Can someone clear this up for me? (1 Viewer)

[luey88]

Hey guys,

just getting a bit confused with working out galvanic cell half equations...

do you go down the right hand side of the standard potentials and read across, then up the left and read across?

or is it down right, then up left?

cheers x

 

[Dreamerish*~]

I don't get what you're trying to ask.

Are you trying to work out the overall voltage, or just the half-equations?

 

[luey88]

well both really, you cant work out the voltage if you arent sure which way to do the half equations?!

ok, for example, in the left hand beaker you have copper nitrate solution and a copper electrode.

right hand beaker, you have silver nitrate solution and a silver electrode.

will it go; Ag(+) + e(-) > Ag E= 0.80 V
Cu > Cu (2+) + 2e(-) E= -0.34 V

CELL VOLTAGE = 0.80 -0.34
= +0.46 V

or will you do Ag the opposite way and Cu the opposite way?

 

[Dreamerish*~]

Well, it's the standard reduction potential.

So you take the voltage value given for the reduction half equation, and for the oxidation half equation, you take the negative of the voltage value.

Either way, you know you're wrong if your overall voltage is less than 0.

I can see I didn't explain it very well.

Just remember, one's oxidation and one's reduction - so you'll have to take the negative of one V value. If you add the two and the overall voltage is less than 0, then you flipped the wrong one. I'm afraid that's the best I can do. You might want to wait for someone who isn't stressing over German Extension this afternoon.

 

[Captain pi]

well both really, you cant work out the voltage if you arent sure which way to do the half equations?!

ok, for example, in the left hand beaker you have copper nitrate solution and a copper electrode.

right hand beaker, you have silver nitrate solution and a silver electrode.

will it go; Ag(+) + e(-) > Ag E= 0.80 V
Cu > Cu (2+) + 2e(-) E= -0.34 V

CELL VOLTAGE = 0.80 -0.34
= +0.46 V

or will you do Ag the opposite way and Cu the opposite way?

For a galvanic cell, the more active metal (in this case, Cu) oxidizes and the less active metal (~ Ag) is reduced. Otherwise, the voltage would be negative <=> non-spontaneous.

Oxidation is loss of electrons which means X --> X+ + e- for a metal, X.