if i had a beaker with .0001M of Cu solution and copper solid (anode) and another beaker with 1M Cu solution and Cu solid (cathode), how would i calculate the half reactions of these two?
would the two equations be:
cu2+(aq) --> Cu(s) + 2e-
and ?
What you have described, here and in your previous post, is a concentration cell. One of the half cells is
Cu(s)==> Cu^+2(?M) + 2e and the other one is
Cu^+2(?M) + 2e ==> Cu(s)
One of the two concentrations is 1 M and the other is 0.001 M (or whatever it was--I'm not looking at it right now). So you use the Nernst equation to determine the voltage of each of the half cells and add the oxidation half to the reduction half to obtain the cell potential.
To calculate the half-reactions for the two beakers, you need to consider the oxidation and reduction processes separately. Let's start with the anode half-reaction (oxidation):
1. Identify the oxidation state of the reactant at the anode, which is Cu(s). In its solid state, copper has an oxidation state of 0.
2. Next, identify the oxidation state of the product at the anode, which is Cu2+(aq). The copper ion has an oxidation state of +2 in an aqueous solution.
3. Write the half-reaction by showing the loss of electrons:
Cu(s) → Cu2+(aq) + 2e-
Now, for the cathode half-reaction (reduction):
1. Identify the oxidation state of the reactant at the cathode, which is Cu2+(aq). As mentioned earlier, copper ion has an oxidation state of +2 in an aqueous solution.
2. Next, identify the oxidation state of the product at the cathode, which is Cu(s). In its solid state, copper has an oxidation state of 0.
3. Write the half-reaction by showing the gain of electrons:
Cu2+(aq) + 2e- → Cu(s)
So, the half-reactions for the anode and cathode in this setup would be:
Anode (oxidation): Cu(s) → Cu2+(aq) + 2e-
Cathode (reduction): Cu2+(aq) + 2e- → Cu(s)
These two half-reactions represent the oxidation and reduction processes that occur during the electrochemical cell setup with the Cu solution and copper solid.