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# 17.9: 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 joule 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 anode reaction can push electrons into the circuit and how strongly the cathode 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 force, 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 $$\PageIndex{1}$$ : Galvanic Cell EMF

When the galvanic 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

$\ce{Pt , Cl_2(g)} \mid \ce{Cl^{–}(1 M)} \parallel \ce{Fe^{2+}(1 M), Fe^{3+}(1 M)} \mid \ce{Pt}$

What is the value of the cell emf?

Solution:

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 $$\PageIndex{2}$$ : Voltmeter Readings

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

$\ce{Cu} \mid \ce{Cu^{2+}(1 M)} \parallel \ce{Zn^{2+}(1 M)} \mid \ce{Zn}$

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

$\ce{Cu + Zn^{2+} \rightarrow Cu^{2+} + Zn}$

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

Example $$\PageIndex{2}$$ 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 atm (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

$\ce{Zn} \mid \ce{Z^{2+}(1 M)} \parallel \ce{H^{+}(1 M)} \mid \ce{H_{2}(1 atm), Pt}\label{4}$ E° = 0.76 V

$\ce{Pt, H_{2}(1 atm)} \mid \ce{H^{+}(1 M)} \parallel \ce{Cu^{2+}(1 M)} \mid \ce{Cu}\label{5}$ E° = 0.34 V

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:

$\ce{Zn \mid Zn^{2+} (1 M)} \parallel \ce{Cu^{2+}(1 M)} \mid \ce{Cu E^{o} = 1.10}\label{6}$

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 reduction 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 $$\PageIndex{1}$$ contains a number of useful standard reduction potentials.

As an example of the use of the table, the entry corresponding to the electrode $$Cu^{2+}(1 M)\mid Cu is + 0.34 V$$. Thus when this electrode is written

TABLE $$\PageIndex{1}$$: Standard Reduction Potentials at 298.15 K.
Acidic Solution Standard Reduction Potential, E° (volts)
$$\ce{F_{2}(g) + 2}e^– \ce{/rightarrow 2F^{–} (aq)}$$
$$2.87$$
$$\ce{Co^{3+}(aq) + }e^– \ce{/rightarrow Co^{2+}(aq)}$$
$$1.92$$
$$\ce{Au^{+}(aq) + }e^– \ce{\rightarrow Au(s)}$$
$$1.83$$
$$\ce{H_{2}O_{2}(aq) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow 4H_{2}O(l)}$$
$$1.763$$
$$\ce{Ce^{4+}(aq) + }e^– \ce{\rightarrow Ce^{3+}(aq)}$$
$$1.72$$
$$\ce{Pb^{4+}(aq) + 2}e^– \ce{\rightarrow Pb^{2+}(aq)}$$
$$1.69$$
$$\ce{PbO_{2}(s) + SO_{4}^{2−}(aq) + 4H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow PbSO_{4}(s) + 6H_{2}O(l)}$$
$$1.690$$
$$\ce{NiO_{2}(s) + 4H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow Ni^{2+}(aq) + 6H_{2}O(l)}$$
$$1.68$$
$$\ce{2HClO(aq) + 2H_\ce{3}O^\ce{+}(aq) + 2}e^– \ce{\rightarrow Cl_{2}(g) + 4H_{2}O(l)}$$
$$1.63$$
$$\ce{Au^{3+}(aq) + 3}e^− \ce{\rightarrow Au(s)}$$
$$1.52$$
$$\ce{MnO_4^{−}(aq) + 8H_{3}O^{+}(aq) + 5}e^– \ce{\rightarrow Mn^{2+}(aq) + 12H_{2}O(l)}$$
$$1.51$$
$$\ce{BrO_{3}^{−}(aq) + 6H_{3}O^{+}(aq) + 5}e^− \ce{\rightarrow \frac{1}{2} Br_{2}(aq) + 9H_{2}O(l)}$$
$$1.478$$
$$\ce{2ClO_{3}^{−}(aq) + 12H_{3}O^{+}(aq) + 10}e^– \ce{\rightarrow Cl_{2}(g) + 18H_{2}O(l)}$$
$$1.47$$
$$\ce{Cr_{2}O_{7}^{2−}(aq) + 14H_{3}O^{+}(aq) + 6}e^– \ce{\rightarrow 2Cr^{3+}(aq) + 21H_{2}O(l)}$$
$$1.36$$
$$\ce{Cl_{2} (g) + 2}e^− \ce{\rightarrow 2Cl^{−}(aq)}$$
$$1.358$$
$$\ce{N_{2}H_{5}^+(aq) + 3H_{3}O^{+}(aq) + 2}e^- \ce{\rightarrow 2NH_{4}^{+}(aq) + 3H_{2}O(l)}$$
$$1.275$$
$$\ce{MnO_{2}(s) + 4H_{3}O^+(aq) + 2}e^– \ce{\rightarrow Mn^{2+}(aq) + 6H_{2}O(l)}$$
$$1.23$$
$$\ce{O_{2}(g) + 4H_{3}O^{+}(aq) + 4}e^– \ce{\rightarrow 6H_{2}O(l)}$$
$$1.229$$
$$\ce{ClO_{4}^{−}(aq) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow ClO_{3}^{−}(aq) + 3H_{2}O(l)}$$
$$1.201$$
$$\ce{IO_{3}^{−}(aq) + 6H_{3}O^+(aq) + 5}e^– \ce{\rightarrow \frac{1}{2} I_{2}(aq) + 9H_{2}O(l)}$$
$$1.195$$
$$\ce{Pt^{2+}(aq) + 2}e^− \ce{\rightarrow Pt(s)}$$
$$1.188$$
$$\ce{Br_{2}(l) + 2}e^− \ce{\rightarrow 2Br^{-}(aq)}$$
$$1.066$$
$$\ce{AuCl_{4}^{−}(aq) + 3}e^- \ce{\rightarrow Au(s) + 4Cl^{−}(aq)}$$
$$1.00$$
$$\ce{NO_{3}^{−}(aq) + 4H_{3}O^{+}(aq) + 3}e^{–} \ce{\rightarrow NO(g) + 6H_{2}O(l)}$$
$$0.96$$
$$\ce{NO_{3}^{−}(aq) + 3H_{3}O^+(aq) + 2}e^– \ce{\rightarrow HNO_{2}(aq) + 4H_{2}O(l)}$$
$$0.94$$
$$\ce{Pd^{2+}(aq) + 2}e^− \ce{\rightarrow Pd(s)}$$
$$0.915$$
$$\ce{2Hg^{2+}(aq) +2}e^− \ce{\rightarrow Hg_{2}^{2+}(aq)}$$
$$0.9110$$
$$\ce{Hg^{2+}(aq) +2}e^− \ce{\rightarrow Hg(l)}$$
$$0.8535$$
$$\ce{SbCl_{6}^{−}(aq) + 2}e^− \ce{\rightarrow SbCl_{4}^{−}(aq) + 2Cl^{−}(aq)}$$
$$0.84$$
$$\ce{Ag^{+}(aq) + }e^− \ce{\rightarrow Ag(s)}$$
$$0.7991$$
$$\ce{Hg_{2}^{2+}(aq) + 2}e^- \ce{\rightarrow 2Hg(l)}$$
$$0.7960$$
$$\ce{Fe^{3+}(aq) + }e^− \ce{\rightarrow Fe^{2+}(aq)}$$
$$0.771$$
$$\ce{[PtCl_{4}]^{2−}(aq) + 2}e^- \ce{\rightarrow Pt(s) + 4Cl^{–}(aq)}$$
$$0.758$$
$$\ce{[PtCl_{6}]^{2−}(aq) + 2}e^− \ce{\rightarrow [PtCl_4]^{2−}(aq) + 2Cl^{–}(aq)}$$
$$0.726$$
$$\ce{O_{2}(g) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow H_{2}O_{2}(aq) + 2H_{2}O(l)}$$
$$0.695$$
$$\ce{TeO_{2}(s) + 4H_{3}O^+(aq) + 4}e^– \ce{\rightarrow Te(s) + 6H_{2}O(l)}$$
$$0.604$$
$$\ce{H_{3}AsO_{4}(aq) + 2H_{3}O^{+}(aq) + 2}e^− \ce{\rightarrow HAsO_{2}(aq) + 4H_{2}O(l)}$$
$$0.560$$
$$\ce{I_{2}(s) + 2}e^− \ce{\rightarrow 2I^{−}(aq)}$$
$$0.535$$
$$\ce{Cu^{+}(aq) + }e^− \ce{\rightarrow Cu(s)}$$
$$0.521$$
$$\ce{[RhCl_{6}]^{3−}(aq) + 3}e^− \ce{\rightarrow Rh(s) + 6Cl^{–}(aq)}$$
$$0.5$$
$$\ce{Cu^{2+}(aq) + 2}e^− \ce{\rightarrow Cu(s)}$$
$$0.340$$
$$\ce{Hg_{2}Cl_{2}(s) + 2}e^− \ce{\rightarrow 2Hg(l) + 2Cl^{−}(aq)}$$
$$0.27$$
$$\ce{AgCl(s) + }e^− \ce{\rightarrow Ag(s) + Cl^{−}(aq)}$$
$$0.222$$
$$\ce{Cu^{2+}(aq) + }e^− \ce{\rightarrow Cu^{+}(aq)}$$
$$0.159$$
$$\ce{SO_{4}^{2−}(aq) + 4H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow H_{2}SO_{3}(aq) + 5H_{2}O(l)}$$
$$0.158$$
$$\ce{Sn^{4+}(aq) + 2}e^− \ce{\rightarrow Sn^{2+}(aq)}$$
$$0.15$$
$$\ce{S(s) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow H_{2}S(aq) + 2H_{2}O(l)}$$
$$0.144$$
$$\ce{AgBr(s) + }e^− \ce{\rightarrow Ag(s) + Br^{−}(aq)}$$
$$0.0713$$
$$\ce{2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow 2H_{2}(g) + 2H_{2}O(l )}$$ (reference electrode)
$$0.0000$$
$$\ce{N_{2}O(g) + 6H_{3}O^{+}(aq) + 4}e^– \ce{\rightarrow 2NH_{3}OH^{+}(aq) + 5H_{2}O(l)}$$
$$– 0.05$$
$$\ce{HgS(s, black) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow Hg(l) + H_{2}S(g) + 2H_{2}O(l)}$$
$$– 0.085$$
$$\ce{Se(s) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow H_{2}Se(aq) + 2H_{2}O(l)}$$
$$– 0.115$$
$$\ce{Pb^{2+}(aq) + 2}e^− \ce{\rightarrow Pb(s)}$$
$$– 0.125$$
$$\ce{Sn^{2+}(aq) + 2}e^− \ce{\rightarrow Sn(s)}$$
$$– 0.1375$$
$$\ce{AgI(s) + }e^− \ce{\rightarrow Ag(s) + I^{−}(aq)}$$
$$– 0.1522$$
$$\ce{[SnF_6]^{2–}(aq) + 4}e^− \ce{\rightarrow Sn(s) + 6F^{−}(aq)}$$
$$– 0.200$$
$$\ce{Ni^{2+}(aq) + 2}e^− \ce{\rightarrow Ni(s)}$$
$$– 0.25$$
$$\ce{Co^{2+}(aq) + 2}e^− \ce{\rightarrow Co(s)}$$
$$-0.277$$
$$\ce{Tl^{+}(aq) + }e^− \ce{\rightarrow Tl(s)}$$
$$– 0.3363$$
$$\ce{PbSO_{4}(s) + 2}e^− \ce{\rightarrow Pb(s) + SO_{4}^{2−}(aq)}$$
$$– 0.3505$$
$$\ce{Cd^{2+}(aq) + 2}e^− \ce{\rightarrow Cd(s)}$$
$$– 0.403$$
$$\ce{Cr^{3+}(aq) + }e^− \ce{\rightarrow Cr^{2+}(aq)}$$
$$– 0.424$$
$$\ce{Fe^{2+}(aq) + 2}e^− \ce{\rightarrow Fe(s)}$$
$$– 0.44$$
$$\ce{2CO_{2}(g) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow (COOH)_{2}(aq) + 2H_{2}O(l)}$$
$$– 0.481$$
$$\ce{Ga^{3+}(aq) + 3}e^− \ce{\rightarrow Ga(s)}$$
$$– 0.53$$
$$\ce{Cr^{3+}(aq) + 3}e^− \ce{\rightarrow Cr(s)}$$
$$– 0.74$$
$$\ce{Zn^{2+}(aq) + 2}e^− \ce{\rightarrow Zn(s)}$$
$$– 0.763$$
$$\ce{Cr^{2+}(aq) + 2}e^− \ce{\rightarrow Cr(s)}$$
$$– 0.90$$
$$\ce{V^{2+}(aq) + 2}e^− \ce{\rightarrow V(s)}$$
$$– 1.13$$
$$\ce{Mn^{2+}(aq) + 2}e^− \ce{\rightarrow Mn(s)}$$
$$– 1.18$$
$$\ce{Zr^{4+}(aq) + 4}e^− \ce{\rightarrow Zr(s)}$$
$$– 1.55$$
$$\ce{Al^{3+}(aq) + 3}e^− \ce{\rightarrow Al(s)}$$
$$– 1.676$$
$$\ce{H_{2}(g) + 2}e^− \ce{\rightarrow 2H^{−}(aq)}$$
$$– 2.25$$
$$\ce{Mg^{2+}(aq) + 2}e^− \ce{\rightarrow Mg(s)}$$
$$– 2.356$$
$$\ce{Na^{+}(aq) + }e^− \ce{\rightarrow Na(s)}$$
$$- 2.714$$
$$\ce{Ca^{2+}(aq) + 2}e^− \ce{\rightarrow Ca(s)}$$
$$– 2.84$$
$$\ce{Sr^{2+}(aq) + 2}e^− \ce{\rightarrow Sr(s)}$$
$$– 2.89$$
$$\ce{Ba^{2+}(aq) + 2}e^- \ce{\rightarrow Ba(s)}$$
$$– 2.92$$
$$\ce{Rb^{+}(aq) + }e^− \ce{\rightarrow Rb(s)}$$
$$– 2.925$$
$$\ce{K^{+}(aq) + }e^− \ce{\rightarrow K(s)}$$
$$– 2.925$$
$$\ce{Li^{+}(aq) + }e^− \ce{\rightarrow Li(s)}$$
$$– 3.045$$

 Basic Solution Standard Reduction Potential, E° (volts) $$\ce{ClO^{–}(aq) + H_{2}O(l) + 2}e^– \ce{\rightarrow Cl^{–}(aq) + 2OH^{–}(aq)}$$ $$0.89$$ $$\ce{OOH^{-}(aq) + H_{2}O(l) + 2}e^– \ce{\rightarrow 3OH^{–}(aq)}$$ $$0.867$$ $$\ce{2NH_{2}OH(aq) + 2}e^– \ce{\rightarrow N_{2}H_{4}(aq) + 2OH^{-}(aq)}$$ $$0.73$$ $$\ce{ClO_{3}^{–}(aq) + 3H_{2}O(l) + 6}e^– \ce{\rightarrow Cl^{–}(aq) + 6OH^{–}(aq)}$$ $$0.622$$ $$\ce{ClO_{3}^{–}(aq) + 3H_{2}O(l) + 6}e^– \ce{\rightarrow Cl^{–}(aq) + 6OH^{–}(aq)}$$ $$0.622$$ $$\ce{MnO_{4}^{–}(aq) + 2H_{2}O(l) + 3}e^– \ce{\rightarrow MnO_{2}(s) + 4OH^{–}(aq)}$$ $$0.60$$ $$\ce{MnO_{4}^{–}(aq) + }e^– \ce{\rightarrow MnO_{4}^{2–}(aq)}$$ $$0.56$$ $$\ce{NiO_{2}(s) + 2H_{2}O(l) + 2}e^– \ce{\rightarrow Ni(OH)_{2}(s) + 2OH^{–}(aq)}$$ $$0.49$$ $$\ce{Ag_{2}CrO_{4}^{–}(s) + 2}e^– \ce{\rightarrow 2Ag(s) + CrO_{4}^{2–}(aq)}$$ $$0.4491$$ $$\ce{O_{2}(g) + 2H_{2}O(l) + 4}e^– \ce{\rightarrow 4OH^{–}(aq)}$$ $$0.401$$ $$\ce{ClO_{4}^{–}(aq) + H_{2}O(l) + 2}e^– \ce{\rightarrow ClO_{3}^{–}(aq) + 2OH^{–}(aq)}$$ $$0.374$$ $$\ce{Ag_{2}O(s) + H_{2}O(l) + 2}e^– \ce{\rightarrow 2Ag(s) + 2OH^{–}(aq)}$$ $$0.342$$ $$\ce{2NO_{2}^{–}(aq) + 3H_{2}O(l) + 4}e^– \ce{\rightarrow N_{2}O(g) + 6OH^{–}(aq)}$$ $$0.15$$ $$\ce{[Co(NH_{3})_{6}]^{3+}(aq) + }e^- \ce{\rightarrow [Co(NH_{3})_{6}]^{3+}(aq)}$$ $$0.058$$ $$\ce{HgO(s) + H_{2}O(l) + 2}e^– \ce{\rightarrow Hg(l) + 2OH^{–}(aq)}$$ $$0.0977$$ $$\ce{O_{2}(g) + H_{2}O(l) + 2}e^– \ce{\rightarrow OOH^{–}(aq) + OH^{–}(aq)}$$ $$0.0649$$ $$\ce{NO_{3}^{-}(aq) + H_{2}O(l) + 2}e^– \ce{\rightarrow NO_{2}^{–}(aq) + 2OH^{–}(aq)}$$ $$0.01$$ $$\ce{MnO_{2}(s) + 2H_{2}O(l) + 2}e^– \ce{\rightarrow Mn(OH)_{2}(s) + 2OH^{–}(aq)}$$ $$- 0.05$$ $$\ce{CrO_{4}^{2–}(aq) + 4H_{2}O(l) + 3}e^– \ce{\rightarrow Cr(OH)_{3}(s) + 5OH^{–}(aq)}$$ $$-0.11$$ $$\ce{Cu_{2}O(s) + H_{2}O(l) + 2}e^– \ce{\rightarrow 2Cu(s) + 2OH^{–}(aq)}$$ $$-0.365$$ $$\ce{FeO_{2}(aq) + H_{2}O(l) + 2}e^– \ce{\rightarrow HFeO_{2}^{–}(aq) + OH^{–}(aq)}$$ $$-0.69$$ $$\ce{2H_{2}O(l) + 2}e^– \ce{\rightarrow H_{2}(g) + 2OH^{–}(aq)}$$ $$-0.8277$$ $$\ce{2NO_{3}^{–}(aq) + 2H_{2}O(l) + 2}e^– \ce{\rightarrow N_{2}O_{4}(g) + 4OH^{–}(aq)}$$ $$-0.86$$ $$\ce{HFeO_{2}^{-}(aq) + 2}e^– \ce{\rightarrow Fe(s) + 3OH^{–}(aq)}$$ $$-0.8$$ $$\ce{SO_{4}^{2–}(aq) + H_{2}O(l) + 2}e^– \ce{\rightarrow SO_{3}^{2–}(aq) + 2OH^{–}(aq)}$$ $$-0.936$$ $$\ce{N_{2}(g) + 4H_{2}O(l) + 4}e^– \ce{\rightarrow N_{2}H_{4}(aq) + 4OH^{–}(aq)}$$ $$-1.16$$ $$\ce{[Zn(OH)_{4}]^{2–}(aq) + 2}e^– \ce{\rightarrow Zn(s) + 4OH^{–}(aq)}$$ $$-1.285$$ $$\ce{Zn(OH)_{2}(s) + 2}e^{–} \ce{\rightarrow Zn(s) + 2OH^{–}(aq)}$$ $$-1.246$$ $$\ce{[Zn(CN)_{4}]^{2–}(aq) + 2}e^– \ce{\rightarrow Zn(s) + 4CN^{–}(aq)}$$ $$-1.34$$ $$\ce{Cr(OH)_{3}(s) + 3}e^– \ce{\rightarrow Cr(s) + 3OH^{–}(aq)}$$ $$-1.33$$ $$\ce{SiO_{3}^{2–}(aq) + 3H_{2}O(l) + 4}e^– \ce{\rightarrow Si(s) + 6OH^{–}(aq)}$$ $$-1.69$$

contacts here

to the right of Pt, H2(1 atm)│H+(1 M), as in Eq. $$\ref{5}$$ 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

$\ce{Pt, H_{2}(1 atm)} \mid \ce{H^{+}(1 M)} \parallel \ce{Zn^{2+} (1 M)} \mid \ce{Zn E^{o} = – 0.76 V}$

Since Eq. $$\ref{4}$$ 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. $$\ref{6}$$ so long as both electrodes are given in the table.

Example $$\PageIndex{3}$$ : Find the standard emf for the cell

$\ce{Hg(l)} \mid \ce{Hg^{2+} (1 M)} \parallel \ce{Br^{–}} \mid \ce{Br_{2}(l), Pt}$

Solution

From Table $$\PageIndex{1}$$ we have;

$\ce{Pt, H_{2}(1 atm)} \mid \ce{H^{+}(1 M)} \parallel \ce{Hg^{2+} (1 M)} \mid \ce{Hg(l)E^{o} = + 0.85 V}$

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

$\ce{Hg(l)} \mid \ce{Hg^{2+} (1 M)} \parallel \ce{H^{+}(1 M)} \mid {H_{2}(1 atm), Pt E^{o} = – 0.85 V}\label{10}$

For the other electrode Table $$\PageIndex{1}$$ gives

$\ce{Pt, H_{2}(1 atm)} \mid \ce{H^{+}(1 M)} \parallel \ce{Br^{–}(1 M)} \mid \ce{Br_{2}(l), Pt E^{o} = +1.07 V}\label{11}$

Adding the cells of Eqs. $$\ref{10}$$ and $$\ref{11}$$, we obtain

$\ce{Hg(l)} \mid \ce{Hg^{2+} (1 M)} \parallel \ce{Br^{–}(1 M)} \mid \ce{Br_{2}(l), Pt E^{o} = (1.07 – 0.85) V =+ 0.22 V}$

The positive value of the standard emf obtained in Example $$\PageIndex{3}$$ indicates that the corresponding cell reaction is spontaneous:

$\ce{Hg(l) + Br_{2}(l) \rightarrow Hg^{2+} (1 M) + 2Br^{–}(1 M)}$

In other words, bromine is a strong enough oxidizing agent to convert mercury metal to mercury(II) ions in aqueous solution, assuming the concentrations of mercury(II) and bromide ions to be 1 mol dm–3. This corresponds to the observations made where liquid mercury combined with liquid bromine to form mercury(II) bromide. Thus the standard reduction potentials in Table $$\PageIndex{1}$$ 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 $$\PageIndex{1}$$ 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.