Write an overall balanced equation for these unbalanced half-reactions:
In --> In3+
Cd2+ --> Cd
I multiplied both sides by integers to get the electrons to be equal on both sides but I don't know what the final overall balanced equation looks like.
I think you're looking for this. Just add the two equations you multiplied to get the electrons equal.
2In + 3Cd^2+ ==> 2In^3+ + 3Cd
i'm not sure, but it wants to combine the two half reactions into a full reaction.
In> In3+ + 3e-
Cd2+ + 2e- >> Cd
Yes, I understand how to split it up with the e-, but what is the final overall balanced equation?
It is balanced in charge, and mass. Why would one want more?
To balance the overall equation for the given unbalanced half-reactions, you need to combine the two half-reactions in a way that ensures the number of electrons gained and lost is equal. Let's start by balancing the half-reactions individually.
1. Balance the half-reaction: In -> In3+
We begin with indium (In) being oxidized to form indium ions (In3+). We need to balance the atoms and charges on both sides of the equation.
In this case, since the number of indium atoms remains the same, we only need to balance the charge. The indium on the left-hand side (In) has a charge of zero, while the indium on the right-hand side (In3+) has a charge of +3. Therefore, we need to add three electrons (e-) to the left side to balance the charges.
The balanced half-reaction is: In -> In3+ + 3e-
2. Balance the half-reaction: Cd2+ -> Cd
This half-reaction involves the reduction of cadmium ions (Cd2+) to form cadmium (Cd). Again, we need to balance the atoms and charges on both sides.
In this case, the cadmium atoms remain the same. However, the charge changes from +2 on the left (Cd2+) to zero on the right (Cd). So, we need to add two electrons (2e-) to the right side to balance the charges.
The balanced half-reaction is: Cd2+ + 2e- -> Cd
Now that both half-reactions are balanced individually, we can adjust the number of electrons to be the same in both reactions. To do this, we multiply the first half-reaction by 2 and the second half-reaction by 3 to get an equal number of electrons:
2(In -> In3+ + 3e-) and 3(Cd2+ + 2e- -> Cd)
Now we add the two balanced half-reactions together to obtain the overall balanced equation:
2(In -> In3+ + 3e-) + 3(Cd2+ + 2e- -> Cd)
This gives us the following overall balanced equation:
2In + 3Cd2+ -> 2In3+ + 3Cd