Question

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ratory Assignment Experiment 15 chemistry (Galvanic Cells) Name Rox couples 2 - -> 2Br le ......... Fe +2 2e ....-> Sn 3e > Cr +1.07 +0.77 -0.14 -1.0V der the cell consisting of the Fe* (IM) Fe**and and Bry(lamy B (IM) redox couples. ile the reduction half-reaction and the corresponding standard reduction potential and indicate which electrode does this occur at. Write the oxidation half reaction and the corresponding oxidation potential indicate which electrode does this occur at. Write the equation of the overall cell reaction and the value of Bºcell? 2. Determine the cell voltage, Eº cell, of a galvanic cell consisting of Sa+2(IM) / Sn and C+3 (1M) Cr redox couples. 3. Use the Nernst Equation to determine the cell potential, Ecell, of the galvanic cell consisting of Sa+20.0001M) /Sn and Cr+ (O.IM/Cr redox couples. 4. State the purpose of the salt bridge in the experiment? _What electrochemical process occurs 5. What is the sign of the anode in a galvanic cell? at the cathode?

Answer 1

2 Br Ot Boat 26 - E = 1.07 E-0 TTV Sot +20 - sn E = -0.14 V E = -10v Fe3+ / Fe²+ & Pral Br Higher the reduction potential, greater the tendency to Reduction half reaction get reduced. Brg + 20 2Br (2) E° = 1:07V occur on Bra/Br electrode . act as note Cathode Oxidiation half reaction Fe2+ - Festt lé (n) E O = -0.77 v occur on Fe2+ / Fest electrode. act as fullmate Anode Equation of the overall reaction. Multiply equation (11) by d 2 Fe2+ - 2Fe 3+ + 2e (6) (c) B8 g +20 overall reaction: 2 Br 2Fe 2 + + Br. - 25e + 288 r consider reduction botential of both electrode 7 Treduction potential) 1.07 0.77 0.zon

Og Galvanic cell consisting of Sn /sn and Cr3+1 or Sn²+/sn act as Anate Cathode Cot/ or act as Anode. Sn²+ + 2e In t = -0.14V Cr3+ +30 - co E = -lov Ecell - E cathode Eanode = -0.14 – (-1.0) = -0.14 +lee = 0.86 V Galvanic cell consisting of a Sn (0.000/M) / son and co2+ (0.1M)/cr. Ecell = 0.86v According to Nernst Equation (from 0.2) Ecell = Ecell - BT ln [c] coza for aA + bB È cc + dD for the given cell- Sn²+ + 2€ - Sn (Reduction half reaction) cr. Cr3+ + 36 (od oxidation half creaction) Multiply equation is by 3 and (ii) by 2 and then add . 3572+ + 20% - 3Sn + 20,3+ value of n= 6 T= 298 de fail = 0.86 -0.054 lang [5] [63 IF 965000 for solid concentration ti

so. E cell = 0.86 - oost log tous **86-0059 g fax - 0186 -0.059 log 100 = 0.86 -0.059 x to log 10 Flogn" = nlogm = 0.86 -0.059x. - 0.86 -0.059 =0.86-0.0098 = 0-8502 v Ans. Use of salt bridge. The salt bridge is an inverted U-tube filled with Concentrated solution of inent electrolyte like kel, KNO3 NHA NO z etc. Important function are - i) salt bridge completes the electrical circuito salt bridge connects the two solution of the half cells and their electrode are connected by means of a wire. Therefore salt bridge completes the circuito ii) Salt bridge maintain electrical neutrality 4 tulo half cell solutions, As in a cell, oxidation takes place at anode. due to which there is excess of positive ions in the solution Similarly at cathode reduction takes blare tue ion from solution get election and reduced solution. pret excess we in present in the

this accumulation of charges causes the cell to Stop working, So salt bridge replace the ions half cells. provides cations and lost or producted in anoins to the two ms. a In Anode galvanic cell have negative sign. reduction reaction takes 1. At Cathode blace,