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? 2Ag()+Fe()2Ag()+Fe()+2+aqssaq

Half-EquationAg()+eAg()+aqsFe()+2eFe()2+aqs
Standard Electrode Potential, 𝐸 (V)+0.79960.447

Q2:

A galvanic cell consists of a Mg electrode in 1 MMg(NO)32 solution and a Ag electrode in 1 MAgNO3 solution. What is the standard cell potential?

Half-Equation Mg()+2eMg()2+aqsAg()+eAg()+aqs
Standard Electrode Potential, 𝐸 (V)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 Au3+/Au and Ni2+/Ni half-cells.

Half-EquationAu()+3eAu()3+aqsNi()+2eNi()+aqs
Standard Electrode Potential, 𝐸 (V)+1.4980.257
  • A1.241 V
  • B1.755 V
  • C1.755 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. Sn()+2Cu()Sn()+2Cu()saqaqaq2+2++

Half-Equation Sn()+2eSn()2+aqsCu()+eCu()2++aqaq
Standard Electrode Potential, 𝐸 (V)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. Cd()|Cd(,0.10M)Ni(,0.50M)|Ni()saqaqs2+2+

Half-EquationCd()+2eCd()2+aqsNi()+2eNi()2+aqs
Standard Electrode Potential, 𝐸()V0.40300.257
  • 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 LaO23 to La metal.

Half-EquationCa()+2eCa()2+aqsAl()+3eAl()3+aqsFe()+2eFe()2+aqsLa()+3eLa()3+aqs
Standard Electrode Potential, 𝐸 (V)2.8681.6620.4472.52
  • AAl only
  • BCa only
  • CFe only
  • DAl and Fe
  • ECa and Al

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?

  • ACu(,1M)Cu()Ag(,1M)Ag()2++aqsaqs
  • BCu(,1M)Cu()Ag()Ag(,1M)2++aqssaq
  • CCu()Cu(,1M)Ag()Ag(,1M)saqsaq2++
  • DCu()Cu(,M)Ag(,M)Ag()saqaqs112++
  • EAg(,1M)Ag()Cu()Cu(,1M)+2+aqssaq

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:

Cl+2e2ClVCrO+14H+6e2Cr+7HOV2272+3+2𝐸=+1.36𝐸=+1.33

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 (<+1.36V), 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 (<+1.36V), 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 (>+1.36V), 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 (>+1.36V), making the oxidation of chloride ions by dichromate ions feasible.

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