Question

When the Cu? concentration is 1.15 M. the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.1937. What is the Zn? concentration? Cu?aq+7(3) Cu()+Zn2+(-) Answer M

Answer 1

Answer The given cell reaction is cult (aq) + 2n (3) Cu (s) + 2 + (aq) The cathode half reaction cu²t (aq) tre cn (s) E = 0.34v The ano de half reaction Zn2+ (aq) + 2e 20 (6) E = 08-0-76 The stadard reduetin potential of the cell E cathode - E Anode 0.34v - 60.76)v cel) = 1:10 v Eºcell log Cell n Now according to nernst equation at 298k 0.059 [277) ㅌ [cu2t given, or a cell = 1:193 v, Icnt?] E l'15W M, n = total no. of electrons tranfer.=2 OQS9 Ton27) RA 1.193 = 1:10 0.059 2027) log [2-1) |cscanned with CamScanner tog 2

ar, (1.193 -110) = 0.059 log [Zn2+ 1:15 2 ar 0.09382 0.059 [212+] -log 1:15 ar 3.153 Zn2+ -log 1.15 w [2n2+) - 3.153 10 1.15 or [2n27) 0.000703 X 115 4. [2n²+] = 8.09 X10 2+ -4 8.09 x 10 so the concentration of an CScanned with CamScanner