CHAPTER XVII

ENERGETICS OF THE VOLTAIC CELL

558*. Source of the Energy of the Current given by a Voltaie Cell. We have considered the question as to how the E.M.F. of a voltaic cell is produced, and now we have to consider more in detail from whence the energy necessary for the maintenance of a current is derived. This energy is evidently, in part at any rate, derived from the energy of the chemical processes which take place in the cell during the time when it is sending a current. In § 228 we have considered the energy which is liberated or absorbed during certain chemical changes, and the question arises as to the connection between the total quantity of energy which is evolved as heat, when the reaction takes place without the production of an electric current, and the energy represented by the current when this is produced. It was thought for some time that the whole of the energy corresponding to any chemical change was converted into electrical energy when the change took place in a voltaic cell, and the fact that the E.M.F. of the Daniell cell, when calculated on this hypothesis from the thermo-chemical data for the chemical changes which take place in this cell, agreed very well with the value as obtained by direct measurement, supported this view.

Thus in § 228 we have seen that when 65 grams (one gram atom) of zinc are dissolved in dilute sulphuric acid according to the equation Zn + H2SO4 = ZnSO4 + H2, 38,066 calories are evolved. Experiment has shown that when 63 grams of copper are converted into copper sulphate in solution in water according to the equation Cu+H,SO1 = CuSO4 + H2 12,500 calories are absorbed, so that 12,500 calories are evolved when CuSO4 is split up into Cu and HSO4. Hence when one equivalent, that is, since zinc is a diad, 65/2 grams of zinc are converted into the sulphate, while at the same time one equivalent of copper (63/2 grams) is deposited from the sulphate, 19033 +6250=25283 calories are on the whole evolved.

Now the reactions considered above are those which go on in the Daniell cell when it is sending a current, and we have seen that the quantities of zinc and copper considered above, are dissolved and precipitated respectively when 96,550 coulombs of electricity pass through the cell. If the E. M.F. of the cell is E volts, then the passage of 96,550 coulombs of electricity will correspond to 96550 × 107 × E ergs, for one

volt is equal to 10 c.g.s. units, and one coulomb is 101 c.g.s. units. If, then, the whole of the energy corresponding to the chemical reaction is converted into electrical, we shall have, since 25,283 calories is equal to 25283 × 4.2 × 107 ergs, or 1068 × 10o ergs,

Now direct measurement has given the value 1.096 volts for the E.M.F. of a Daniell cell, so that in this case it would seem that the electrical energy of the cell is equal to the chemical energy corresponding to the reactions which go on in the cell during the passage of the current.

When, however, the same method of calculation came to be applied to other forms of cells it was found that the E.M.F.'s calculated on this hypothesis differed from the observed values by more than could be accounted for by errors of experiment. The reason for these differences was shown by Helmholtz to be due to the fact that the hypothesis that the electrical and chemical energies were in all cases exactly equal was not true. He showed that this was only true in the case of cells in which the E.M.F. does not vary with the temperature, the Daniell being a cell of this kind.

In order to see the reason for this, we may consider the case of a reversible cell in which all the chemical changes that take place when the cell is allowed to send a current can be reversed when a current is sent through the cell in the reverse direction.

Suppose that when the temperature of the cell is T, (on the absolute scale) the E. M. F. is E1, and that when the temperature is reduced to 71⁄2 the E.M.F. falls to E. If now, when the temperature of the cell is 71, it be allowed to send a current till 2 units of electricity have passed, the work done is QE1, and is represented by the area of the rectangle ABMO (Fig. 530). Now let the temperature of the cell be reduced to T2, so that the E.M.F. is E, and let a quantity of electricity be passed through the cell in the reverse direction. The work which will E have to be done will be QE, and is represented by the rectangle CDOM. During the passage of this electricity in the reverse direction, the chemical changes which took place in the cell during the time when it was sending a current will be exactly reversed, so that if the cell be now heated up to the temperature 7, it will be in exactly the same condition as that in which it was at the start, that is, it will have been carried through a cycle of operations (§ 260).

T

A

B

Tg

C

X

Ο

FIG. 530.

Since, as the cell is a reversible cell, none of the products of the chemical changes which go on during the passage of a current escape from the cell, and also since the elements carry their atomic heats into their compounds, it will require the same quantity of heat to raise the temperature of the cell from T2 to T1 as it did to cool it from T1 to T, and hence these operations exactly balance one another and need not be considered. Also any external work done on account of change of volume will be negligible. A consideration of the diagram shows that an amount of external work represented by the rectangle ABCD has been done during the cycle, and since the chemical state of the cell is the same at the end as at the start, it follows that this work cannot have been done at the expense of chemical energy, but must have been derived from some other source. In other words, the electrical energy of the cell when it was sending a current must have been greater than the chemical energy corresponding to the chemical changes which took place during the passage of this

current.

Next suppose that the E. M.F. of the cell decreases as the temperature increases, so that when the temperature is 7, the E.M.F. is represented by OD, while when the temperature is 7, the E.M.F. is OA. Then, if when the cell is at a temperature T1, it is allowed to send a current till Q units of electricity have passed, the work done will be represented by DCMO, while the work which must be done to drive Q units in the reverse direction when the temperature is lowered to T is represented by ABMO. Hence in this case more work has to be done by the external source than is done by the cell when such a cycle is traversed, that is, the electrical energy which the cell supplies is not as great as would be expected from thermo-chemical data.

Since energy can neither be created nor destroyed, it follows that in the first case considered, namely, when the E. M.F. of the cell decreased with decrease of temperature, since more work is done by the cell when sending the current than is supplied by the chemical changes which take place, the extra energy will have to be supplied at the expense of the heat of the cell, so that if no outside heat is supplied the cell will get cooler as the current passes, or to keep it at a constant temperature heat must be supplied. In the second case, where increase of temperature causes decrease of E. M.F., the opposite is the case, and the cell will get hotter when sending a current, and to keep its temperature constant heat must be abstracted.

By means of the second law of thermo-dynamics it is possible to calculate the quantity of heat which must be supplied or abstracted in this way. Consider the first case, where the temperature coefficient of the cell is positive, that is, where increase of temperature is accompanied by increase in the E.M.F. Here heat has to be supplied while the cell is passing from the condition represented by the point A to that represented by B, and abstracted, since we are now sending the current in

the reverse direction, while passing from C to D. Let H1 be the heat (measured in ergs) supplied at the temperature T1, and H, the heat abstracted at the temperature T2, then, by the second law of thermodynamics (§ 261),

H1-HT-T

Ti

But H1-H is the heat used during the cycle, and is equal to the rectangle ABCD, which we have seen is equal to Q(E1-E).

Hence

But

E-E 2 is the rate of change of the E.M.F. of the cell with temperaT1 - 72 ture, that is, the temperature coefficient, so that if we represent this by SE

and if, further, we take Q as one unit, we have that the quantity of 819 heat converted into electrical energy during the passage of the unit quantity of electricity is

h=7&E
ST

If the E.M.F. decreases with increase of temperature, so that the
temperature coefficient is negative, we have simply to change the sign
SE
ST

of

Hence if his the total quantity of heat produced by the chemical changes which go on in a cell during the passage of unit quantity of electricity, and E is the E.M.F. of the cell, the following equation will hold :-

Thus, in order to be able to calculate the E.M.F. of a cell from thermochemical data, it is necessary to know the temperature coefficient of the cell.

Since in the case of the Daniell cell the E.M.F. calculated without taking account of the effect of the temperature coefficient agrees with the observed value, it is evident that in the case of this cell the E.M.F. can only vary very little with temperature, and experiment has shown that this is the case, the temperature coefficient being +0.000034.

559*. Experimental Verification of the Helmholtz Formula.— The direct experimental verification, by means of thermal measurements, of the correctness of Helmholtz's expression for the difference

between the electrical energy available in a cell and the chemical energy corresponding to the processes which go on in the cell during the time it is sending a current, has been undertaken by Jahn. The cell to be examined was placed within a Bunsen's ice-calorimeter (§ 212), while there were two external circuits each containing a galvanometer. One of these circuits, AGB (Fig. 531), was of low resistance, and the other, AGB, of very high resistance. This being so, practically the whole of the current sent by the cell passes through the galvanometer G1, and this reading of the galvanometer serves to measure the current passing. The very small current which passes through the high resistance galvanometer, G2, is proportional to the difference in potential between the points A and B, so that the deflection of this galvanometer serves to measure the difference of potential between A and B.

FIG. 531.

E

Let C be the current sent by the cell and e the difference in potential between the points A and B when this current is passing. Of course e will be less than the E.M.F. E of the cell on open circuit, since the cell itself has resistance as well as the wires connecting the poles to the points A and B, and therefore, according to Ohm's law, there will be a fall of potential when a current is passing. The heat developed in the branch AG,B in the time will be equal to eCt in electrical units or aeCt in calories, where a is the value of one joule in calories, that is, 0.2387. If r is the resistance of the wires connecting the terminals of the cell with the points A and B, between which e is measured, the energy spent in these wires during a time will be by Joule's law r1Ct joules, or ar C't calories. Hence the total energy expended by the cell on the portions of the circuit outside the calorimeter is a C(e+ Cr1)t.

In addition to the heat developed in the external circuit of the cell, there will be heat developed within the cell itself, owing to the passage of the current through the electrolyte, and if r, is the resistance of the cell, the quantity of heat developed in this way will be art calories. The total heat developed in the circuit will thus be

aC(e+ Cr1)t+ar Ct,

and this represents the total quantity of energy transformed during the passage of Ct units of electricity through the cell. We have seen in $550 that if E is the E.M.F. on open circuit of a cell of which the internal resistance is r, then the E.M.F. between the terminals, when it is sending a current C, is E-rC, so that the difference of potential between the points PN (Fig. 531) is E-rC, where E is the E.M.F. of the cell measured on open circuit. Since the resistance of the two wires PA and BN is 7, there will be a further decrease in the difference