Higher Tier: electrode half equations conserve atoms and charge
| English | 中文 | Pinyin |
|---|---|---|
| electrode half equation | 电极半反应方程式 | diàn jí bàn fǎn yìng fāng chéng shì |
| discharge/dɪsˈtʃɑːdʒ/ | 放电析出 | fàng diàn xī chū |
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?
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.
Match each technical term to its precise meaning.
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?
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.
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?
The result is 15 electrons. 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.
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: 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.
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.
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.
An atom- and charge-balanced equation for one electrode reaction: write the technical term.
electrode half equation means An atom- and charge-balanced equation for one electrode reaction.