[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"doc-detail-207963-en":3,"doc-seo-207963-105":30,"detail-sidebar-cat-0-en-105":89},{"code":4,"msg":5,"data":6},0,"success",{"doc_id":7,"user_id":8,"nickname":9,"user_avatar":10,"doc_module":4,"category_id":11,"category_name":12,"doc_title":13,"doc_description":14,"doc_content":15,"file_id":16,"file_url":17,"file_type":18,"file_size":19,"view_count":4,"is_deleted":4,"is_public":20,"is_downloadable":20,"audit_status":20,"page_count":21,"language":22,"language_code":23,"site_id":24,"html_lang":23,"table_of_contents":25,"faqs":26,"seo_title":27,"seo_description":14,"update_tm":28,"read_time":29},207963,549768064778,"Felix Montgomery","https://ap-avatar.wpscdn.com/davatar_6f874abed73319feea01a86fa6f0fab8",4,"Exam","Electrolysis - Electrolytic cells, molten salts, and aqueous solutions","Electrolysis covers how electrolytic cells convert electrical energy into chemical energy through non-spontaneous reactions. It explains Faraday’s relationships linking charge and moles of electrons using the Faraday and Avogadro constants, and describes cathode (reduction, electron gain) versus anode (oxidation, electron loss) processes. Separate sections analyze molten salts and aqueous solutions, including electrode potentials, hydrogen versus metal deposition, oxygen versus halogen evolution, and how dilution changes products. Worked calculations demonstrate determining stoichiometric charge effects and calculating Avogadro’s constant from experimental data.","Electrolysis  \nElectrolytic cells convert electrical energy to chemical energy, by bringing about non-spontaneous processes.  \nF = Le connects the Faraday constant, the Avogadro constant and the charge on the electron  \nF= Faraday constant ( a measure of the charge of 1 mole of electrons = 96320 coulombs  \nL = Avogadro constant = 6.022 x 1023 e = charge of on electron = 1.60 x 10-19 coulombs.  \nElectrolysis of molten salts  \nWhen a simple ionic compound is electrolysed in the molten state using inert electrodes, the salt splits and the metal ion moves to the negative electrode and the negative ion moves to the positive electrode  \ne.g. if molten lead bromide is electrolysed, the lead will form at the negative electrode and bromine will form atthe positive electrode  \nAt the negative electrode (cathode)  \nAt the negative electrode, positively charged ions gain electrons to become metal atoms.  \nThis is classed as reduction. Reduction is gaining electrons  \nNa+ (l )+ e- 􏃠 Na (s) (sodium ions become sodium atoms)  \nCu2+ (l )+ 2e- 􏃠 Cu (s)  \nAl3+ (l )+ 3e- 􏃠 Al (s)  \nAt the positive electrode (anode)  \nAt the positive electrode, negatively charged ions lose electrons.  \nThis is classed as oxidation. Oxidation is losing electrons 2Cl- (l) 􏃠 Cl2 (g)+ 2e- (chloride ions becomes chlorine)  \n2Br- (l) 􏃠 Br2 (l)+ 2e- (bromide ions becomes bromine)  \n2I- 􏃠 I2 + 2e- (Iodide ions becomes Iodine)  \n2O2- (l) 􏃠 O2 (g)+ 4e- (oxide ions becomes oxygen)  \nOIL RIG can help remember that Oxidation is Loss of electrons: Reduction is Gain of electrons  \nElectrolysing aqueous solutions  \nIf an aqueous solution is electrolysed, using inert electrodes, the ions discharged depend on the electrode potentials of the ions involved.  \nThe negative electrode  \nIn aqueous solutions there is a mixture of ions: H+ and OH- ions are present in addition to the ions from the salt [e.g. in](e.g. in) copper chloride solution there are H+ ,OH- (from water) Cu2+ ,Cl- (from the salt)  \nAt the negative electrode (cathode) in aqueous mixtures the cation with the more positive electrode potential discharges.  \nIn aqueous solutions where the metal is more negative in the electrochemical series than hydrogen, the metal will not be evolved at the cathode. (e.g. sodium chloride, calcium fluoride) Hydrogen gas will be evolved at the cathode instead.  \n2H+ (aq) + 2e- 􏃠 H2 (g)  \nIn aqueous solutions where the metal has a more positive E° value than hydrogen in the electrochemical series (e.g. copper chloride or silver fluoride,) the metal will be evolved at the cathode.  \nCu2+ (aq )+ 2e- 􏃠 Cu (s)  \nThis happens because in the aqueous solution water molecules break down producing hydrogen ions and hydroxide ions that are discharged.  \nThe positive electrode  \nAt the positive electrode (anode), oxygen is produced unless the solution contains halide ions when the halogen is produced.  \n4OH- (aq)→ O2 (g)+ 2H2O (l) + 4e- or 4OH- – 4e- → O2 + 2H2O  \nIf a halide ion is present then the halogen is produced e.g. 2Cl- (aq)􏃠 Cl2 (g)+ 2e-  \nThe concentration of the negative ion can change the product evolved. A concentrated solution of sodium chloride, would give mostly chlorine gas. With increasingly dilute solutions, less chlorine and more oxygen will be evolved  \n[N Goalby chemrevise.org](N Goalby chemrevise.org) 1  \nCalculations with Electrolysis  \nVarious questions can be asked using the following equations and constants (found in data book)  \ncharge passed (in Coulombs) = current x time (in secs)  \nMoles of electrons = charge (in C) / F  \nNumber of electrons passed = charge (in C) / electronic charge e  \nExample  \nA chromium salt was electrolysed. The chromium metal is deposited on the cathode, according to the following equation. Crn+(aq) + ne– → Cr(s)  \nA current of 1.2 A was passed for 50 minutes through a solution of the chromium salt. 0.647 g of chromium was deposited. Calculate the value of n in the above equation .  \ncharge passed (in Coulombs) = current x time (in secs) charge p","cbCaingWZRgZTZre","https://ap.wps.com/l/cbCaingWZRgZTZre","pdf",71658,1,2,"English","en",105,"# Electrolysis fundamentals\n## Faraday’s constants and electron charge relationships\n# Electrolysis of molten salts\n## Cathode: reduction and metal deposition\n## Anode: oxidation and halogen/oxygen formation\n# Electrolysing aqueous solutions\n## Cathode: cation discharge rules\n## Anode: oxygen or halogen evolution\n# Calculations with electrolysis\n## Charge, moles of electrons, and product amount\n## Worked example: finding n for chromium deposition\n## Worked example: calculating Avogadro’s constant","[{\"question\":\"What do Faraday’s relationships connect in electrolysis calculations?\",\"answer\":\"They connect charge passed to moles of electrons using the Faraday constant, and they relate electron amount to the Avogadro constant via the number of particles corresponding to electrons.\"},{\"question\":\"How do cathode products differ between molten salts and aqueous solutions?\",\"answer\":\"In molten salts, metal cations form at the cathode through gain of electrons. In aqueous solutions, the discharged cation depends on electrode potentials; hydrogen evolves at the cathode when the metal is more negative than hydrogen in the electrochemical series.\"},{\"question\":\"Why can oxygen or halogens be produced at the anode in aqueous electrolysis?\",\"answer\":\"Oxygen is produced at the anode from hydroxide ions unless halide ions are present, in which case the corresponding halogen gas is formed. Dilution can shift the relative amounts of chlorine and oxygen formed.\"}]","Electrolysis - Electrolytic cells, molten salts, and aqueous solutions | PDF",1788601922,5,{"code":4,"msg":31,"data":32},"ok",{"site_id":24,"language":23,"slug":33,"title":13,"keywords":34,"description":14,"schema_data":35,"social_meta":84,"head_meta":86,"extra_data":88,"updated_unix":28},"electrolysis-electrolytic-cells-molten-salts-and-aqueous-solutions","",{"@graph":36,"@context":83},[37,52,66],{"@type":38,"itemListElement":39},"BreadcrumbList",[40,44,47,50],{"item":41,"name":42,"@type":43,"position":20},"https://docshare.wps.com","Home","ListItem",{"item":45,"name":46,"@type":43,"position":21},"https://docshare.wps.com/document/","Document",{"item":48,"name":12,"@type":43,"position":49},"https://docshare.wps.com/document/exam/",3,{"item":51,"name":13,"@type":43,"position":11},"https://docshare.wps.com/document/electrolysis-electrolytic-cells-molten-salts-and-aqueous-solutions/207963/",{"url":51,"name":13,"@type":53,"author":54,"headline":13,"publisher":56,"fileFormat":59,"inLanguage":23,"description":14,"dateModified":60,"datePublished":60,"encodingFormat":59,"isAccessibleForFree":61,"interactionStatistic":62},"DigitalDocument",{"name":9,"@type":55},"Person",{"url":41,"name":57,"@type":58},"DocShare","Organization","application/pdf","2026-09-05",true,{"@type":63,"interactionType":64,"userInteractionCount":4},"InteractionCounter",{"@type":65},"ViewAction",{"@type":67,"mainEntity":68},"FAQPage",[69,75,79],{"name":70,"@type":71,"acceptedAnswer":72},"What do Faraday’s relationships connect in electrolysis calculations?","Question",{"text":73,"@type":74},"They connect charge passed to moles of electrons using the Faraday constant, and they relate electron amount to the Avogadro constant via the number of particles corresponding to electrons.","Answer",{"name":76,"@type":71,"acceptedAnswer":77},"How do cathode products differ between molten salts and aqueous solutions?",{"text":78,"@type":74},"In molten salts, metal cations form at the cathode through gain of electrons. In aqueous solutions, the discharged cation depends on electrode potentials; hydrogen evolves at the cathode when the metal is more negative than hydrogen in the electrochemical series.",{"name":80,"@type":71,"acceptedAnswer":81},"Why can oxygen or halogens be produced at the anode in aqueous electrolysis?",{"text":82,"@type":74},"Oxygen is produced at the anode from hydroxide ions unless halide ions are present, in which case the corresponding halogen gas is formed. Dilution can shift the relative amounts of chlorine and oxygen formed.","https://schema.org",{"og:url":51,"og:type":85,"og:title":13,"og:site_name":57,"og:description":14},"article",{"robots":87,"canonical":51},"index,follow",{"doc_id":7,"site_id":24},{"code":4,"msg":5,"data":90},[91,95,99,102,106,111,116,121,126,129,133],{"id":20,"doc_module":4,"doc_module_name":46,"category_name":92,"show_sort_weight":93,"slug":94},"Story & Novel",90,"story-novel",{"id":21,"doc_module":4,"doc_module_name":46,"category_name":96,"show_sort_weight":97,"slug":98},"Literature",80,"literature",{"id":11,"doc_module":4,"doc_module_name":46,"category_name":12,"show_sort_weight":100,"slug":101},70,"exam",{"id":29,"doc_module":4,"doc_module_name":46,"category_name":103,"show_sort_weight":104,"slug":105},"Comic",60,"comic",{"id":107,"doc_module":4,"doc_module_name":46,"category_name":108,"show_sort_weight":109,"slug":110},6,"Technology",50,"technology",{"id":112,"doc_module":4,"doc_module_name":46,"category_name":113,"show_sort_weight":114,"slug":115},7,"Healthcare",40,"healthcare",{"id":117,"doc_module":4,"doc_module_name":46,"category_name":118,"show_sort_weight":119,"slug":120},8,"Research & Report",30,"research-report",{"id":122,"doc_module":4,"doc_module_name":46,"category_name":123,"show_sort_weight":124,"slug":125},9,"Religion & Spirituality",20,"religion-spirituality",{"id":124,"doc_module":4,"doc_module_name":46,"category_name":127,"show_sort_weight":124,"slug":128},"World Cup","world-cup",{"id":130,"doc_module":4,"doc_module_name":46,"category_name":131,"show_sort_weight":130,"slug":132},10,"Lifestyle","lifestyle",{"id":134,"doc_module":4,"doc_module_name":46,"category_name":135,"show_sort_weight":29,"slug":136},19,"General","general"]