Question

Revised 7/25/19 Name Station Date CHM 112 Determination of an Equivalent Mass by Electrolysis Pre Lab Questions 1- In the following reactions determine a: which element is oxidized, b: which element is reduced, c: oxidizing agent, d: reducing agent: A- Fe2+ + H+ + Cr2O7 Felt + C + H2O B- Fe2S3 + 12HNO3 - 2Fe(NO3)3 + 35 + 6NO2 + 6H20 C-1 + MnO +H* - 12 + MnO2 + H20 2- Balance the following equations: A- MnO4 + Br - Mn? + Br? (acidic solution) B- Fe?' + Cr20,- Felt + Cr+ (acidic solution) C- CN + Fet - CNO + Fe2+ (basic solution) 3- How many gram of aluminum metal produces by the electrolysis of molten AICI3 for 3.25 hr with an electrical current of 15.0 A.

Answer 1

oxidation ? Loss of electrons on increase in oxidation State of an atom is called Oxidation. The atom of which oxidation state is increased, is considered as oxidised atom, oxidising agent =- The Substance/comparend which oxidises the other is called oxidising agent. Oxidising ageist undergoes reduction Reduction - Gain of elections as addition of electrons op decrease in oxidation state of an atom is called reduction : The atom of which oxidation State in decreased is considered as Reduced atom. Reducing agent. The compound which reduces the other is called Reducing agent. Reducing agent that Lendergoes oxidation. 1. A Fet2 + H+ + C220-22 Fets + Cat3 tho * Oxidation number (state) of Pe in feta_ * oxidation number 06 H in H+ = +1 * oxidation number of fe in fets - +3 * Oxidation number of or in cxt3 - +3. * In Ho, oxidation number of H=+1 and for o'= -2 * c2507² oxidation number of o'=-2 in cro=2 ... (2[69] +7[0]) = -2 2(67) +7(-2) =-2 -207-04-2) oxidation numbersCr= +6] In Cr2052

(42) Fe CD (+6) (-2) + H+ + Crom (+3), (+3) rets + Cats + 6+1)(-2) ho Increase in boxidation number of Fe COXIDATION) Decrease in Oxidation numberdcr. CREDUCTION) .. a) The element that is Oxidised - Fe te Fe is Oxidised (Increase in oxidation number from +2 to +3) o the element that is reduced = cr ie. Cris Reduced (Decrease in oxidatio nuomber from to to . O oxidising agent is csont Fetis oxidised by criso 2 ::.C5oz is oxidising d) Reducing agent is fet?:- crgot? is reduced by Fet? ... Fetz as Reducing agent. agent Fey S3 + 12 14NO3 - > 2FeCNO3)3 + 35+6NO2 + 6H2O * Fe 3 oxidation State of Pe=+3 Oxidation State of Ss-2 HNO3 Oxidation State of H=+1 Oxidation state of 0 = -2 Oxidatico State of N = +5 HNO3 (H) +CD+360)=0 (H+N+36-2)=0 Oxidation state of N= +5 * FeCNOS: Oxidation State of Fe=+3 CFet3: N03) oxidation State of a Let oxidation State of N = 2 (Fe)+3(N)+960) = 0 34 -(18-3) -15 . (+3) + 30 +96-2) -0 → 2015 - oxidation state 03 (N = 45

* S-oxidation State of S=0 * NO2: Let oxidation state of N= . . oxidation state of o=-2 .-.WN+200)]=0 2+26-2)=0 > 2=+4 oxidation State of N=+4. * H20oxidation State of- Hz+1 oxidaclion state of 0 -2 63362) 41245)62): Fes3 + 12 HNO3 - (+3)65)62) ) 64X-2) 4042 > 2FeCNO3), +35+6N0216 cxidation increase in oxidation numbers Reduction (Decrease in Oxidation ase in oxidatremberg a) The a element that' is oxidised - S The element that is Reduced EN c) oxidising agent is HNO3 d) Reducing agent is Fez$3 © I. + Mnox + 4* - Tg + Mnoz + HO ak oxidation number do I in I I- * oxidation number of Hin H+ = +7 * Mino Let oxidation number of Mnase oxidation number of 0= -2 [CMD) +4 (0)} – 1 6+46-2)]--| 2=(8-1)= +7 ... oxidation number of Mn= +7

* oxidation state of I in * Mnoz: oxidation state of o=-2 Let Be' is oxidation state of Mn. (Mn) +260) -0. 2+26-2)=0 -> 121=+41 : oxidation state of Mn in Mnogo +41 * H₂O oxidation state of H = +1 oxidation state of O=-2 (-) I (+7) (-2) (+) + Mno4 +H (0) (+4)(2) GD (+2) Iz + Mno, + 10 Coxidation Increase in oxidation number : Decrease in oxidation numbert (Reduction) a) The element that is oxidised: I Coxidation nuomber increased from - 1) The element that is Reduced: Mn. Coxidation number decreased foon +7 to 14) c) oxidising agent is mnou because I is oxidised by MOOL d) Reducing agent is I Because mnou is Reduced by I