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10: Nernst Equation

  • Page ID
    204134
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    Objectives

    • To determine the Nernst equation experimentally
    • To determine the \(\ce{K_{sp}}\; \text{for}\; \ce{AgCl}, \ce{AgBr}, \ce{AgI}\)

    Introduction

    Last week, you measured Eo values with various combinations of metals and metal ions. This week, you will learn about how E depends on concentration of ions in a concentration cell.

    In a concentration cell, both electrodes are made of the same metal and the solutions in the cells are of the same ion as the electrodes. The difference between the cells lies in the metal ion concentration. In this case, the chemical reaction is given below.

    \[\ce{Ag}(s) + \ce{Ag^{+}} (conc) \ce{<=>} \ce{Ag}(s) + \ce{Ag^{+}} (dil) \label{1}\]

    \(E^{o}\) is zero for a concentration cell because the half reactions are identical in both directions. The Nernst equation becomes the following.

    \[ E= \frac{-(0.0591V)}{n} * log_{10} \frac{\ce{Ag_{dil}^{+}}}{\ce{Ag_{conc}^{+}}} \label{2}\]

    The spontaneous direction for the reaction is from left to right when \(\ce{Ag_{conc}^{+}}\)(C1) is greater than \(\ce{Ag_{dil}^{+}}\)(C2). In cell 1, ions accept electrons from the metal, plate out on the electrode, and lower the concentration of the ions. In cell 2, atoms in the metal leave electrons behind and enter the solution, thus raising the ion concentration. The reaction occurs in such a direction as to equalize the concentrations of the two solutions.

    Pre-Lab Assignment

    • Look up the \(\ce{K_{sp}}\)values for \(\ce{AgCl}, \ce{AgBr},\; \text{and}\; \ce{AgI}\)

    Pre Lab Video


    Procedure

    Safety and Waste Disposal

    • The solutions must be disposed of in the hazardous waste container. 
    • Filter paper must be disposed of in the solid hazardous waste container. 

    Part I: Determination of Experimental Nernst Equation

    Step 1

    Construct a data table consisting of columns for E, C1, and C2. Leave additional space as you will be adding columns later.


    Step 2

    Part I Step 2.jpgLabel a paper towel to place your well plate on. You will need two 0.10 M, and one each of 0.010, 0.0010, 0.00010, and 0.000010 M.


    Step 3

    Part I Step 3a.jpgUPart I Step 3b.jpgsing gloves, make the above solutions diluting the 0.10 M \(\ce{AgNO_{3}}\) solution and making serial dilutions as we did in the strong and weak acids lab. You will only need to make 10 mL for each solution.


    Step 4

    Part I Step 4.jpgTransfer approximately 2 mL of the prepared solutions to your well plate. Be careful not to contaminate the adjacent wells.


    Step 5

    Part I Step 5.jpgCut a piece of filter paper into 5 thin strips and wet with 1.0 M \(\ce{NaNO_{3}}\) This will be your salt bridge. The pieces of filter paper need not be large. They just need to connect the wells you will be testing.


    Step 6

    Attach two \(\ce{Ag}\) wires to the alligator clips on the labquest.


    Step 7

    Part I Step 7.jpgPlace one of the wires in the first 0.10 M concentration. The second wire will be placed in each of the remaining solutions.


    Step 8

    For your table, C1 will be 0.10 M for all of the trials. The other solutions will be C2.


    Step 9

    Measure the voltage for each of the five combinations.


    Step 10

    Measure the temperature of the room. Why is this information important?


    Step 11

    Empty all waste into the waste container. Be sure to clean your well plate extremely well to avoid contamination.


    Part II: Determination of \(\ce{K_{sp}}\) of Very Insoluble Silver Halides

    In this portion of the lab, you will determine the concentration of \(\ce{Ag^{+}}\) ion in a saturated solution of silver halides (\(\ce{AgCl}, \ce{AgBr},\; \text{and}\; \ce{AgI}\)) by measuring the voltage of a concentration cell in which the silver halide solution is paired with 0.10 M \(\ce{Ag^{+}}\). 

    Step 1

    Part II Step 1.jpgPut 0.10 M \(\ce{NaCl}\) solution into one of your wells and 0.10 M \(\ce{AgNO_{3}}\) in the other.


    Step 2

    Part II Step 2.jpgTo the vial containing \(\ce{NaCl}\), add 1 drop of 0.10 M \(\ce{AgNO_{3}}\). You should observe the formation of a small amount of the white precipitate. If you don’t, add another drop. What is this precipitate?


    Step 3

    Part II Step 3.jpgCreate a salt bridge by using a strip of filter paper and 1.0 M \(\ce{NaNO_{3}}\).


    Step 4

    Part II Step 4.jpgAgain, using silver wires for both the cathode and anode, place each electrode in one of the two solutions.


    Step 5

    Measure the voltage of the cell.


    Step 6

    Repeat the above procedure using \(\ce{NaBr}\) and \(\ce{NaI}\).


    Step 7

    Dispose of all waste into the waste container.


    Observations

    E (V) C1 (M) C2 (M) log (C2/C1)
           
           
           
           
           

     

    Test E (V)

    \(\ce{AgCl}\)

     

    \(\ce{AgBr}\)

     

    \(\ce{AgI}\)

     

    Calculations

    Part I:

    Step 1

    Calculate log (C2/C1)


    Step 2

    Graph voltage vs. log (C2/C1)


    Step 3

    Determine the equation of the line as well as the \(R^{2}\)


    Step 4

    Based on equation 2 above, what should the theoretical value for your slope be? How does your value compare?


    Part II:

    Step 1

    Using the equation of the line from part one, determine \(\ce{[Ag_{dil}^{+}]}\). What is your \(\ce{[Ag_{conc}^{+}]}\)?


    Step 2

    What is your \(\ce{[Cl^{-}]}\)? Very little of the \(\ce{AgNO_{3}}\) is added to your \(\ce{NaCl}\) solution, so \(\ce{[Cl^{-}]}\) remains essentially unchanged.


    Step 3

    Calculate \(\ce{K_{sp}}\) using \(\ce{K_{sp}} = \ce{[Ag_{dil}^{+}]}\ce{[Cl^{-}]}\).


    Step 4

    Repeat the calculation for \(\ce{AgBr}\) and \(\ce{AgI}\)?


    Step 5

    How would using the theoretical value of -0.0591V rather than the value from your graph affect your results? Calculate \(\ce{K_{sp}}\) for all three trials using the theoretical slope.


    Step 6

    Compare the theoretical value of \(\ce{K_{sp}}\) that you looked up for prelab to the \(\ce{K_{sp}}\) values you calculated.


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