For the following balanced equation: 2 Ag+ (aq) + Cu(s) Cu2+ (aq) + 2 Ag(s)
Which letter corresponds to the correct cell notation at standard state conditions?
A. Cu(s) Cu2+(aq) Ag+(aq) Ag(s)
B. Cu2+(aq) Cu(s) Ag(s) Ag+(aq)
C. Ag+(aq) Cu2+(aq) Cu(s) Ag(s)
D. 2Ag+(aq) Cu2+(aq) 2Ag(s) Cu(s)
E. Cu(s) Cu2+(aq) 2Ag+(aq) 2Ag(s)
What reaction is occurring at the anode?
A. Ag+ + e– Ag
B. Cu2+ + 2e– Cu
C. Cu Cu2+ + 2e–
D. Ag Ag+ + e–
Where is the arrow. I can't tell between reactants and products.
Two notes here:
1. For question #2, note that oxidiation occurs at the anode. Oxidation, you will recall, is the loss of electrons.
2. For #1, you need to show the separtion of phases as phase 1|phase two followed by || to show the salt bridge then Phase 2|phase 1.
I still don't understand are the reactants on one side of the || and on the other side is products?
for the second one there are 2 answers that could possibly work here
To determine the correct cell notation at standard state conditions, we need to identify the anode and cathode and write them in the correct order.
In the given balanced equation, we have:
2 Ag+(aq) + Cu(s) -> Cu2+(aq) + 2 Ag(s)
The species on the left side of the equation (2 Ag+(aq) and Cu(s)) are the anode, where oxidation occurs, while the species on the right side (Cu2+(aq) and 2 Ag(s)) are the cathode, where reduction occurs.
Now, let's look at the options provided:
A. Cu(s) Cu2+(aq) Ag+(aq) Ag(s)
B. Cu2+(aq) Cu(s) Ag(s) Ag+(aq)
C. Ag+(aq) Cu2+(aq) Cu(s) Ag(s)
D. 2Ag+(aq) Cu2+(aq) 2Ag(s) Cu(s)
E. Cu(s) Cu2+(aq) 2Ag+(aq) 2Ag(s)
Considering that the anode (oxidation) is represented on the left side of the cell notation and the cathode (reduction) is represented on the right side, we can see that option D. 2Ag+(aq) Cu2+(aq) 2Ag(s) Cu(s) correctly represents the anode and cathode species in the given balanced equation. So, the correct cell notation is option D.
Now, let's determine the reaction occurring at the anode. In option D, we have 2Ag+(aq) -> 2Ag(s) + 2e-. This indicates that Ag+ ions are being reduced to Ag atoms, which is the reduction half-reaction. Therefore, the answer is A. Ag+ + e– Ag.