# Worksheet: The Nernst Equation

In this worksheet, we will practice using the Nernst equation to calculate reduction potentials under nonstandard concentrations.

Q1:

A battery is dead when it has no cell potential. Consider a battery with the following overall reaction. The standard electrode potentials for the half-cells in this battery are given in the table.

Half-Equation Standard Electrode Potential, 𝐸()⦵V Cu()+2eCu()2+–aqs Ag()+eAg()+–aqs +0.340 +0.7996

To 2 significant figures, what is the value of when this battery is dead at 298.15 K?

• A
• B
• C
• D
• E

If a particular dead battery is found to have = 0.11 M, what is the concentration of silver ions?

• A M
• B M
• C0.11 M
• D M
• E0.22 M

Q2:

Calculate to 2 significant figures the equilibrium constant at for the following reaction.

Note that each standard electrode potential is expressed per mole of the half-reaction shown in the table.

Half-Equation Standard Electrode Potential, 𝐸()⦵V 2H()+2eH()+–2aqg 2HO()+2eH()+2OH()2–2–lgaq 0.000 −0.8277
• A
• B
• C
• D
• E

Q3:

Using the standard electrode potentials shown in the table, calculate to 2 significant figures the equilibrium constant at 373 K for the following reaction.

Half-Equation Standard Electrode Potential, 𝐸⦵ (V) Hg()+2eHg()2+–aql [HgBr]()+2eHg()+4Br()42–––aqlaq +0.851 +0.21
• A
• B
• C
• D
• E

Q4:

Using the standard electrode potentials shown in the table, calculate to 2 significant figures the equilibrium constant at 373 K for the following reaction.

Half-Equation Standard Electrode Potential, 𝐸⦵ (V) Cd()+2eCd()2+–aqs CdS()+2eCd()+S()ssaq–2– –0.4030 –1.17
• A
• B
• C
• D
• E

Q5:

Using the standard electrode potentials shown in the table, calculate to 2 significant figures the equilibrium constant at 298.15 K for the following reaction.

Half-Equation Standard Electrode Potential, 𝐸⦵ (V) Ag()+eAg()+–aqs AgCl()+eAg()+Cl()ssaq–– +0.7996 +0.22233
• A
• B
• C
• D
• E

Q6:

In the half-cells of an electrochemical cell, 1.00 M aqueous bromide ions are oxidized to 0.110 M bromine and 0.0230 M aluminum ions are reduced to aluminum metal. Using the standard electrode potentials shown in the table, calculate to 3 decimal places the cell potential for the cell at 298.15 K. Note that standard electrode potentials are measured using 1.00 M solutions of the reacting ions.

Half-Equation Standard Electrode Potential, 𝐸⦵ (V) Br()+2e2Br()2––aqaq Al()+3eAl()3+–aqs +1.0873 −1.662

Q7:

The half-cells of a galvanic cell consist of an aluminum electrode in a 0.0150 M aluminum nitrate solution and a nickel electrode in a 0.250 M nickel(II) nitrate solution. Using the standard electrode potentials shown in the table, calculate to 2 decimal places the cell potential for the galvanic cell at 298.15 K. Note that standard electrode potentials are measured using 1.00 M solutions of the reacting ions.

Half-Equation Standard Electrode Potential, 𝐸()⦵V Al()+3eAl()3+–aqs Ni()+2eNi()2+–aqs –1.662 –0.257

Q8:

Using the standard electrode potentials shown in the table, calculate to 2 decimal places the cell potential at 298.15 K for the cell with the following overall reaction.

Half-Equation Standard Electrode Potential, 𝐸()⦵V HgS()+2eHg()+S()slaq–2– Ag()+eAg()+–aqs –0.70 +0.7996

Q9:

Using the standard electrode potential data in the table, calculate the standard cell potential for the following reaction at 298 K.

Half-Equation Standard Electrode Potential, 𝐸()⦵V Co()+2eCo()2+–aqs Fe()+2eFe()2+–aqs −0.28 −0.447

Q10:

Calculate to 3 significant figures the cell potential for the following reaction at 298 K.

Half-Equation Standard Electrode Potential, 𝐸()⦵V Al()+3eAl()3+–aqs Cu()+2eCu()2+–aqs –1.662 +0.340

Q11:

A natural spring containing manganese is found to be highly acidic. Using the diagram shown, what form of iron would you predict to be most abundant? • A
• B
• C
• D
• E