Answer the following questions based on the following electrochemical reaction:
Al(s) + Cu 2+(aq) <====> Cu(s) + Al3+ (aq)
Write the balanced chemical equation for this cell.
I'm confused.. how would this be balanced?
mass is balanced on each side, as well as electrons. Each Cu ion has to receive two electrons, and each Al atom donate three
2Al(s) + 3Cu 2+(aq) <====> 3Cu(s) + 2Al3+ (aq)
so, six electrons were donated, and six received, and the mass is the same on each side.
To balance the chemical equation for this cell, you need to make sure that the number of atoms of each element is the same on both sides of the equation. Here's how you can balance it step-by-step:
1. Determine the oxidation numbers for each element. In this reaction, aluminum (Al) is being oxidized from an oxidation state of 0 to +3, while copper (Cu) is being reduced from an oxidation state of +2 to 0.
2. Write the half-reactions for each electrode. The half-reaction for the oxidation of aluminum is:
Al(s) ---> Al3+(aq) + 3e-
The half-reaction for the reduction of copper is:
Cu2+(aq) + 2e- ---> Cu(s)
3. Balance the atoms and charges in each half-reaction. In the oxidation half-reaction, we can balance the atoms by adding 3 electrons (e-) to the left side. In the reduction half-reaction, the number of electrons on both sides is already balanced.
Al(s) ---> Al3+(aq) + 3e-
Cu2+(aq) + 2e- ---> Cu(s)
4. Multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3 to make the number of electrons equal in both reactions:
2Al(s) ---> 2Al3+(aq) + 6e-
3Cu2+(aq) + 6e- ---> 3Cu(s)
5. Combine the balanced half-reactions to form the overall balanced equation. Cancel out the electrons on both sides:
2Al(s) + 3Cu2+(aq) ---> 2Al3+(aq) + 3Cu(s)
So, the balanced chemical equation for this cell is:
2Al(s) + 3Cu2+(aq) ---> 2Al3+(aq) + 3Cu(s)