Question

can someone explain how can we get 125g?

Eºred= -0.441 V Fe2+ + 2e Cd²+ + 2e Fe cd Eºred= -0.403 V A galvanic cell based on these half-reactions is set up under standard conditions where each solutions is 1.00 L and each electrode weighs exactly 100.0 g. How much will the Cd electrode weigh when the non-standard potential of the cell is 0.03221 V? 125 g Computer's answer now shown above. You are correct. Your receipt no. is 168-935 Previous Tries

Answer 1

cell reaction: Fe(s) + Cd2+(aq) -----> Fe^2+(aq) + Cd(s)

Fe2+ + 2e- ---> Fe(s)    E0 = - 0.441 v

Cd2+ + 2e- ----> Cd(s) E0 = -0.403 v

E0cell = E0cathode - E0anode

        = (-0.403)-(-0.441)

        = 0.038 v

Ecell = E0cell - (0.0591/n)log([Fe2+]/[Cd2+])

During the reaction.

[Fe+2] concentration increases, [Cd2+] concentration decreases.

Hence, [Fe+2] = 1+x , [Cd2+] = 1 -x

0.03221 = 0.038 - (0.0591/2)log((1-x)/(1+x))

x = 0.222 M

concentration of Cd+2 solution decreases = x = 0.22 M

mass of Cd produced/deposited = M*M.wt

                              = 0.222*112.41

                              = 25 g

Total mass of Cd after reaction = 100+25 = 125 g