electrocatalysis Chemistry

5 个科学问题 · 化学 Chemistry

本页目录

氢燃料电池里为什么阴极必须要用铂催化?

Why must the cathode of a H2 fuel cell use platinum catalysis?

氧还原反应在常温下极慢,铂能大幅降低其过电位、加快反应,这正是电池成本与寿命的关键瓶颈。
Oxygen reduction is very slow at room temperature; platinum sharply lowers its overpotential, making it the key cost and durability bottleneck.
难度 专业 · 适合 大众 CC BY 4.0 · 已核验 · Verified 2026-08-23
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Housecroft & Sharpe,《Inorganic Chemistry》第5版,Pearson,2018

C. E. Housecroft and A. G. Sharpe, Inorganic Chemistry, 5th ed., Pearson, 2018

#燃料电池 #铂 #氧还原

为什么极少量硫化物就能毒化燃料电池?

Why does trace sulfur poison fuel cells so badly?

硫会强烈吸附在铂电极上占据活性位点,使氧还原反应受阻,对贵金属催化剂极为敏感,百万分之一级就能造成伤害。
Sulfur strongly adsorbs on platinum sites, blocking the oxygen-reduction reaction; even parts-per-million levels damage the precious-metal catalyst.
难度 专业 · 适合 大众 CC BY 4.0 · 已核验 · Verified 2026-08-23
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Housecroft & Sharpe,《Inorganic Chemistry》第5版,Pearson,2018

C. E. Housecroft and A. G. Sharpe, Inorganic Chemistry, 5th ed., Pearson, 2018

#燃料电池 #硫中毒 #铂

电解水制氢的两半反应是什么?

What are the two half-reactions of water electrolysis?

阴极发生 HER(2H⁺+2e⁻→H₂),阳极发生 OER(2H₂O→O₂+4H⁺+4e⁻);OER 慢是能耗瓶颈,需要好催化剂。
Cathode HER (2H⁺+2e⁻→H₂) and anode OER (2H₂O→O₂+4H⁺+4e⁻); the slow OER is the energy bottleneck needing good catalysts.
难度 专业 · 适合 大众 CC BY 4.0 · 已核验 · Verified 2026-08-23 · 2 条独立证据
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傅献彩 等,《物理化学》第5版,高等教育出版社,2005

Fu Xiancai et al., Physical Chemistry, 5th ed., Higher Education Press, 2005

#电解水 #HER #OER

为什么析氢反应在酸性和碱性电解液中的速率决定步骤可能不同?

Why can the rate-determining step of the hydrogen evolution reaction differ between acidic and alkaline electrolytes?

酸性液中质子充足,Volmer 步骤直接由 H+ 在活性位上的吸附放电主导;碱性液中水是唯一质子源,必须先断裂 O–H 键,水的解离往往成为额外的能垒。因此同一催化剂在两种介质中的 Tafel 斜率和火山图位置都会移动。这也是 NiMo 等合金在碱性中优于 Pt 的一个解释角度。
In acid, protons are abundant and the Volmer discharge of H+ onto active sites usually controls the rate. In alkali, water is the only proton source, so O–H bond cleavage adds a distinct kinetic barrier that often becomes rate-determining. Consequently the same catalyst shows different Tafel slopes and volcano positions in the two media, which partly explains why NiMo-type alloys can even outperform Pt in alkaline conditions.
难度 进阶 · 适合 初中(12-15) CC BY 4.0 · 来源待核验 · Source pending review
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出处备注:著录含书名,但缺版次 / 出版社 / 年份中的一部分,需编辑补全后才能做独立比对。

Nørskov 等 1998 年 J. Electrochem. Soc. 论文《Origin of the overpotential for hydrogen evolution on a transition metal surface》(145卷28页)

J. K. Nørskov, A. Rossmeisl, A. Logadottir, R. Sumida, T. F. Jaramik, "Origin of the overpotential for hydrogen evolution on a transition metal surface", Journal of The Electrochemical Society, 1998, 145, 28-34

#析氢反应 #电催化机理

水的理论分解电压只有1.23 V,实际电解水时阳极析氧为什么非要额外付出那么大的过电位?

Water splitting needs only 1.23 V thermodynamically — why does the oxygen evolution anode in real electrolysis demand such a hefty extra overpotential?

析氧要走完四电子路径,依次经过*OH、*O、*OOH等含氧中间体,它们的吸附能受标度关系捆绑,*OOH与*OH的自由能之差被经验地钉在约3.2 eV,而热力学理想值仅为2×1.23≈2.46 eV。这约0.74 eV的盈余平摊到两个耦合步骤,折合每电子约0.37 V的理论最小过电位。该内在损耗叠加溶液欧姆降与气泡遮蔽,共同推高了工作电位。改用双活性位点或借助晶格氧参与机制去解耦标度关系,是当前降低能耗的核心策略。
Four electrons must pass through *OH, *O and *OOH intermediates whose binding energies are chained by scaling relations: the *OOH-to-*OH free-energy gap is empirically pinned near 3.2 eV against a thermodynamic ideal of 2 x 1.23 = 2.46 eV. That ~0.74 eV surplus, shared over the two coupled steps, amounts to a theoretical minimum overpotential of about 0.37 V per electron. This intrinsic tax, added to ohmic losses and bubble shadowing, pushes the operating potential upward. Decoupling the scaling through dual active sites or lattice-oxygen participation now defines the main route to cheaper electrolysis.
难度 进阶 · 适合 初中(12-15) CC BY 4.0 · 已核验 · Verified 2026-11-04
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I. C. Man 等,《Universality in Oxygen Evolution Electrocatalysis on Oxide Surfaces》,ChemCatChem 3: 1159,2011 (DOI: 10.1002/cctc.201000397)

I. C. Man et al., 'Universality in Oxygen Evolution Electrocatalysis on Oxide Surfaces', ChemCatChem 3: 1159, 2011, DOI: 10.1002/cctc.201000397

#析氧反应 #吸附能标度