Electromotive Force of Galvanic Cells

Using a voltmeter to measure the electrical potential difference (commonly called voltage) between two electrodes provides a quantitative indication of just how spontaneous a redox reaction is. The potential difference is measured in volts (V), an SI unit which corresponds to one jouleThe SI unit of energy (or heat or work); equal to 0.239 calories, or a kg m2/s2 per ampere-second (1V = 1 J A^–1 s^–1). The voltage indicates the tendency for current to flow in the external circuit, that is, it shows how strongly the anodeThe electrode in an electrochemical cell where oxidation occurs. The positively charged electrode in a vacuum tube. reaction can push electrons into the circuit and how strongly the cathodeThe electrode in an electrochemical cell where reduction occurs; the negatively charged electrode in a vacuum tube. reaction can pull them out. The potential difference is greatest when a large electrical resistance in the external circuit prevents any current from flowing. The maximum potential difference which can be measured for a given cell is called the electromotive forceThe electrical potential difference that can develop across a voltaic (galvanic) cell, used as a measure of the spontaneity of a reaction., abbreviated emf and represented by the symbol E.

By convention, when a cell is written in shorthand notation, its emf is given a positive value if the cell reaction is spontaneous. That is, if the electrode on the left forces electrons into the external circuit and the electrode on the right withdraws them, then the dial on the voltmeter gives the cell emf. On the other hand, if the half-cell on the right side of the shorthand cell notation is releasing electrons, making the right-hand terminal of the voltmeter negative, the cell emf is minus the reading of the meter. This corresponds to a nonspontaneous cell reaction, written in the conventional way.

EXAMPLE 1 When the galvanic cellAn electrochemical cell in which a spontaneous reaction occurs. Such a cell can be used to generate electricity. Also called voltaic cell. shown in Fig. 2 from Galvanic Cells is connected to a voltmeter, the reading is 0.59 V. The shorthand notation for this cell is

      Pt, Cl₂(g)│Cl^–(1 M)║ Fe²⁺(1 M), Fe³⁺(1 M)│Pt

What is the value of the cell emf?

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. We have already seen that this cell as written corresponds to a nonspontaneous reaction. Therefore the emf must be negative and E = – 0.59 V.

EXAMPLE 2 If the voltmeter in Fig. 17.5 reads 1.10 V, what is the emf for the cell

      Cu│Cu²⁺(1 M)║Zn²⁺‌‌(1 M)│Zn

Solution In this case the shorthand notation corresponds to the reverse of Eq. (1) in Galvanic Cells; that is, it refers to the nonspontaneous cell reaction

      Cu + Zn²⁺ → Cu²⁺ + Zn

Consequently the emf for this cell must be negative and E = –1.10 V.

Example 2 shows that if the cell notation is written in reverse, the cell emf changes sign, since for the spontaneous reaction shown in Eq. (2) from Galvanic Cells the emf would have been +1.10 V.

Experimentally measured cell emf's are found to depend on the concentrations of species in solution and on the pressures of gases involved in the cell reaction. Consequently it is necessary to specify concentrations and pressures when reporting an emf, and we shall only consider cells in which all concentrations are 1mol dm^–3 and all pressures are 1 atmAbbreviation for atmosphere, a unit of pressure equal to 101.325 kPa or 760 mmHg. (101.3 kPa).

The emf of such a cell is said to be its standard electromotive force and is given the symbol E°.

The electromotive forces of galvanic cells are found to be additive. That is, if we measure the emf’s of the two cells

            Zn│Zn²⁺ (1 M)║H⁺(1 M)│H₂(1 atm), Pt      E° = 0.76 V      (1)

Pt, H₂(1 atm)│H⁺(1 M)║Cu²⁺(1 M)│Cu                  E° = 0.34 V      (2)

the sum of the E° values corresponds to the measured emf for a third cell with which we discuss in the section on Cell Notation and Conventions:

      Zn│Zn²⁺ (1 M)║ Cu²⁺(1 M)│Cu E° = 1.10 V      (3)

Whenever the right-hand electrode of one cell is identical to the left-hand electrode of another, we can add the emf's in this way, canceling the electrode which appears twice. This additivity makes it possible to store a large amount of emf data in a small table. By convention such data are tabulated as standard reductionThat part of a chemical reaction in which a reactant gains electrons; simultaneous oxidation of a reactant must occur. potentials. These refer to the emf of a cell whose left-hand electrode is the hydrogen-gas electrode and whose right-hand electrode is the electrode whose emf is being sought. Table 1 contains a number of useful standard reduction potentials.

As an example of the use of the table, the entry corresponding to the electrode Cu²⁺(1 M)│Cu is + 0.34 V. Thus when this electrode is written

TABLE 1 Standard Reduction Potentials at 298.15 K.

to the right of Pt, H₂(1 atm)│H⁺(1 M), as in Eq. (2) above, the E° is + 0.34 V. For the Zn²⁺│Zn redox couple, we find E° = – 0.76 V in Table 17.1. This means that for the cell

      Pt, H₂(1 atm)│H⁺(1 M)║ Zn²⁺ (1 M)│ZnE° = – 0.76 V

Since Eq. (1) shows this cell in reverse, we change the sign of E°, obtaining + 0.76 V. Thus we can combine standard reduction potentials from Table 1 to obtain emf's for cells like Eq. (3) so long as both electrodes are given in the table.

EXAMPLE 3 Find the standard emf for the cell

      Hg(l)│Hg²⁺ (1 M)║Br^–│Br₂(l), Pt

Solution From Table 1 we have

      Pt, H₂(1 atm)│H⁺(1 M)║Hg²⁺ (1 M)│Hg(l)E° = + 0.85 V

Since we want to be the left-hand electrode, this must be reversed and the sign of E° must be changed:

      Hg(l)│Hg²⁺ (1 M)║H⁺(1 M)│H₂(1 atm), Pt E° = – 0.85 V      (4)

For the other electrode Table 1 gives

Pt, H₂(1 atm)│H⁺(1 M)║Br^–(1 M)│Br₂(l), Pt E° = +1.07 V      (5)

Adding the cells of Eqs. (4) and (5), we obtain

      Hg(l)│Hg²⁺ (1 M)║Br^–(1 M)│Br₂(l), Pt E° = (1.07 – 0.85) V =+ 0.22 V

The positive value of the standard emf obtained in Example 3 indicates that the corresponding cell reaction is spontaneous:

      Hg(l) + Br₂(l) → Hg²⁺ (1 M) + 2Br^–(1 M)

In other words, bromine is a strong enough oxidizing agentA chemical species that accepts electrons in order to oxidize another species. In the process the oxidizing agent is itself reduced. to convert mercury metalAn element characterized by a glossy surface, high thermal and electrical conductivity, malleability, and ductility. to mercury(II) ions in aqueousDescribing a solution in which the solvent is water. solution, assuming the concentrations of mercury(II) and bromide ions to be 1 mol dm^–3. This corresponds to the observations made where liquidA state of matter in which the atomic-scale particles remain close together but are able to change their positions so that the matter takes the shape of its container mercury combined with liquid bromine to form mercury(II) bromide. Thus the standard reduction potentials in Table 1 can be used to predict whether a particular reaction will take place, just as Table 1 in Redox Couples was used in our earlier discussion of redox reactions. The advantage of Table 1 is that it gives quantitative as well as qualitative information. It not only tells us that Br₂(l) is a stronger oxidizing agent than Hg²⁺ (1 M) [because Br₂(l) is above Hg²⁺ (1 M), but it also tells us how much stronger, in terms of the cell emf of + 0.22 V.