51 Standard Cell Potentials, E° (M18Q5)

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.

Table 1. Selected Standard Reduction Potentials at 25 °C
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.

Zn/Cu: Measured Voltage=1.08 V; Calculated Voltage=1.10 V; Zn is anode, Cu is cathode

Ag/Cu: Measured Voltage=(-)0.45 V; Calculated Voltage=0.46 V; Cu is anode, Ag is cathode

Ag/Zn: Measured Voltage=1.53 V; Calculated Voltage=1.56 V; Ag is anode, Zn is cathode

Cu/Ni: Measured Voltage=0.09 V*; Calculated Voltage=0.59 V; Ni is anode, Cu is cathode

Zn/Pb: Measured Voltage=0.61 V; Calculated Voltage=0.64 V; Zn is anode, Pb is cathode

Ni/Pb: Measured Voltage=0.38 V*; Calculated Voltage=0.12 V; Pb is anode, Ni is cathode

Ag/Pb: Measured Voltage=0.92 V; Calculated Voltage=0.92 V; Pb is anode, Ag is cathode

* – 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

  1. Use Table 1 to choose the better reducing agent:
    1. Ag(s) or Sn(s)
    2. Br–(aq) or Cl–(aq)
    3. Zn(s) or Co(s)
  2. Use Table 1 to determine which metal will react spontaneously with Ag+ ions, but not with Zn2+ ions: Cu, Au, Al, or Mg.
  3. 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+.
  4. According to Table 1, what metal will reduce Ni2+ to Ni(s): Fe(s), Cu(s), Ag(s), or Au(s)?

Answers to Chemistry End of Section Exercises

    1. Sn(s)
    2. Br-(aq)
    3. Zn(s)
  2. Cu
  3. Co2+
  4. Fe(s)

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