an electrochemical cell uses Al and Al+3 in one compartment and Ag and Ag+ in the other
write a balanced equation for the rxn that will occur spontaneously in this cell
would that just be Al + Al+3 + 2e --> Ag + Ag+ ?
Not quite.
The half equations you want are as follows:
Al ==> Al^+3 + 3e
Ag^+ + e ==> Ag
=================
Multiply equation 1 by 1 and equation 2 by 3 (to make the electrons equal).
Al ==> Al^+3 + 3e
3Ag^+ + 3e ==> 3Ag
===================
Now add them.
Al + 3Ag^+ + 3e ==> Al^+3 + 3e + 3Ag
Now cancel the electrons (since there are 3 on each side). That gives you the final equation.
To write a balanced equation for the reaction that occurs spontaneously in this electrochemical cell, we need to consider the redox reactions happening at the two electrodes.
At the anode (the negative electrode), aluminum (Al) reacts with aluminum ions (Al+3) to form aluminum ions with a lower positive charge and release electrons. This can be represented as:
Al(s) → Al+3(aq) + 3e-
At the cathode (the positive electrode), silver ions (Ag+) accept electrons and get reduced to form silver (Ag) metal. This can be represented as:
Ag+(aq) + e- → Ag(s)
Since the number of electrons involved in the oxidation and reduction half-reactions needs to be equal, we multiply the reduction half-reaction by 3:
3Ag+(aq) + 3e- → 3Ag(s)
Now, we can combine the two half-reactions to get the overall balanced equation for the spontaneous reaction in the electrochemical cell:
Al(s) + 3Ag+(aq) → Al+3(aq) + 3Ag(s)
So, the balanced equation for the reaction occurring spontaneously in this electrochemical cell is:
Al(s) + 3Ag+(aq) → Al+3(aq) + 3Ag(s)