Identify the species so that we can identify which half reactions are taking place. It is clear that in one compartment, there are both copper ions (Cu
2+) and copper metal (Cu). Hence, the equation Cu
2+ + 2 e
- ⇌ Cu
(s) will prove useful.
In the other compartment, there are both silver ions (Ag
+) and silver metal (Ag). Hence, the equation Ag
+ + e
- ⇌ Ag
(s) will prove useful.
Next, identify which is getting reduced and which is getting oxidised.
On the
data sheet, it is clear that the equation involving copper is placed (somewhere) above the equation involving silver. Hence, copper must be the more reactive substance and is the one to lose electrons (get oxidised). This leaves us with silver ions being what gains electrons.
Hence, the reactions we are interested in are:
Ag
+ + e
- → Ag
(s) (not flipped, as per data sheet)
Cu
(s) → Cu
2+ + 2 e
- (flipped, as per data sheet)
The equations also make it abundantly clear that electrons are flowing from the copper compartment, to the silver compartment. Thus we can draw the electron flow from left (through the voltmeter) to right