Write the balanced overall reaction from these unbalanced half-reactions
ln--->ln^3+
Cd^2+--->Cd
In ==> In^3+ + 3e
tough eh!
Cd^2+ + 2e ==> Cd
To write the balanced overall reaction from the unbalanced half-reactions, you need to make sure that the number of electrons transferred is the same in both reactions.
Let's start by balancing the individual half-reactions:
Half-reaction 1: ln ⟶ ln^3+
The number of electrons in the half-reaction is not specified, so we need to determine it. Looking at the oxidation states, we see that the oxidation state of ln changes from 0 to +3. Since the oxidation state increases by 3, it means that 3 electrons are being lost.
Therefore, we can write the balanced half-reaction as:
ln ⟶ ln^3+ + 3e^-
Half-reaction 2: Cd^2+ ⟶ Cd
Again, the number of electrons transferred is not specified. But looking at the oxidation states, we see that the oxidation state of Cd changes from +2 to 0. Since the oxidation state decreases by 2, it means that 2 electrons are being gained.
Therefore, we can write the balanced half-reaction as:
Cd^2+ + 2e^- ⟶ Cd
To balance the overall reaction, we need to make sure that the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. In this case, the oxidation half-reaction loses 3 electrons, and the reduction half-reaction gains 2 electrons. To equalize the electrons transferred, we need to multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2:
2ln ⟶ 2ln^3+ + 6e^-
3Cd^2+ + 6e^- ⟶ 3Cd
Now, we can combine the half-reactions to form the balanced overall reaction:
2ln + 3Cd^2+ ⟶ 2ln^3+ + 3Cd