M18Q5: Standard Cell Potentials, E°
Learning Objectives
- Use standard reduction potentials to calculate a standard cell potential, E°cell, and predict whether a reaction is product-favored or reactant-favored.
- Use a table of standard reduction potentials to rank the strengths of oxidizing and reducing agents, to predict which substances can oxidize or reduce another species, and to predict whether a reaction will be reactant-favored or product-favored.
| Key Concepts and Summary | Key Equations | End of Section Exercises |
Interpreting E°cell values
Under standard conditions (1 bar for gases; 1 M for solutes, and pure solids), the equation is written as:
E°cell = Eºcathode – Eºanode
Remember that all E° are given for reduction half-reactions and that the negative sign in the above equation is present to account for oxidation reaction occurring at the anode.
If E°cell is greater than zero volts, the redox reaction will occur spontaneously toward products. If E°cell is less than zero volts, the redox reaction is nonspontaneous and reactant-favored. In a Galvanic cell, only the spontaneous reaction will occur, but an equation can be written either way leading to both spontaneous and nonspontaneous reactions.
Using a Standard Reduction Potential Table
Tables like this (Table 1) make it possible to determine the standard cell potential for many oxidation-reduction reactions.
| Half-Reaction | E° (V) |
|---|---|
| F2(g) + 2 e– → 2 F–(aq) | +2.87 |
| PbO2(s) + SO42-(aq) + 4 H+(aq) + 2 e– → PbSO4(s) + 2 H2O(ℓ) | +1.69 |
| Au3+(aq) + 3 e– → Au(s) | +1.52 |
| MnO4–(aq) + 8 H+(aq) + 5 e– → Mn2+(aq) + 4 H2O(ℓ) | +1.51 |
| Cl2(g) + 2 e– → 2 Cl–(aq) | +1.36 |
| O2(g) + 4 H+(aq) + 4 e– → 2 H2O(ℓ) | +1.23 |
| Pt2+(aq) + 2 e– → Pt(s) | +1.19 |
| Br2(aq) + 2 e– → 2 Br–(aq) | +1.07 |
| Ag+(aq) + e– → Ag(s) | +0.80 |
| Hg22+(aq) + 2 e– → 2 Hg(ℓ) | +0.80 |
| Fe3+(aq) + e– → Fe2+(aq) | +0.77 |
| MnO4–(aq) + 2 H2O(ℓ) + 3 e– → MnO2(s) + 4 OH–(aq) | +0.58 |
| I2(s) + 2 e– → 2 I–(aq) | +0.54 |
| NiO2(s) + 2 H2O(ℓ) + 2 e– → Ni(OH)2(s) + 2 OH–(aq) | +0.49 |
| Cu2+(aq) + 2 e– → Cu(s) | +0.34 |
| Hg2Cl2(s) + 2 e– → 2 Hg(ℓ) + 2 Cl–(aq) | +0.27 |
| AgCl(s) + 2 e– → Ag(s) + Cl–(aq) | +0.22 |
| Sn4+(aq) + 2 e– → Sn2+(aq) | +0.15 |
| 2 H+(aq) + 2 e– → H2(g) | 0.00 |
| Pb2+(aq) + 2 e– → Pb(s) | −0.13 |
| Sn2+(aq) + 2 e– → Sn(s) | −0.14 |
| Ni2+(aq) + 2 e– → Ni(s) | −0.25 |
| Co2+(aq) + 2 e– → Co(s) | −0.28 |
| PbSO4(s) + 2 e– → Pb(s) + SO42-(aq) | −0.35 |
| Cd2+(aq) + 2 e– → Cd(s) | −0.40 |
| Fe2+(aq) + 2 e– → Fe(s) | −0.44 |
| Cr3+(aq) + 3 e– → Cr(s) | −0.74 |
| Zn2+(aq) + 2 e– → Zn(s) | −0.76 |
| Mn2+(aq) + 2 e– → Mn(s) | −1.18 |
| Zn(OH)2(s) + 2 e– → Zn(s) + 2 OH–(aq) | −1.25 |
| Al3+(aq) + 3 e– → Al(s) | −1.68 |
| Mg2+(aq) + 2 e– → Mg(s) | −2.36 |
| Na+(aq) + e– → Na(s) | −2.71 |
| Ca2+(aq) + 2 e– → Ca(s) | −2.84 |
| Ba2+(aq) + 2 e– → Ba(s) | −2.92 |
| K+(aq) + e– → K(s) | −2.92 |
| Li+(aq) + e– → Li(s) | −3.04 |
Voltaic cells have positive cell potentials. In order to achieve this, the reduction reaction at the cathode will have the most positive (or least negative) E° and the oxidation reaction, at the anode, will have the least positive (or most negative) E°.
Example 1
Cell Potentials from Standard Reduction Potentials
What is the standard cell potential for a galvanic cell that consists of Au3+/Au and Ni2+/Ni half-cells? Identify the oxidizing and reducing agents.
Solution
Using Table 1, the reactions involved in the galvanic cell, both written as reductions, are
Au3+(aq) + 3 e– → Au(s) E°Au3+/Au = +1.52 V
Ni2+(aq) + 2 e– → Ni(s) E°Ni2+/Ni = -0.25 V
The most positive E° is for Au3+/Au so that is the reduction (cathode) reaction. The least positive E° is for Ni2+/Ni so that is the oxidation (anode) reaction.
E°cell = Eºcathode – Eºanode = +1.52 V – (-0.25 V) = 1.77 V
Check Your Learning
A galvanic cell consists of a Mg electrode in 1 M Mg(NO3)2 solution and a Ag electrode in 1 M AgNO3 solution. Calculate the standard cell potential at 25 °C.
Answer:
3.16 V
Demonstration: Electrochemical Cells
Set up. Different electrochemical cells are assembled, using a variety of metal and 1.0 M metal aqueous solutions. The metal electrodes are connected with a voltmeter and a salt bridge is used to complete the circuit. The cells presented are: Zn/Cu, Ag/Cu, Ag/Zn, Cu/Ni, Zn/Pb, Ni/Pb, and Ag/Pb.
Prediction. Using the table of standard reductions, predict what the standard cell voltage should be in each cell along with assigning which half cell is the anode and which is the cathode.
Explanation: In each cell, when the measured voltage is positive, the anode is connected to the black wire and the cathode is connected to the red wire. For Ag/Cu, the negative voltage indicates that the anode is connected to the red wire and the cathode is connected to the black wire.
To calculate the cell voltage, E°cell = E°cathode – E°anode, using the values from the standard reduction potential table. For example, in the Zn/Cu cell, the Zn is connected to the black wire (anode) and the Cu is connected to the red wire (cathode). For copper, the E° = 0.34 V. For zinc, the E° = -0.76 V. Therefore,
E°cell = E°Cu2+/Cu – E°Zn2+/Zn = 0.34 – (-0.76) = 1.10 V
Below is each cell, its calculated and measured voltage, and its anode/cathode assignment.
| Measured Voltage | Calculated Voltage | Anode | Cathode | |
| Zn/Cu | 1.08 V | 1.10 V | Zn | Cu |
| Ag/Cu | (-)0.45 V | 0.46 V | Cu | Ag |
| Ag/Zn | 1.53 V | 1.56 V | Ag | Zn |
| Cu/Ni | 0.09 V* | 0.59 V | Ni | Cu |
| Zn/Pb | 0.61 V | 0.64 V | Zn | Pb |
| Ni/Pb | 0.38 V* | 0.12 V | Pb | Ni |
| Ag/Pb | 0.92 V | 0.92 V | Pb | Ag |
*nickel metal is easily oxidized, which can cause the measured voltage to be quite different than the calculated voltage.
Key Concepts and Summary
Drawing on our knowledge from the previous section, given two half-cells, we can use the equation, E°cell = Eºcathode – Eºanode, to calculate the standard cell potential generated when these electrodes are connected. If E°cell > 0, the cathode and anode have been assigned correctly. If E°cell < 0, the cathode and anode have been assigned incorrectly, and what was thought to be the cathode is actually the anode, and vice versa.
Key Equations
- E°cell = Eºcathode – Eºanode
Chemistry End of Section Exercises
- Use Table 1 to choose the better reducing agent:
- Ag(s) or Sn(s)
- Br–(aq) or Cl–(aq)
- Zn(s) or Co(s)
- Use Table 1 to determine which metal will react spontaneously with Ag+ ions, but not with Zn2+ ions: Cu, Au, Al, or Mg.
- Use Table 1 to determine which metal ion will oxidize Zn(s) to Zn2+(aq), but not oxidize Pb(s) to Pb2+(aq): Al3+, Au3+, Co2+, or Mg2+.
- According to Table 1, what metal will reduce Ni2+ to Ni(s): Fe(s), Cu(s), Ag(s), or Au(s)?
- In the standard reduction potentials shown below, what is the strongest reducing agent?
Half-Reaction E° (V) Al3+(aq) + 3 e– → Al(s) -1.66 AgBr(s) + e– → Ag(s) + Br–(aq) +0.07 Sn4+(aq) + 2 e– → Sn2+(aq) +0.14 Fe3+(aq) + e– → Fe2+(aq) +0.77 - For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions. A table of standard reduction potentials can be found in Appendix K.
- Mg(s) + Ni2+(aq) → Mg2+(aq) + Ni(s)
- 2 Ag+(aq) + Cu(s) → Cu2+(aq) + 2 Ag(s)
- Mn(s) + Sn(NO3)2(aq) → Mn(NO3)2(aq) + Sn(s)
- 3 Fe(NO3)2(aq) + Au(NO3)3(aq) → 3 Fe(NO3)3(aq) + Au(s)
- A voltaic cell is created with one cell containing a zinc metal electrode in a 1.0 M Zn(NO3)2 solution and the other cell containing a copper metal electrode in a 1.0 M Cu(NO3)2 solution. The cell are connected with a salt bridge and an external wire with a volt meter. A table of standard reduction potentials can be found in Appendix K. Use this information to answer the following questions:
- Write the half reactions for each cell and indicate which reaction is occurring at the anode and which is occurring at the cathode.
- Write the net ionic reaction for this voltaic cell.
- What is the cell potential?
- Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the cell in which cadmium metal is oxidized to 1 M cadmium(II) ion and a half-cell consisting of an aluminum electrode in 1 M aluminum nitrate solution. Is the reaction spontaneous at standard conditions?
- Consider an electrochemical cell constructed from the half-cells: Fe electrode in 1.0 M FeCl2, and Sn electrode in 1.0 M Sn(NO3)2. A KCl salt bridge links the half-cells. The relevant standard reduction potentials are
Fe2+(aq) + 2 e– → Fe(s) E° = -0.44 V
Sn2+(aq) + 2 e– → Sn(s) E° = -0.14 VWhen the cell is operating spontaneously, which statement below is correct?
- The tin electrode loses mass and the tin electrode is the cathode.
- The tin electrode gains mass and the tin electrode is the cathode.
- The iron electrode gains mass and the iron electrode is the anode.
- The iron electrode loses mass and the iron electrode is the cathode.
- The iron electrode gains mass and the iron electrode is the cathode.
- Consider the following electrochemical cell
Pt(s) | Pu3+(aq), Pu4+(aq) || Cl2(g), Cl–(aq) | Pt(s)
Given that the standard cell potential is 0.35 V and the standard reduction potential of chlorine (Cl2/Cl–) is 1.36 V, what is the standard reduction potential for Pu4+/Pu3+?
Answers to Chemistry End of Section Exercises
- (a) Sn(s); (b) Br-(aq); (c) Zn(s)
- Cu
- Co2+
- Fe(s)
- Al(s)
- (a) +2.106 V (spontaneous)
(b) +0.4591 V (spontaneous)
(c) +1.0425 V (spontaneous)
(d) +0.749 V (spontaneous) - (a) Oxidation @ anode: Zn(s) ⟶ Zn2+(aq) + 2 e– E°anode = -0.763 V
Reduction @ cathode: Cu2+(aq) + 2 e– ⟶ Cu(s) E°cathode = +0.34 V
(b) Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
(c) E°cell = 1.103 V - 3 Cd(s) + 2 Al3+(aq) → 3 Cd2+(aq) + 2 Al(s); −1.273 V; nonspontaneous
- B
- 1.01 V
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