For this equation
2 Ag+ (aq) + Cu(s) <--> Cu2+ (aq) + 2 Ag(s)
Which occurs at the anode?
A. Cu --> Cu2+ + 2e–
B. Cu2+ + 2e– --> Cu
C. Ag+ + e– -->Ag
D. Ag --> Ag+ + e–
A, right because it's oxidation?
Oxidation occurs at the anode. Which is losing electrons.
So Cu--> Cu2+ +2e- is losing electrons, right? Because its becoming positive
That's right and the equation of Cu ==> Cu^2+ + 2e shows it losing two electrons.
Yes, you are correct. Option A, Cu --> Cu2+ + 2e–, represents the oxidation half-reaction occurring at the anode.
To determine which species is being oxidized (and therefore occurring at the anode), you need to identify the half-reaction involving the loss of electrons. In this case, the copper metal (Cu) is losing two electrons to form copper ions (Cu2+), which indicates oxidation.
Remember, oxidation refers to the loss of electrons, while reduction refers to the gain of electrons.