Question

Electrolytic cells use electricity to cause a nonspontaneous redox reaction to occur. An electrolytic cell is constructed using the following components: . a power source, such as a battery, • the substance that will undergo electrolysis, and • two inert electrodes (usually platinum), which serve as the electrical connection between the power source and the substance undergoing electrolysis. As with any cell, oxidation occurs at the anode and reduction occurs at the cathode. The oxidation of water, which produces oxygen (O2) gas, will compete with the oxidation of the anion, and the reduction of water, which produces hydrogen (H2) gas, will compete with the reduction of the cation. Only the reaction with the higher (more positive) potential will occur. Keep in mind that the potentials for the oxidation reaction have the opposite sign as those for the corresponding reduction reactions. For example, the oxidation reaction Ag(s) +Ag + e would have an Eox of -0.80 V.

Answer 1

Anode (Positive Electrolytic cell Mile i In Electrolytic cell - Cathode I CNegatiel), Anode ! * the charge . * oxidation will happen Celectron loss) * Anions will move towards Na' Bo Holt it. 1 Cathode - Always write mode it-ue Charge reaction first 1* reduction will happen Celectron gain) mode reaction . it cations will move towards There is possibilidy that two anions can oxidise - - - - - at Anode ( Bo or -) but only that Anion will oxidise whose oxidation, polential will be greater so in OH and Br- OH oxidation polential is more so oh will oxidise OH = -0.40V(oxiddion potential) Br=(-1.099 oxidation reaction 401 so - 029) +X H2O + 4e - Oa will produce at Anode

Cathode Reaction two locations can reduce here either Not or H* but only that cation will reduce whose reduc potential will greater, so in Nat and Ht, * reduction potential is Higher so it will reduce A' = 0oor reduction polen Hall reduction polen Hall Nate - 9.71MR:P) tu Ht Agt OH' - reduction reaction 2H+ + 20 --H2 (9) so. Ha gas will produce at Cathode Drosewer; - 02 (9), 12 (9) Holk Electrolysis oy Aq. Ag F At Anode oxidation reaction oxidation potential of OH'= -0.40:1 oxidation potential of E' = -2.87 So OH- will oxidise at Anode. 40H5 O2(g) +2 H3 0 + 4e - O2 will produce at Anode At Cathode reduction reaction Rip Of H7 50.00V Rip ¢ Ag+ = 0.801 Rip of Ag+ is more so Agt will reduce Agite - og se as a rescut Ag will deposite of ташої » O2 (9) 43 cs) ile Cathode.