Question

you have a Pb-Zn electrochemical cell. The Pb electrode is immersed in an electrolyte solution containing .9 molar Pb +2 ions and the Zn electrode is immersed in an electrolyte solution containing 1.3 molar Zn+2 ions. At a temperature of 30 degrees C, what is the voltage of the cell?

Answer 1

The equation of the cell will be:

Pb(s)/ Pb⁺² (0,9M)// Zn⁺²(1,3M)/Zn

We can use the equation of Nerst to calculate the Voltage

E = E⁰ -RT/nF LnQ

where E is cell potencial E⁰ standard potencial, R gas ideal constant, T temperature, n electrons transfered, F faraday constant, Q cocient reaction

E⁰cell = E⁰ande -E⁰cathode

According to the cell disposition Pb is the anode and Zn is the cathode. Pb oxidice and loses electrons and the zinc reduces and absorbes electrons. The standard reaction potential for this 2 reactions are:

Zn+2 + 2e ----- Zn(s) E⁰= -0,76V

Pb +2 + 2e ------ Pb(s)  E⁰= -0,13V

In the standar reaction we have both reduction reactions, but in our case the Pb is oxidicing so we have to change the value of E⁰.

So for Pbwe will have

Pb(s) ----------Pb +2 + 2e  E⁰= 0,13V

E⁰​cell = E⁰anode -E⁰cathode = -0,76V - (+0,13V) = -0,89V

Now we substitute in nerst equation

E= -0,89V - (8,31J/Kmol * 303K/2*96500C) Ln (0,9/1,3)

E=  -0,89V + 0,005V = -0,885V