Question

Consider the following reactions and determine which specie will be oxidized and which one will be reduced, also, determine the reducing agent, the oxidizing agent, which specie will be found at the anode and which one will be found at the cathode. How many moles of electrons are transferred and how many Faradays of energy will be required? 1. MnO, + NO, Mn + NO, Species oxidized Reduced Reducing Found at the Oxidizing Agent Agent Found at the anode cathode #moles of Electrons transferred #Faradays used Anodic Equation Cathodic Equation 2, NO + H ---à NH, Species oxidized Reduced Reducing Found at the Oxidizing Agent Agent Found at the anode cathode fi moles of Electrons transferred # Faradays used Anodic Equation Cathodic Equation

Answer 1

Balance Equation 27 20 MnO 4 TONO & Gout - 2017 + + dono (1) oxidized - Noz oxidized to No +3720 . (2) Reduced mnog peduced to Mut (3 oxidizing Agent Mnoa (4) Reducing Agent Noz (6) mole of electron G faradajs used C2, Qunf Mno 4+se- Mut 2+ for 20 Mnou in Balanced equation = 10of 20x5 = lou election Ans = 10of Anodic Equation 80 NO +60nt 8ONO + 3H20 Camolic Eyrration 20mnou - 20 Mn ² + (Anode – oxidation | Cathode Reduction) + NO3 + 1out + de Ninu +3420 (1) oxidized – NiL (2) Reduced = Nog to Nhu ② oxidizing Agent = Nož ( Reducing agent - Nil

o found at Carmode - Nhy. de- Toumd at Anode - NIL mole of electron transferred - fenadogs we - of M Anodic Reaction – NiL. + Camodic Reaction - No3 + 10n + de nu +3420 look in a question we have only Reduction half Reaction so we are conable to write Anodic Reaction, so you can write NiL or NA UUUUU Thank you very much Hope this will help you. Good Luck