I don't think B is the correct answer for the first part nor C for the second part.
Check me out on this. In the reaction, Cu is going to Cu^+2 so that half cell is
Cu==> Cu^+2 + 2e. (-0.34 v).That is oxidation (which answers the second question).
The other half reaction is
Ag^+ + e ==> Ag (+0.8 v)which is the reduction half. and is the cathode.
So the cell notation, placing the anode on the left, and moving from left to right from solid to solution to salt bridge to solution to solid is
Standard cell notation moves from left to right through the salt bridge as
solid(anode)/anode soln||cathode soln/solid cathode (unless of course we have Pt, H2, and those things).
Let's see about the process of elimination. For part 1,
A. can't be right. Cu^+2 can't dip into solid Cu. It's the other way around.
B. can't be right for two reasons. #1 is we never use coefficients from the balanced equation and #2 is Ag doesn't dip into Cu soln.
C. can't be right. Ag soln and Cu soln don't mix.
E. would be ok IF we allowed coefficients from the balanced equation; but we don't so E is out.
I think D is the only possibility. And it follows all the rules, I think. The correct answer to this gives the almost automatic answer for part 2.
Check my thinking.
so part 2 will be B?
Yes, oxidation occurs at the anode and Cu(s) is being oxidized to Cu^+2 and giving up 2e.
Sorry, I hit the post button twice.
I think the answer to part 1 is (E): Cu(s) Cu2+(aq) 2Ag+(aq) 2Ag(s)
nvm, PART D is correct.
ok i put Cu ---> Cu2+ + 2eľ for #2 and it was wrong.
kris vang was here
Consider the following reaction:
2Ag+(aq) + Cu(s) ↔ Cu2+(aq) + 2Ag(s)
Step 1: Write the two half-reactions.
Ag+(aq) + e- ↔ Ag(s)
Cu(s) ↔ Cu2+(aq) + 2e-
Step 2: Determine the cathode and anode.
Anode: Cu(s) ↔ Cu2+(aq) + 2e-
Cathode: Ag+(aq) + e- ↔ Ag(s)
Step 3: Construct the Diagram.
Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s)
i need help with a chemistry problem
consider the reaction 2ag+ (aq) + Cu (S) --> Cu2+ (aq) + 2ag (S). the half reactions for this reaction would
A) show the oxidation and reduction of the reactants as separate reactions.
B)separate out the charges on each side of the equation.
C)show the loss of electrons in the reduction half-reaction.
D)rebalance the charge values on either side of the equation.