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).
Separate the reaction below into balanced oxidation and reduction half-
reactions:
Cd(𝑠) + CuCl2 (𝑠) → CdCl2 (𝑠) + Cu(𝑠)
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