1. 2004, #6 Tony Liu

2. Question:

3. Part A  Question: In the diagram above, label the electrode that is the cathode. Justify your answer.  Based on the description, we know that when the switch is closed, the Sn electrode increases in mass. Therefore, Sn2+ ions are reduced.  Since we also know that reduction happens at the cathode, the Sn electrode is the cathode.

4. Part B  Question: In the diagram above, draw an arrow indicating the direction of the electron flow in the external circuit when the switch is closed.  Since we know that the Sn electrode is the cathode, and that reduction happens at the cathode, electrons flow to the cathode.  Thus, arrows should be drawn from the right electrode to the left one.

5. Part C  Question: If the standard cell potential, Ecell, is +.60V, what is the standard reduction potential, in volts for the X3+/X electrode?  Since we don’t know the volts for the X electrode, we will assign it the variable Z. Since X is the anode, its volts is really –Z.  The equation to find the Ecell is Ecell(cathode)- Ecell(anode).  Therefore, the equation becomes.60=-.14-Z. Z=-.74

6. Part D  Question: Identify metal X.  Simply look at the Standard Reduction potentials chart to see which metal will be reduced with -.74V.  Answer is Chromium.

7. Part E  Question: Write a balanced net-ionic equation for the overall chemical reaction occurring in the cell.  To do this, simply balance the equations until the electrons are equal.  Next, flip the Chromium equation because it is oxidized, not reduced.  Finally, simply add the equations.  The answer is: 3Sn2+(aq) + 2X(s) ->3Sn(s) + 2X3+(aq)

8. Part F  There are two parts to F, so we’ll take a look at the background information first and then do each part separately.  Background: In the cell, the concentration of Sn2+ is changed from 1.0M to 0.50M and the concentration of X3+ is changed from 1.0M to.10M.

9. Part F, i  Question: Substitute all the appropriate values for determining the cell potential, Ecell, into the Nernst equation (Do Not do any calculations).  This is a fairly simple question, as all we need to know is the Nernst equation, which is:  Ecell=E^cell – (RT)/(nF)lnQ, where:  E^cell is the expected cell potential  R is the gas constant, in this case 8.31 volt*couloumb/mol*K  T is temperature in Kelvin  n is amount of mols of electrons involved in the reaction  F is Faraday's constant  Q is the equilibrium constant

10. Part F, i, Cont.  Thus we can simply plug in the values they gave us into the equation.  E^cell is, as we know, is +.60V.  R is 8.31, T is 273K (because it’s standard), F is 96,500.  Since we balanced the equation already, we know 6 electrons are involved so n=6.  Q, as the equilibrium constant, [X3+]/[Sn2+], so it is (.1/.5).

11. Part F, i, Cont.  Thus, Ecell=.60 – (8.31*273)/(6*96500)ln(.1/.5)

12. Part F, ii  Question: On the basis of your response in part (f) (i), will the cell potential, Ecell, be greater than, less than, or equal to the original Ecell? Justify your answer.  If we solve the equation out, we see that the cell potential will be greater than the original Ecell, because (8.31*273)/(6*96500)ln(.1/.5) is a negative number.