Orientation of external cell when voltage is supplied by it to voltaic cell so as to oppose its cell potential

Question

Given below is a page from our Chemistry Textbook:

As it can be seen, the direction of $E_{ext}$ in figure $3.2(c)$ is opposite to the direction in figures $3.2(a)$ and $3.2(b)$.

I understand that $E_{ext}$ is supposed to oppose potential of the voltaic cell in figures $3.2(a)$ and $3.2(b)$. But in figure $3.2(c)$ since $E_{ext}$ is flipped, isn't $E_{ext}$ supporting the electrical potential of the voltaic cell?

Is there is reason for the flipping of the direction, or is it a mistake on part of the writer?

Answer 1

The figures in the OP's post are nicely drawn, as expected in a modern textbook, but figure 3.2(c) has the external battery reversed, which is incorrect. Here is how it works, without the needless complication of the potentiometer.

First, start with a standard Daniell cell with standard assumptions, i.e., negligible internal resistance, unimolar concentrations, and so on. Then the open circuit voltage is 1.100 V and, under light load, essentially the same. This is shown in Fig. 1 below:

As shown, electron flow is from the zinc anode to the copper cathode, via the external load resistor. Note that DMM means digital multimeter, used in voltmeter mode, and DVM means digital voltmeter. The current flow is $11 \mu A $.

Now cut the wire to the cathode and insert an external DC voltage supply that is turned on, but set to supply zero volts between its terminals. This is shown Fig. 2 below:

The external DC supply is schematically depicted as a battery supplying 0.000 V between its terminals. Its internal impedance is assumed negligible. Effectively, this 0 V battery is the same as a piece of wire: the situation is the same as in Fig. 1. The current flow is still $11 \mu A $.

Now start making it interesting. First, set the external DC supply to produce +0.500 V, as shown in Fig. 3 below:

Note how the external DC supply is connected: its positive terminal connects to the copper cathode. The Daniell cell potential is opposed by the external supply voltage and the DMM shows that the difference, which is +0.600 V, is across the load resistor. Therefore, the current flow is only $6 \mu A $.

Next, set the external DC supply to produce +1.100 V, as shown in Fig. 4 below:

No current flows because the external DC supply voltage nulls (exactly opposes) the Daniell cell voltage. Both ends of the resistor are at -1.100 V with respect to the copper electrode, so no current flows and there is no oxidation or reduction taking place in the cell reservoirs. There is no anode or cathode, just electrodes. This is the dividing line between voltaic cell operation and electrolytic cell operation.

Finally, set the external DC supply to produce +3.000 V, as shown in Fig. 5 below:

Now the electron flow is from the negative terminal of the external DC supply, through the load resistor and into the zinc electrode, where reduction will take place in that cell reservoir. This is the electrolysis mode of operation. Note the voltage across the resistor is -1.900 V, i.e., -3.000 V minus -1.100 V. So the zinc electrode is now the cathode and the copper electrode is the anode.

Answer 2

Suppose two identical Daniell cells are connected with the zinc electrodes electrically connected through an amperemeter, and the two copper plates also electrically connected. Of course no currant will be measured by the ammeter. But the tension between the plate is independent on the connections. Now if one of the cell has a higher nominal voltage, maybe because of aa lower zinc ion concentration, this "stronger cell" imposes its capacity of producing electrons to the other one. Electrons will be flying through the ammeter. So the weakest cell has to reverse the sense of its reactions on its plates. The cell will work as in an electrolysis. Zinc ions $\ce{Zn^{2+}}$ will be reduced into metallic zinc in this cell. Slowly the two zinc ion concentrations will tend to equalize themselves in both cells.