E^o = volts

2. Write the equilibrium expression for the reaction in Question 1. (Remember how we handle substances in the solid phase.)

K = (Remember this expression is also equal to Q)

3. When we prepare an electrochemical cell using standard conditions the concentration of ions is 1 M at 298 K. When we measure the voltage the reaction is not at equilibrium. Calculate the value of Q for the reaction when the electrochemical half cells is first connected and the concentrations of all ions is 1.00 M.

Q =

4. Based on the value of E^o is this reaction thermodynamically favored? (Y/N)

5. What is the value of K for the reaction in Question 1.

(G^o = -nFE^o where F = 96,500 J volt^-1

and G^o = -RT ln K where R = 8.314 J mol^-1 K^-1

so E^o = RT/nF ln K or E^o = 0.0257/n ln K

n is the number of electrons transferred in the balanced chemical equation)

K =

6. Suppose we prepared an electrochemical cell based on the same reaction, but instead of 1 M Zn²⁺ we made the [Zn²⁺]=0.1 M and we'll leave the [Cu²⁺]=1.0 M. Use Dr. Greenbowe's Flash animation to measure the cell potential for this electrochemical cell.

E = volts

7. Use the equation

E_cell = E^o - (0.0592/n) log [Zn²⁺]/[Cu²⁺]

(NOTE: n = # of electrons transferred)

calculate E_cell for the reaction. How does your calculated value compare to the measured value.