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

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?

• You are correct: figure 3.2(c) should not have the battery reversed. Whomever drew that figure messed up.
– Ed V
Jun 17 at 17:12
• (+1) I am giving you an upvote to get you 10 rep points: anyone who throws down a 50 bounty, when they are barely over a 100, deserves respect.
– Ed V
Jun 17 at 19:47
• I am going to be blunt. This is NCERT, what else do you expect? That's the only reason I never read this book even after how many times I was forced to read it by the school. There arrangement works but obviously it's not the arrangement the text is referring to. Surely the diagram making team knew nothing about chemistry and so when they had to reverse the direction of current they foolishly also reversed the direction of the batter. They must have thought that they are smart. Jun 18 at 4:22
• @NisargBhavsar What's more frustrating is that reference book I decided to use blindly follows NCERT. This same mistake is in Modern's ABC Chemistry. Jun 18 at 9:14
• Thanks for the green checkmark and the bounty! Happy to be of assistance!
– Ed V
Jun 25 at 19:25

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.

• Thanks. You have a minus before 1.100V in the second last line which is unnecessary. I can't suggest an edit because your answer is so good that I can't find the other 5 characters required to do so. Jun 18 at 1:06
• The DMM reads -1.900 V because the voltage across the resistor is -3.000 V - (-1.100 V) = -1.900 V. The voltages are with respect to the copper anode. In the Daniell cell itself, as in Fig. 1, the DMM probes would be black (reference) on the zinc and red (input) on the copper, so the cell voltage would be +1.100 V.
– Ed V
Jun 18 at 1:10
• It's a shame the book is wrong. It's a staple book used by almost every student in my country. The mistake has been around for a long time and still hasn't been rectified in the latest edition. Jun 18 at 1:11
• Sorry. My bad. I missed the minus before 3.000V Jun 18 at 1:15
• Yeah, textbook mistakes happen and are often hard to catch and fix. I think this one was caused by someone getting a little tricky with the potentiometer. Had they simply Thevenized the battery and pot, it would have reduced to the Thevenin equivalent voltage, i.e., the voltage divider voltage, in series with the Thevenin equivalent resistance, which is just the parallel resistance of the two pot arms. By design, to have a stiff voltage source, the Thevenin equivalent resistance would be negligible compared with the load resistor I used.
– Ed V
Jun 18 at 1:17

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.

• I must admit that figure $3.1$ has something puzzling. There is an arrow coming out of the copper plate. This arrow has no meaning. Nothing is coming out of this plate. On the contrary, some ions $\ce{Cu^{2+}}$ are attracted by the copper plate and gets discharged as metallic copper. The only thing that is leaving the copper plate are the sulfate ions $\ce{SO4^{2-}}$ which are pushed through the slat bridge. But this ions are not leaving the plate ! The same stupid arrow is reported on the zinc anode, as if something was deposited on the zinc plate. In the contrary, Zinc metal is dissolved. Jun 17 at 17:02
• Right now, you have one up vote, which I gave you previously. There is nothing wrong with your answer, but I will double check. ;-) I think the drawings are a little too cute and the use of the potentiometer just reduces clarity. It would have been better, in my opinion, to just show a simple opposing DC voltage source.
– Ed V
Jun 17 at 19:40
• @Maurice The yellow arrows show the direction of electron flow. Electrons released by zinc's oxidation flow through the wire, through the copper electrode into the solution, which are used by $\ce{Cu^{2+}}$ ions to undergo reduction. Jun 18 at 0:49
• @Alpha Delta. Electrons do not go into solution on the copper plate !!! They do not go out of the solution in the zinc compartment !!! Jun 18 at 7:53
• @Alpha Delta. We are all doing mistakes. Even a book may be wrong ! Jun 18 at 10:11