For the following redox reaction, give the complete reduction half reaction.
Na(s) + Al+3(aq) Na+1(aq) + Al(s)
Please help asap!!
To determine the reduction half-reaction in a redox reaction, you need to identify the species that is undergoing reduction, which means it is gaining electrons.
In the given reaction:
Na(s) + Al+3(aq) -> Na+1(aq) + Al(s)
In this reaction, the sodium atom (Na) is getting oxidized, and the aluminum ion (Al+3) is getting reduced. So, we need to write the reduction half-reaction for the Al+3 ion.
The oxidation state of aluminum (Al) changes from +3 to 0, which means it is gaining 3 electrons. Therefore, the reduction half-reaction for Al+3 can be written as follows:
Al+3(aq) + 3e- -> Al(s)
In this reduction half-reaction, 3 electrons (3e-) are shown on the left side, indicating the gain of electrons by the aluminum ion to form solid aluminum (Al).
Hence, the complete reduction half-reaction is:
Al+3(aq) + 3e- -> Al(s)