Question

How is the electrode potential of a galvanic cell determined?I need the equation....

Answer 1

@thomaster

Answer 2

What?

Answer 3

??? No one?

Answer 4

You mean you need the reaction equation? For example for the Daniell-element?

Answer 5

yah....

Answer 6

Well I just write some stuff then:¨
Lets consider the Daniell-element (a galvanic cell)
We can write it as a cell-diagram:
\[\Large \sf Zn(s)lZn ^{2+}(aq)l l Cu ^{2+}(aq)l Cu(s)\]
As half cells we get:
\[\Large \sf Zn ^{2+}(aq) + 2e ^{-} \to Zn(s)\]
\[\Large \sf Cu ^{2+}(aq)+2e _{-} \to Cu(s)\]
Cell potential: Right cell minus left cell:
\[\Large \sf Zn(s)+ Cu ^{2+}(aq) \to Zn ^{2+}(aq)+Cu(s)\]

Answer 7

thanks.

Answer 8

In general: In a galvanic cell the cathode has a higher potential than the anode: the species undergoing reduction, withdraws electrons from its electrode, so leaving a relative positive charge on it (corresponding to a high potential).

Answer 9

Wait a sec:
You are going to determine electrode potential. Stupid me.
We use Nernst equation:
\[\Large E=E ^{\Theta}-\frac{ RT }{ zF }\ln \left( \frac{ \left[ \sf red \right] }{ \left[ \sf ox \right] } \right)\]

Answer 10

ow.lmao its all good:) lol thnks:)

Answer 11

Where \(E^{\Theta}\) is the standard electron potential.

Answer 12

electrode*