# Worksheet: Electrochemical Cell Potential

In this worksheet, we will practice calculating the potential of an electrochemical cell from the half-cell potentials of the anode and cathode.

Q1:

What is the standard electrode potential for the galvanic cell with the following overall reaction?

Half-Equation Standard Electrode Potential, 𝐸⦵ (V) Ag()+eAg()+–aqs Fe()+2eFe()2+–aqs +0.7996 −0.447

Q2:

A galvanic cell consists of a electrode in 1 M solution and a electrode in 1 M solution. What is the standard cell potential?

Half-Equation Standard Electrode Potential, 𝐸⦵ (V) Mg()+2eMg()2+–aqs Ag()+eAg()+–aqs −2.372 +0.7996

Q3:

Using the standard electrode potentials in the table below, calculate the standard cell potential for a galvanic cell consisting of / and / half-cells.

Half-Equation Standard Electrode Potential, 𝐸⦵ (V) Au()+3eAu()3+–aqs Ni()+2eNi()+–aqs +1.498 −0.257
• A V
• B1.755 V
• C V
• D1.241 V

Q4:

Using the standard electrode potentials in the table below, calculate the standard cell potential for a galvanic cell with the following overall reaction.

Half-Equation Standard Electrode Potential, 𝐸⦵ (V) Sn()+2eSn()2+–aqs Cu()+eCu()2+–+aqaq −0.1375 +0.153

Q5:

Using the standard electrode potentials shown in the table, calculate, to 3 decimal places, the cell potential for the following electrochemical cell.

Half-Equation Standard Electrode Potential, 𝐸()⦵V Cd()+2eCd()2+–aqs Ni()+2eNi()2+–aqs −0.4030 −0.257
• A0.167 V
• B0.146 V
• C0.164 V
• D0.155 V
• E0.125 V

Q6:

Using the standard electrode potentials shown in the table, determine which of the following metals are capable of reducing to metal.

 Half-Equation Standard Electrode Potential, 𝐸⦵ (V) Ca()+2eCa()2+–aqs Al()+3eAl()3+–aqs Fe()+2eFe()2+–aqs La()+3eLa()3+–aqs −2.868 −1.662 −0.447 −2.52
• A only
• B only
• C only
• D and
• E and

Q7:

The figure shows the setup of a voltaic cell. What one of the following is the correct expression of the setup in cell notation?

• A
• B
• C
• D
• E

Q8:

Concentrated hydrochloric acid is not used to acidify the dichromate ions in the oxidation of a primary alcohol to a carboxylic acid. In practice, the dichromate oxidizes the chloride ions, so sulfuric acid is used instead.

The half-equations and standard electrode potentials for this oxidation are shown:

What would the standard cell potential for a galvanic cell consisting of these two half-cells be?

Is the reaction feasible according to previous calculation? How does this explain the choice of acid in the experiment discussed?

• AThe reaction is feasible and concentrated hydrochloric acid would have a more negative electrode potential (), making the oxidation of chloride ions by dichromate ions also feasible.
• BThe reaction is not feasible; however, concentrated hydrochloric acid would have a more negative electrode potential (), making the oxidation of chloride ions by dichromate ions feasible.
• CThe reaction is feasible and concentrated hydrochloric acid would have a more positive electrode potential (), making the oxidation of chloride ions by dichromate ions also feasible.
• DThe reaction is not feasible; however, concentrated hydrochloric acid would have a more positive electrode potential (), making the oxidation of chloride ions by dichromate ions feasible.