Photograph 46155
At Cathode: Copper ions (Cu²⁺) are reduced to copper metal. Half-equation: Cu²⁺ + 2e⁻ → Cu(s) At Anode: Hydroxide ions (OH⁻) are…
photograph · rule, not yet confirmedClassified by size, repetition and page coverage. Rules help classify and never discard: the picture stays recorded either way.native pixelsThe image stream embedded in the document, stored as it was.282 × 179 px · jpegappears in 1 place
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chemistry igcse acapprinted 26 · PDF 26supplied fileSourceLive pagehttps://aitchison.edu.pk/portal/resources/data/chemistry-igcse-acap.pdf
Caption in this contextAt Cathode: Copper ions (Cu²⁺) are reduced to copper metal. Half-equation: Cu²⁺ + 2e⁻ → Cu(s) At Anode: Hydroxide ions (OH⁻) are oxidized to form oxygen gas. Half-equation: 4OH⁻ → O₂(g) + 2H₂O + 4e⁻ Products: Copper metal at cathode, oxygen gas at anode.
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- photograph, rule, not yet confirmed (confidence 0.60)
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- native pixels · 9.0 KB
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- 282 × 179 px
- File title
- image b0c3a04372f9
- Description (written by a model, not a caption)
- This diagram illustrates an electrolysis setup using a power source to split sodium chloride (NaCl) solution, with hydrogen gas (H₂) produced at the negative electrode (cathode) and chlorine gas (Cl₂) at the positive electrode (anode), while ions like Na⁺, Cl⁻, H⁺, and OH⁻ move within the solution.
- Words visible in the picture (read by a model)
- negative electrode (cathode) power source positive electrode (anode) H₂ Cl₂ H⁺ OH⁻ Na⁺ Cl⁻ NaCl solution
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