Consider the electrochemical cell that can be built with Cu and Zn. -If you were
ID: 911525 • Letter: C
Question
Consider the electrochemical cell that can be built with Cu and Zn. -If you were to build this cell, which metal would be reduced? Which would be oxidized? -Write an equation for the reaction that occurs at the anode and the reaction that occurs at the cathode. -Predict what the voltage of this cell should be under standard conditions.
ANSWER
Zn(s) + Cu^2+(aq) -----> Cu(s) + Zn^2+(aq)
anode : oxidation half cell : Zn-electrode (
Zn(s) -----> Zn^2+(aq) + 2e-
cathode : redduction half cell : Cu-electrode
Cu^2+(aq) + 2e- ------> Cu(s)
E0cell = Ecathode - Eanode
= 0.34-(-0.76)
= 1.1 V
Question1:
-Repeat this procedure for two other metals Cu2+ + 2e- Cu +0.34 V Fe2+ + 2e- Fe -0.44 V Zn2+ + 2e- Zn -0.76 V Al3+ + 3e- Al -1.66 V Mg2+ + 2e- Mg
-Use the Nernst equation to determine the cell potential, Ecell, of the Cu/Zn galvanic cell if the copper sulfate solution is diluted to 0.10M. How does diluting this solution change the predicted voltage?
Explanation / Answer
Ecell = E0cell-0.0591/nlog[Zn^+2]/[Cu^2+]
Ecell = 1.1-0.0591/2log(1/0.1)
= 1.07 V
generally on decreasing the concentration of ions, in the solution Ecell is decreasing.