[{"data":1,"prerenderedAt":102},["ShallowReactive",2],{"cheatsheet-electrochemistry":3},{"sheet":4},{"sheetSlug":5,"topicSlugs":6,"primaryTopic":7,"title":8,"subtitle":9,"sections":10},"electrochemistry",[5],26,"Electrochemistry","Galvanic cells, E°cell, the Nernst equation, electrolysis, and the E-ΔG-K triangle.",[11,23,31,39,47,55,63,72,81,89],{"heading":12,"kind":13,"items":14},"Building a galvanic cell","rules",[15],{"ref":16,"mode":19,"provenance":20,"text":21,"source_hash":22},{"type":17,"topic":7,"id":18,"field":17},"rule","EC-R001","transform","owned_workbook","Reverse the half-reaction with the more negative standard reduction potential (it becomes the oxidation/anode) so E°cell comes out positive.","0f08f7d778e47575164ca5269142615aefca4c31ec7dd1ac41720ddb25956fa6",{"heading":24,"kind":13,"items":25},"E°cell",[26],{"ref":27,"mode":19,"provenance":20,"text":29,"source_hash":30},{"type":17,"topic":7,"id":28,"field":17},"EC-R002","E°cell = E°(cathode) − E°(anode), both as reduction potentials. Multiplying a half-reaction by a coefficient does NOT change its potential.","c5fd805f4cfed98bb7df3efa821c4cd8c4dcf20e268adae11f11c9ca8ec23b38",{"heading":32,"kind":13,"items":33},"Cell notation",[34],{"ref":35,"mode":19,"provenance":20,"text":37,"source_hash":38},{"type":17,"topic":7,"id":36,"field":17},"EC-R003","anode | anode ion || cathode ion | cathode. Oxidation on the left; no coefficients. Use an inert Pt electrode when neither side is a conducting solid.","6be4839d74994febc185312cca36b92ae557dc0ce42eaaa8446baaee82e6bc21",{"heading":40,"kind":13,"items":41},"Nernst equation",[42],{"ref":43,"mode":19,"provenance":20,"text":45,"source_hash":46},{"type":17,"topic":7,"id":44,"field":17},"EC-R005","E = E° − (0.0592/n) log Q at 25 °C (n = electrons transferred, Q = products/reactants). Use it for any non-standard concentration or pressure.","c5aff0493c37e756494774914af3884f67a79c2ae23d816a36d29e82092f10ca",{"heading":48,"kind":13,"items":49},"pH from cell voltage",[50],{"ref":51,"mode":19,"provenance":20,"text":53,"source_hash":54},{"type":17,"topic":7,"id":52,"field":17},"EC-R007","Put the measured E into the Nernst equation, solve for [H⁺], then pH = −log[H⁺].","7712b9905bf1fa63f37ebe56c54e28dea60ef0eb0b8ec2ad7a8252a14cb7d64c",{"heading":56,"kind":13,"items":57},"Charge & Faraday",[58],{"ref":59,"mode":19,"provenance":20,"text":61,"source_hash":62},{"type":17,"topic":7,"id":60,"field":17},"EC-R008","Charge (C) = current(A) × time(s) (convert h/min to s). Moles of e⁻ = charge / F, with F = 96,485 C/mol e⁻.","ebe72495c09960975da5ece4429a13b96e7695a08eee55cb81e6d1126bb8283b",{"heading":64,"kind":65,"items":66},"Electrolysis stoichiometry","steps",[67],{"ref":68,"mode":19,"provenance":20,"text":70,"source_hash":71},{"type":17,"topic":7,"id":69,"field":17},"EC-R009","For Mⁿ⁺ + ne⁻ → M: mol metal = mol e⁻ / n. Convert that to mass with molar mass to finish a deposition problem.","5f9227e03117cbd0d8364d909703a50725710da5603f565c41a9ecb690aee351",{"heading":73,"kind":13,"items":74},"ΔG° = −nFE°cell",[75],{"ref":76,"mode":19,"provenance":78,"text":79,"source_hash":80},{"type":17,"topic":7,"id":77,"field":17},"EC-R011","original","ΔG° = −nFE°cell. E°cell > 0 ↔ ΔG° \u003C 0 (spontaneous galvanic); E°cell \u003C 0 ↔ ΔG° > 0 (electrolytic if driven). n = e⁻ per balanced reaction.","aaeed456731cdbc139f6c7df314317e752b9eba17223a9ab185de9dbdebf1d99",{"heading":82,"kind":13,"items":83},"K from E°cell",[84],{"ref":85,"mode":19,"provenance":78,"text":87,"source_hash":88},{"type":17,"topic":7,"id":86,"field":17},"EC-R012","ln K = nFE°cell/(RT), or log K = nFE°cell/(2.303RT). E°cell > 0 → K > 1. Together with ΔG° = −RT ln K this closes the E°cell ↔ ΔG° ↔ K triangle.","32ad28a31e9ca932956469b73e5a9f038aea09955b9cf3d69c5ae6784dbfff65",{"heading":90,"kind":13,"items":91},"Concentration cells",[92,97],{"ref":93,"mode":19,"provenance":78,"text":95,"source_hash":96},{"type":17,"topic":7,"id":94,"field":17},"EC-R013","Same chemistry, different [ion], so E° = 0 and E = −(0.0592/n) log Q. Write Q from the balanced overall reaction (Mⁿ⁺: Q = [anode]/[cathode]). At equal concentrations Q = 1, E = 0.","25b5417e13d157ac057f5b8767ad337cb55f2ad2d78d243d6f88a95c58254320",{"ref":98,"mode":19,"provenance":78,"text":100,"source_hash":101},{"type":17,"topic":7,"id":99,"field":17},"EC-R014","The low-[ion] half-cell is the anode (its concentration rises); the high-[ion] half-cell is the cathode (falls). Electrons flow anode → cathode until concentrations equalize.","58b21305924772000c1e1d4980ae084f01a216ce85980c6f38d4754f1c57d702",1787246033133]