Question

A voltaic cell utilizes the following reaction and operates at 298 K. 3 Cet (aq) + Cr(s) 3 Ce3 (aq) + Cr3 (aq) (a) What is the emf of this cell under standard conditions? v (b) What is the emf of this cell when [Ce4+]-1.9 M, [ Ce 3 + ] = 0.010 M, and [Cr3+]-0.012 M? (c) What is the emf of the cell when [Ce4+] = 0.52 M, [Ce3+]- 0.89 M, and [Cr3+]-1.2 M? V

Answer 1

The answer is

31 The given chemical equation is 3 Ceutraq) +68(S) 3 cestcaq) + c d baq Dividing the above equation in two haits ( ce+4 + - - - cet3 ) X3 - cart34 36 3 cety +36 3003+ cr - Cat3 + 3 3 ce ty+c8 3 603 + +c8t3 . Here Cety +36- 3 ces is the reduction half C cathode ceil S co- crt3 te is the oxidation halt Canode) cell EcoB 8 = -0.74 VS Ecet ce to l. 61v. 1. Eco Ecathode E anode E 1.61V-(-0.741) - 1.6lvto.74V 235 V : Em f of the cell at Standand condition Forell = 2.350

(b) we know that (..Na noof electrons) Ecell - Erp-0.054) log I anode I cathode] ce () we know that Eccell Eren -0.059) logk Here na noooye - K = [213] [ce 1373 [cery]3 Ecen - Elcen 0.0591 eogterth (ce [Cety J3 1:. [crt3] = 0·01217 I [ce 137=0.01M = 2.350-0.0591 eog 0.012 x10.013 [cety] a loqm (log B3 = 2.355-0.0591 log 12 x 10 x 10 € 0 6.859 & 2.35 V-0.0196 log 12 8109 6.859 = 2.35V-0.0196 log (1-74 x 1090 2.35V -0.0196 X - 8.75 52.351 +001715V - 2.52V & Font of the cell is 2.52v

. (C): when [ce ] = 0.52 M [ceta] = 0.89 m) I crt3] = 1.2m then ean become Ecell focen -0.0591 log [corts lice [Ceth 3 = '2:35V - 0.0196 log ( 12 x (0.89) 0.5293 = 2.35V-0.0196 log(6.01) 2-351 – 0.0196 X0.77 2.351 -0.01V = 2.340 For = 2.340