 2021-04-18 16:01:14
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