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. 2Ag+(aq) Cu2+(aq) 2Ag(s) Cu(s)
B. Cu(s) Cu2+(aq) Ag+(aq) Ag(s)
C. Ag+(aq) Cu2+(aq) Cu(s) Ag(s)
D. Cu2+(aq) Cu(s) Ag(s) Ag+(aq)
E. Cu(s) Cu2+(aq) 2Ag+(aq) 2Ag(s)

What reaction is occurring at the anode?

A. Cu Cu2+ + 2e–
B. Cu2+ + 2e– Cu
C. Ag+ + e– Ag
D. Ag Ag+ + e–

I think it's B, A but I'm not sure

Jessie, you have no arrows. How do you know the reactants from the products? Second, the correct answer is not listed for number one. You must space the active phases with a | mark.

Your answers of B and A are correct if you place the arrow and the | at the correct place(s).

I didn't know how to type those in, but I got it. Thanks!

I've seen arrows done three ways.

>> or the reverse of <<.
With the dash and greater than sign as in --> or <--.
I use the equal sign and the greater than sign as in ==> or <== and if I want to show --> and <-- so show equilibrium, I can write it as <--> or <<>> or <==>.

To determine the correct cell notation at standard state conditions, we need to keep in mind the general format of cell notation:

Anode | Anode Solution || Cathode Solution | Cathode

In this case, the anode represents the oxidation half-reaction (losing electrons), while the cathode represents the reduction half-reaction (gaining electrons).

Let's analyze the given options:

A. 2Ag+(aq) Cu2+(aq) 2Ag(s) Cu(s)
This option shows the oxidation reaction (loss of electrons) happening at the anode, which is the Ag+ ion being reduced to form Ag(s). However, the Cu(s) on the cathode side is not consistent with the reduction reaction occurring there.

B. Cu(s) Cu2+(aq) Ag+(aq) Ag(s)
This option correctly represents the copper electrode (Cu(s)) undergoing oxidation to form Cu2+(aq) at the anode. The reduction of Ag+ ions to Ag(s) is represented at the cathode. So, this cell notation is correct.

C. Ag+(aq) Cu2+(aq) Cu(s) Ag(s)
Here, the reaction at the anode involves the reduction of Cu2+ ions instead of Ag+ ions. Hence, this option is not correct.

D. Cu2+(aq) Cu(s) Ag(s) Ag+(aq)
This option contains the copper electrode losing electrons (oxidation) at the anode. However, the silver electrode being reduced to form Ag+ ions does not match the reaction occurring at the cathode. Therefore, this cell notation is not correct.

E. Cu(s) Cu2+(aq) 2Ag+(aq) 2Ag(s)
Here, the reaction at the anode involves the oxidation of copper (Cu(s)) to Cu2+(aq). However, the reduction of Ag+ to form Ag(s) is not consistent with the reaction at the cathode. Thus, this option is not correct.

Based on the analysis, the correct cell notation at standard state conditions is B. Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s).

Now, let's determine the reaction occurring at the anode. The anode is where oxidation takes place, so we need to find the half-reaction that involves the loss of electrons. Looking at the options:

A. Cu Cu2+ + 2e–
This corresponds to the copper electrode (Cu) losing two electrons (2e–) to become Cu2+. Hence, this reaction occurs at the anode.

B. Cu2+ + 2e– Cu
This represents the reduction of Cu2+ ions to form Cu. It does not occur at the anode.

C. Ag+ + e– Ag
This reaction involves the reduction of Ag+ ions to Ag and does not occur at the anode.

D. Ag Ag+ + e–
This corresponds to the oxidation of Ag to Ag+ and occurs at the anode.

Based on the analysis, the correct reaction occurring at the anode is A. Cu → Cu2+ + 2e–.