Question

An electrochemical cell is made of a left-hand compartment consisting of nickel wire in contact with a 1.0 M Ni(ClO), solution and a right-hand compartment of copper wire immersed in 1.0 M Cu(Cl) 1. What is the overall spontaneous cell reaction for this electrochemical cell? What is the standard cell potential for this cell? Show your work for full credit. 2. 3. What is the direction of electron flow (from which electrode to which)? Which substance is oxidized? 4. If the [Cu ] in the right-hand compartment is changed to 4.0 X 10 5. age of the electrochemical cell? Show your work for full credit. M, what is the expected volt-

Answer 1

Cell represeilalion presenlation Quen Ni24/ Ni solution on left side (anode viu endergo oridation & Cu2+1 eu on right side (Cathode) will undergo reduchion . Doridailion (At anode) Nil le - Ni 2+ - Eifti : -0.2% Reduchon Cu2+ + de - Cue - Bou?e/e4 = 0.999 a) Ner cell non Eqn + Eqn Ni + Cu 2+ Ni 2+ + Cel Eceu a E Calho de - Eanode para pisarnab Harresh Eceu = Ecazulov - Enim Mu' feel = 0.339 V-40-236 ) v = (0.339 +0-236) Eeeu = 0. 575 V mob. pour d) from the net reaction we see nickel å getting onidized to niet as there is lose of 2 eo Oxidized product Ni 2+ 0:- Electrone Flows from ano de lo cathode uprave .. e) Merst Equation for the above Reaction would be no de flors Eceu = Ecev - 0.0591 log {Nd2+] * no log [Ni 2+] . here=n=2 Ecell Ao Teu zaje = 0.575V 0.05m fog 2x10-3] Twirl-IM, [0,20%= 4x1034. Ecell :0. 5751 -0.0591 x 2.398 [cu].Ni] : 1 (constant Ecell = 0 575 V Scanned with 0.0708 0.0708 Eeeus ineo. 5042

E_{cell = 0.5042V}