La(s) �¨ La(OH)3(aq) (basic solution)
Complete and balance the following half-reactions
This CAN'T be that hard.
La + 3OH^- ==> La(OH)3 + 3e
To complete and balance the given half-reactions, we need to follow a few steps:
Step 1: Identify the element undergoing oxidation and the element undergoing reduction in the reaction equation.
In this case, the reaction involves the element La (lanthanum). To determine whether it's getting oxidized or reduced, we need to look at its oxidation state in each half-reaction.
Step 2: Determine the oxidation states of the elements in the given half-reactions.
Since you've only provided one half-reaction, we'll focus on that.
The given half-reaction is:
La(OH)3(aq) → La(OH)3(aq)
In this reaction, the element La is present as La3+. To find its oxidation state, we need to look at the charges of the other elements involved. Hydroxide (OH-) has a charge of -1, and since there are three OH- ions, it makes the total charge on La3+. Therefore, the oxidation state of La is +3.
Step 3: Write the unbalanced half-reaction based on the oxidation states.
Since there is no change in the oxidation state of La in the given reaction, we conclude that it's not undergoing any oxidation or reduction. Therefore, we cannot write a half-reaction for it.
In summary, the half-reaction for La(OH)3(aq) (basic solution) cannot be determined as there is no change in the oxidation state of the lanthanum ion.