The question is as follows:

An aqueous solution containing each of $\ce{Au^{+3}, Cu^{+2}, Ag^{+}\;and\;Li^{+}}$ ions each at the concentration of $1 \mathrm M$, is being electrolysed using inert electrodes. The value of the standard potentials are: $$E^{\circ}_{\ce{Ag^{+}|Ag}}=0.80\,\text{V} = E_1 \\ E^{\circ}_{\ce{Cu^{+2}|Cu}}=0.34\,\text{V} = E_2 \\ E^{\circ}_{\ce{Au^{+3}|Au}}=1.50\,\text{V} = E_3 \\ E^{\circ}_{\ce{Li^{+}|Li}}=-3.03\,\text{V} \\ $$ With increasing voltage, the sequence of deposition of metals on the cathode will be:

Ans: $\ce{Au, Ag, Cu}$

I understand with the increasing value of the standard reduction potential (SRP), the tendency of the metal to get reduced increases, and so obviously, the metal with the highest SRP is deposited first. So we get the increasing order as the answer. But, when I tried to justify this by trying to calculate the specific voltages when electrodeposition should start, I ran into some problems. What follows is my attempt at that. (The following part has been added to the question by me as an edit).

An example of electrolysis of silver ions solution

Let us assume we use a battery of $V$ volts, with the polarity as shown in the diagram. Edit 2: The $V$ here is the minimum voltage required for electrodeposition to start. Now, as the negative end of the battery supplies electrons, so the silver will deposit on that electrode, and so (using the comment by f") oxygen is evolved at the other electrode. We have the electrode reactions as: $$\ce{O2 + 2H2O + 4e^- -> 4OH^-}\qquad E^\circ = V_1 = 0.401\,\mathrm{V}$$ $$\ce{Ag^+ + e^- -> Ag} \qquad E^\circ = V_2 = 0.80\,\mathrm{V}$$ So I reasoned as follows: the hydroxide ions are in the solution, and according to the equation (and its standard reduction potential), the solution is at a higher potential than the the "sink" of electrons, that is the anode. Similarly, the solution is at a lower potential compared to the source of electrons, that is the anode. As both $V_1$ and $V_2$ are positive, using Kirchhoff's Law, we have: $$V+V_1+V_2=0$$ So here we have a contradiction. Both $V_1$ and $V_2$ are positive, but so should be $V$, because of our explicit assumption of how the terminals are connected. But this equation tells us that $V$ is negative!

So what is wrong with this line of reasoning?

Thank you.

  • 2
    $\begingroup$ A metal starts depositing when the applied voltage plus the SRP of the metal exceeds the SRP of the other reaction (presumably generating oxygen gas in this case). $\endgroup$
    – f''
    Aug 2, 2016 at 5:13
  • $\begingroup$ @f'' That validates the answer... Could you explain why that is the case though? $\endgroup$ Aug 2, 2016 at 5:16
  • $\begingroup$ @f'' Also that means gold will deposit on the cathode without applying any external potential difference? (Oxidation potential of oxygen is 1.229 volts) $\endgroup$ Aug 2, 2016 at 5:25
  • $\begingroup$ The standard reduction potential is the electrical reduction potential relative to a platinum/hydrogen electrode. $\endgroup$
    – A.K.
    Aug 2, 2016 at 13:49

3 Answers 3

  1. We can't write a relation for V, V1 and V2, because we can apply virtually any V, and still have V1 and V2 constant :) Though, if V signifies the minimum applied potential for starting reactions, then it's ok.

  2. Instead, V may be compared to total potential of cell, which is V2-V1 or V1-V2 (see [3] below about the sign), but not V1+V2, as you assumed. From wikipedia: E°cell = E°cathode − E°anode.

  3. The exact sign of (V1-V2) is hard to remember and a subject to mess, so I would better remember the processes involved: positive voltage means lack if electrons, i.e. oxidation. But (citation) since the electrode potentials are conventionally defined as reduction potentials (literally: ability to be reduced), the sign of the potential for the metal electrode being oxidized must be reversed when calculating the overall cell potential.

  4. In order to have a spontaneous reaction (ΔG° < 0), E°cell must be positive (see link above). The gold would deposit on the cathode spontaneously if you short external circuit and provide conditions for anode reaction with lower E°. This may be oxidation of OH- (alkaline solution) to O2 on inert electrode. Why do you wonder? Au3+ is powerful oxidizer.

  5. Other note: for high-ionic mix (1M + 1M + 1M + 1M + anions, probably such a mixture is not possible at all) the real redox kinetic will easily be different from simple potential calculation because of polarization effects and other factors. For example, 1-molar Au3+ is probably stable only at low pH, which shifts redox potentials.

  • $\begingroup$ Your point 2 is important, and that could be the source of my error. But I still don't understand where my reasoning is going wrong! That is, I should have derived the same result wherein V1 and V2 had opposite signs...could you help me there? Also, for #1, I forgot to mention that V signifies the minimum applied potential for successful electrodeposition. $\endgroup$ Sep 28, 2016 at 19:22
  • $\begingroup$ If you insist to add potentials using Kirchhoff's Law, then it's ok, but you must add real potentials, not standard ones. Specifically, you are wrong about potential of anode reaction. You have oxygen released, not consumed, as assumed for standard potential. Potential of this process (not a standard potential, which is always for reduction reaction) is negative: E (not E°) = -0.401 V. $\endgroup$
    – sa7
    Sep 29, 2016 at 6:45
  • $\begingroup$ I did consider that, but nevermind. I have spotted my error, your answer (specifically #2) helped a lot! So you can have that bounty... $\endgroup$ Sep 29, 2016 at 7:15
  • $\begingroup$ I revisited this question after a long time, and I realized why we could not understand each other. From chemguide, I had learnt to read positive electrode potentials as favouring forward reaction and vice-versa. So in my mind, I had modelled the electrode potential as the potential difference between the two sides of a reaction. I was essentially trying to model what you knew as a fact! Nevertheless, thank you for your answer! :) $\endgroup$ Jan 4, 2017 at 12:31
  • 1
    $\begingroup$ Yes, electrode potential is just a voltage between electrode and solution, while the difference between "two sides of a reaction" is something meaningless. Chemguide mentions "difference between the negativeness of the metal and the positiveness of the solution around it", I don't like it, because this imply the charges, their absolute values etc, which are unnecessary here. Instead, think of electrode (single electrode) potential as a standalone pump which "creates" a voltage (i.e. pressure). If you manage to short-circuit it in ANY way (you guess how), then current will flow. $\endgroup$
    – sa7
    Jan 4, 2018 at 11:44

So, reading the answer by @sa7, I finally realized what was wrong with my approach. The mistake was in the equation for the silver reduction reaction. It should have been $-V_2$ in the Kirchhoff's Law equation, as in: $$ V + V_1 -V_2=0$$ For the half reaction, $$\ce{Ag^+ + e^- -> Ag} \qquad E^\circ = V_2 = 0.80\,\mathrm{V}$$ $V_2$ is the potential difference $V_{RHS}-V_{LHS}$. I made the mistake of assuming the solution to be $LHS$, seeing $\ce{Ag^+}$ to be on the $LHS$. Actually, I should consider the position of the electrons to decide the $LHS$ and $RHS$. So, the electrode being the source of electrons, is actually the $LHS$. (We can think that the $\ce{Ag}$, being in contact with the solution, has the same potential as the solution, because there is no intervening reaction between the solution and the solid silver. The ions are reacting with the electrode. Is this correct though? Or I could just focus on the electrons and forget everything else...) Thus the term to be added to the Kirchhoff's Law equation is $V_{LHS}-V_{RHS}$, which is equal to $-V_2$.

Carrying out the calculations: $$V=V_2-V_1=0.80\,\mathrm{V} - 0.401\,\mathrm{V}=0.399\,\mathrm{V}$$

(In case you are confused, by $LHS$ I mean "the things that are on the left hand side of the half cell equation", and also their voltage. Same with $RHS$.)


Your own answer, which you marked as "answer" for bounty, is ok, though confusing. Are V1 and V2 real voltages or E° as in your initial question, or what? Also, you introduced "RHS/LHS", but this is not needed as it's implied in definition of voltage as directed potential difference. Kirchhoff's equation is about the directed sum (not difference) of voltages:

Vanode + Vcathode + Vexternal = 0,


Vcathode = E°cathode,
Vanode = -E°anode

I.e., voltage on anode is reversed standard potential, not on cathode, as you write. Though this is not very important, as I stated in #3.

Answering your last question, you can't say "Ag, being in contact with the solution, has the same potential as the solution", this is obviously wrong: electrode always has non-zero potential in solution.

Let's resume. You asked "what is wrong with this line of reasoning". Your initial error in reasoning was that the solution is necessary at a higher potential than anode or lower potential than cathode. This is correct only if external voltage is higher that cell's own potentials. In your example, Vanode is negative, not positive. If this is correct, and considering you wrote "you can have that bounty", then mark my first answer as answer.

  • $\begingroup$ Please allow me some time to think it over and understand it fully and absorb it. Cell potentials, this entire topic has always confused me. I can award the bounty only after three hours, and I'll do it then, surely. Thank you for your response and your time, I appreciate it! $\endgroup$ Sep 29, 2016 at 9:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.