Separate the reaction below into balanced oxidation and reduction half-

reactions:

Cd(𝑠) + CuCl2 (𝑠) → CdCl2 (𝑠) + Cu(𝑠)

1 answer

Cd(𝑠) + CuCl2 (𝑠) → CdCl2 (𝑠) + Cu(𝑠)

Cd(s) ==> Cd^2+ + 2e
CuCl2+ 2e ==> Cu(s) + 2Cl^-
----------------------------------------------
Cd(s) + CuCl2 + 2e == Cu(s) + 2Cl^- +Cd^2+ + 2e
2e cancel on each side and Cd^2+ pairs with 2Cl^- to give
Cd(𝑠) + CuCl2 (𝑠) → CdCl2 (𝑠) + Cu(𝑠)
So I've done what you asked but note that this reaction will not occur unless the CuCl2(s) is placed in solution as CuCl2(aq).