Высокий уровень: объедините полуреакции водородного топливного элемента
Введено| English | Значение |
|---|---|
| fuel-cell anode/ˈfjuːəl sel ˈænəʊd/ | анод топливного элемента |
| fuel-cell cathode/ˈfjuːəl sel ˈkæθəʊd/ | катод топливного элемента |
What would explain this observation?
- Higher Tier: A hydrogen fuel cell separates oxidation and reduction at two electrodes. The same number of electrons must be released at one electrode and accepted at the other.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- For the explicitly acidic-electrolyte model, hydrogen oxidation is 2H₂ → 4H⁺ + 4e⁻. Oxygen reduction is O₂ + 4H⁺ + 4e⁻ → 2H₂O. Adding the half equations cancels four hydrogen ions and four electrons, leaving 2H₂ + O₂ → 2H₂O. Oxidation means electron loss and reduction electron gain. These half equations are the embedded Higher-only part of the source section.
- fuel-cell anode 燃料电池阳极: The electrode where hydrogen oxidation releases electrons in the stated generating cell; fuel-cell cathode 燃料电池阴极: The electrode where oxygen reduction accepts electrons in the stated generating cell.
Какое кислотное полуреакционное уравнение водородного электрода показывает окисление?
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Окисление отдаёт электроны, поэтому полуреакция водорода: H2 → 2H⁺ + 2e⁻, уравнена по атомам (2:2) и заряду (0 с каждой стороны). Электроны слева описывали бы восстановление, а кислород без видов водорода не может образовать воду.
Сопоставьте каждый технический термин с его точным определением.
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Используйте определения для различения смежных величин и процессов.
Choose evidence that can test it
- Check atoms and charges separately. The hydrogen half equation has four hydrogen atoms on each side and zero net charge because +4 from H⁺ balances −4 from electrons. The oxygen half equation has two oxygen atoms and four hydrogen atoms on each side, again zero net charge. In this generating cell electrons leave the hydrogen anode through the external circuit and reach the oxygen cathode. The anode is negative and the cathode positive while the cell supplies electricity.
- Begin with one consistent electrolyte model. Balance hydrogen with H⁺, oxygen with water and charge with electrons, then scale to equal electron numbers and cancel. Do not mix an acidic half equation with an alkaline one containing OH⁻. The electrode names identify oxidation at the anode and reduction at the cathode; signs must be interpreted for a generating fuel cell, rather than copied from an externally powered electrolytic cell.
Какие две привычки делают расследование или модель в данном случае более обоснованными?
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Указание условий и допущений позволяет другим точно воспроизвести работу, а фиксация подтверждающих данных даёт возможность проверить вывод. Сокрытие противоречащих результатов устраняет доказательства, способные оспорить ошибочный вывод, а одно измерение никогда не может обосновать более широкий claim.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: starting from H₂ → 2H⁺ + 2e⁻, multiply by two to match the four electrons consumed by O₂ + 4H⁺ + 4e⁻ → 2H₂O. Two represented H₂ molecules supply four electrons and one O₂ molecule accepts four. If three O₂ molecules react in the same scaled model, twelve electrons pass through the external circuit and six H₂O molecules form. Electrons cancel from the overall equation.
Each represented H₂ molecule releases two electrons in the stated half equation. How many are released by eight H₂ molecules? Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
Каждая молекула H₂ в указанном полуреакции отдает два электрона. Сколько электронов отдают восемь молекул H₂?
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Результат составляет 16 электронов. Известно: каждая молекула H2 отдаёт 2 электрона в указанной полуреакции. Для восьми молекул H2: 8 × 2 = 16 электронов отдано, протекая через внешнюю цепь к кислородному электроду.
Check the conclusion and its limits
- Do not put electrons on the reactant side for hydrogen oxidation, allow them to remain in the overall reaction, or say they pass through the electrolyte like they do through a metal wire. Common-tier students still need the overall reaction and fuel-cell evaluation, but these half equations and charge reasoning are Higher-only.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Electrons remain as a product of the complete balanced fuel-cell reaction. This claim is false: Electrons travel through the EXTERNAL circuit and cancel when the half equations combine: H2 releases exactly what O2 consumes. The overall reaction 2H2 + O2 -> 2H2O contains no electrons.
Higher Tier: combine hydrogen fuel-cell half equations: Check atoms and charges separately. The hydrogen half equation has four hydrogen atoms on each side and zero net charge because +4 from H⁺ balances −4 from electrons. The oxygen half equation has two oxygen atoms and four hydrogen atoms on each side, again zero net charge. In this generating cell electrons leave the hydrogen anode through the external circuit and reach the oxygen cathode. The anode is negative and the cathode positive while the cell supplies electricity.
Электроны остаются в качестве продукта полной сбалансированной реакции топливного элемента.
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Электроны проходят через ВНЕШНЮЮ цепь и сокращаются при объединении полуреакций: H2 отдаёт ровно столько, сколько потребляет O2. Общая реакция 2H2 + O2 → 2H2O не содержит электронов.
Электрод, на котором окисление водорода высвобождает электроны в указанном генерирующем элементе: запишите технический термин.
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аноде топливного элемента означает Электрод, на котором окисление водорода высвобождает электроны в указанном генерирующем элементе.