Higher Tier: electrode half equations conserve atoms and charge
Diperkenalkan| English | Makna |
|---|---|
| electrode half equation/ɪˈlektrəʊd hɑːf ɪˈkweɪʒn/ | persamaan setengah reaksi elektroda |
| discharge/dɪsˈtʃɑːdʒ/ | pelepasan |
What would explain this observation?
- Higher Tier: At a cathode, copper ions become copper by accepting electrons. At an anode, chloride ions become chlorine by releasing electrons. The two electrode changes need separate equations.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Cathode reduction examples are Cu²⁺ + 2e⁻ → Cu, Al³⁺ + 3e⁻ → Al and 2H⁺ + 2e⁻ → H₂. Anode oxidation examples are 2Cl⁻ → Cl₂ + 2e⁻ and 4OH⁻ → O₂ + 2H₂O + 4e⁻. Electrons appear on the reactant side when gained and the product side when lost. The OH⁻ equation conserves both atoms and total charge.
- electrode half equation 电极半反应方程式: An atom- and charge-balanced equation for one electrode reaction; discharge 放电析出: Conversion of an ion at an electrode through electron transfer.
Which equation correctly represents chloride oxidation?
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Oxidation at the anode is electron LOSS: 2Cl- -> Cl2 + 2e- balances both atoms (2:2) and charge (-2 each side). Electrons on the left would be reduction, and one electron leaves the charge unbalanced.
Match each technical term to its precise meaning.
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Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- For 4OH⁻ → O₂ + 2H₂O + 4e⁻, oxygen count is four on each side and hydrogen count four on each side. Charge is −4 on both sides. For 2Cl⁻ → Cl₂ + 2e⁻, chlorine is diatomic and the electron total balances the two negative charges. A correct formula is needed before balancing; changing Cl₂ to Cl would misidentify the product.
- Write the species and intended products first, then balance atoms, then charge with electrons. Check electrode sign and oxidation/reduction identity. Multiply a half equation only when matching overall electron transfer or the question’s required amount. Distinguish the basic metal/halide molten examples from water-derived aqueous examples and the industrial reacting carbon anode.
Which two habits make the investigation or model in this case more defensible?
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Stating the conditions and assumptions lets others repeat the work exactly, and recording the supporting evidence lets them check the conclusion. Hiding disagreeing results removes the evidence that could challenge a wrong conclusion, and one measurement can never establish a wider 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: three represented Cu²⁺ ions require six electrons to form three Cu atoms. Four represented Al³⁺ ions require twelve. Six represented Cl⁻ ions form three Cl₂ molecules and release six electrons. This count supports the HT equation model; it does not require calculating electrical charge, current efficiency or Faraday’s constant outside the GCSE objectives.
How many electrons reduce five Al³⁺ ions to aluminium atoms? Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
How many electrons reduce five Al³⁺ ions to aluminium atoms?
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The result is 15 electrons. Known: each Al3+ needs 3 electrons to become Al. For five ions: 5 x 3 = 15 electrons. The half-equation Al3+ + 3e- -> Al supplies the ratio.
Check the conclusion and its limits
- Electrons do not disappear from charge accounting just because they are not atoms. Anode oxidation is electron loss in this powered electrolysis model. Positive-ion discharge does not mean gaining positive electrons. The entire 4.4.3.5 section and the half-equation requirement across 4.4.3 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.
The hydroxide anode equation can omit water while still conserving hydrogen atoms. This claim is false: Water must appear in the hydroxide anode equation because the hydrogen atoms come from it: 4OH- -> O2 + 2H2O + 4e-. Omitting the water loses four hydrogen atoms and breaks conservation.
Higher Tier: electrode half equations conserve atoms and charge: For 4OH⁻ → O₂ + 2H₂O + 4e⁻, oxygen count is four on each side and hydrogen count four on each side. Charge is −4 on both sides. For 2Cl⁻ → Cl₂ + 2e⁻, chlorine is diatomic and the electron total balances the two negative charges. A correct formula is needed before balancing; changing Cl₂ to Cl would misidentify the product.
The hydroxide anode equation can omit water while still conserving hydrogen atoms.
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Water must appear in the hydroxide anode equation because the hydrogen atoms come from it: 4OH- -> O2 + 2H2O + 4e-. Omitting the water loses four hydrogen atoms and breaks conservation.
An atom- and charge-balanced equation for one electrode reaction: write the technical term.
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electrode half equation means An atom- and charge-balanced equation for one electrode reaction.