Cell Notation and Conventions

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 15:33


Rather than drawing a complete diagram like the figures in the Galvanic Cells section, it is convenient to specify a galvanic cell in shorthand form. The two cells we have just described would be written as

                   Zn│Zn2+‌‌(1 M)║ Cu2+(1 M)│Cu                         (1)


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


The components of the cell are written in order, starting with the left-hand and moving across the salt bridgeA connection that permits ions to pass but that restricts the flow of solution between the anode half cell and the cathode half cell in an electrochemical cell. to the right. A single vertical line indicates a phase boundary, such as that between the solidA state of matter having a specific shape and volume and in which the particles do not readily change their relative positions. Zn electrodeIn an electrochemical cell, a surface on which oxidation or reduction occurs; an electrode conducts electric current into or out of a cell. and Zn2+(aq), or between Cl2(g) and Cl(aq). The double vertical line represents a saltAn ionic compound that can be formed by replacing the hydrogen ion of an acid with a different cation. bridge. Spectator ions, like SO42–(aq) in the Zn–Cu cell, are usually omitted.

By convention, the electrode written to the left of the salt bridge in this cell notation is always taken to be the anodeThe electrode in an electrochemical cell where oxidation occurs. The positively charged electrode in a vacuum tube., and the associated half-equation is always written as an oxidation. The right-hand electrode is therefore always the cathodeThe electrode in an electrochemical cell where reduction occurs; the negatively charged electrode in a vacuum tube., and the half-equation is always written as a reduction. This is easy to remember, because reading from left to right gives anode and cathode in alphabetical order. The cell reaction corresponding to a given shorthand description is obtained by summing the half-equations after multiplying by any factors necessary to equalize the number of electrons lost at the anode with the number gained at the cathode. Unless otherwise stated, standard conditions of 1 M are usually implied. However, non-standard cells can be created.



EXAMPLE 1 Write the half-equations and cell reactions for each of the following cells:

a) Ag│Ag+║H+│H2, Pt

b) Pt│Cr2O72–, Cr3+, H+║Br│Br2(l), Pt


SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture.


a) Since the half-equation at the left electrode is taken by convention to be oxidation, we have

Ag → Ag+ + e


At the right-hand electrode, then, we must have a reduction:


2H+ + 2e → H2


Multiplying the first half-equation by 2 and summing gives the cell reaction


2Ag + 2H+ → 2Ag+ + H2(g)


b) Following the same procedure as in part a, we obtain


             2Cr3+ + 7H2O → Cr2O72– + 14H+ + 6e        Left electrode

               (Br2(l) + 2e → 2Br) × 3                           Right electrode

2Cr3+ + 7H2O + 3Br2(l) → Cr2O72– + 14H+ + 6Br      Cell reaction


Note: The procedures described in other sections for balancing redox equations were used in arriving at the above results.



EXAMPLE 2 Describe in shorthand notation a galvanic cell for which the cell reaction is

Cu(s) + 2Fe3+(aq) → Cu2+(aq) + 2Fe2+(aq)


Solution First divide the cell reaction into half-equations:


Oxidation:             Cu(s) → Cu2+(aq) + 2e

Reduction:      Fe3+ + e → Fe2+


Then write the oxidation as the left-hand electrode and the reduction on the right:


            Cu│Cu2+║ Fe2+, Fe3+│Pt


(Since both Fe2+ and Fe3+ are in solution, a Pt electrode is used.)



The conventions we have developed can be used to decide whether the cell reaction is actually spontaneousCapable of proceeding without an outside source of energy; refers to a reaction in which the products are thermodynamically favored (product-favored reaction).. If it is, an oxidation will release electrons to the external circuit at the left-hand electrode. If a voltmeter is placed in the circuit, these electrons will make its left-hand terminal negative. Since the right-hand electrode corresponds to reduction, electrons will be withdrawn from the right-hand terminal of the voltmeter. This is shown for the Zn–Cu cell in Figure 1 from the Galvanic Cells section. You can readily confirm that the spontaneous cell reaction (Eq. (1) from Galvanic Cells) corresponds to the shorthand cell notation of Eq. (1).

For the cell shown in Fig. 1 in Galvanic Cells, the shorthand notation is


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


According to the conventions we have just developed, this corresponds to the cell reaction


            2Cl(aq) + 2Fe3+(aq) → Cl2(g) + 2Fe2+(aq)


But this is the of the spontaneous cell reaction we described before (Eq. (2) from Galvanic Cells). Since the cell reaction is nonspontaneous, electrons will not be forced into the external circuit at the left-hand electrode, and they will not be withdrawn at the right. In fact the reverse will actually occur. Thus if a voltmeter is connected to this cell, its right-hand terminal will become more negative and its left-hand terminal will become more positive. This is shown in Fig. 2 from Galvanic Cells.

In general, if a galvanic cell is connected to a voltmeter, the electrode connected to the negative terminal of the meter must be the anode. If our shorthand cell notation shows that electrode on the left, then the corresponding cell reaction must be spontaneous. Electrons will be released by the oxidation half-equation on the left and accepted by the reduction on the right. If, on the other hand, the voltmeter shows that the right-hand electrode is releasing electrons, then we must have written our shorthand notation backwards. This means that the reverse of the cell reaction obtained by our rules must actually be occurring, and it is that reverse reaction which is spontaneous. Thus by simply observing which electrode in the cell releases electrons and which accepts them, that is, by finding which electrode is negative and which positive, we can determine whether the cell reaction is spontaneous.