Separate the reaction below into balanced oxidation and reduction half-
reactions:
Cd(𝑠) + CuCl2 (𝑠) → CdCl2 (𝑠) + Cu(𝑠)
Cd(𝑠) + CuCl2 (𝑠) → CdCl2 (𝑠) + Cu(𝑠)
Cd(s) ==> Cd^2+ + 2e
CuCl2+ 2e ==> Cu(s) + 2Cl^-
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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).
To separate the given reaction into balanced oxidation and reduction half-reactions, we need to identify which element is being oxidized and which is being reduced.
In this reaction:
Cd(s) + CuCl2(s) → CdCl2(s) + Cu(s)
The Cd(𝑠) is being oxidized, as its oxidation state increases from 0 to +2 by losing 2 electrons:
Cd(s) → Cd^2+(aq) + 2e^-
The CuCl2(s) is being reduced, as the Cu in CuCl2 goes from an oxidation state of +2 to 0 by gaining 2 electrons:
Cu^2+(aq) + 2e^- → Cu(s)
Now we can balance the charges and atoms in each half-reaction by multiplying the half-reactions as needed:
Oxidation Half-Reaction:
Cd(s) → Cd^2+(aq) + 2e-
Reduction Half-Reaction:
Cu^2+(aq) + 2e^- → Cu(s)
By balancing the coefficients, the overall balanced reaction becomes:
Cd(s) + CuCl2(s) → CdCl2(s) + Cu(s)
To separate the given reaction into balanced oxidation and reduction half-reactions, we need to identify the elements that undergo changes in oxidation states. In this reaction, we can see that cadmium (Cd) loses electrons and its oxidation state increases from 0 in solid form to +2 in CdCl2. On the other hand, copper (Cu) gains electrons, changing its oxidation state from +2 in CuCl2 to 0 in solid form.
To write the oxidation and reduction half-reactions, we follow these steps:
1. Write the overall reaction:
Cd(s) + CuCl2(s) → CdCl2(s) + Cu(s)
2. Identify the changes in oxidation state for each element:
- Cadmium (Cd) changes from 0 to +2 (loses 2 electrons)
- Copper (Cu) changes from +2 to 0 (gains 2 electrons)
3. Write the balanced oxidation and reduction half-reactions:
Oxidation half-reaction (loss of electrons):
Cd(s) → Cd^2+(aq) + 2e^-
Reduction half-reaction (gain of electrons):
Cu^2+(aq) + 2e^- → Cu(s)
Note: In the half-reactions, the symbol (aq) represents the species in aqueous solution, and (s) represents the species in solid form.
By separating the given reaction into balanced oxidation and reduction half-reactions, we illustrate the electron transfer that occurs during the redox reaction.