Question

Consider an electrochemical cell consisting of two vessels connected by a porous separator. One vessel contains 0.500 M HCl solution and an Ag wire electrode coated with AgCl solid. The other vessel contains 2.50 M MgCl2 solution and a Mg wire electrode. First determine the net reaction, and then calculate the emf of this cell. Enter your answer in volts.

Answer 1

Answer: 2.614V

Step-by-step explanation:

At the anode : Mg (s) --> Mg2+(aq) +2e- , Eo = -2.38 V

At the cathode : AgCl (s) + e- ---> Ag(s) + Cl- , Eo = 0.222 V

net reaction :Mg(s) + 2AgCl (s) --------------> Mg2+(aq) + 2Ag(s) +2Cl-(aq)

E cell= Ecathode- E anode- 0.0591/n log Q= [Mg2+] [Cl-]^2

Eocell = 0.222 + 2.38 = 2.602 V

Q = [Mg2+] [Cl-}^2

Q = (2.5) (0.500)^2

Q = 1.5625

Ecell = Eocell - 0.0591 / 2 * log Q

= 2.602 -.0-0591/ 2 * log 1.5625= 0.1938

= 2.602-0.0591/2x0.1938

= 2.62 --0.005726

Ecell = 2.614V