Based on the information provided in the table below, which of the following reactions
can be used to build an electrolytic cell?
A. Ni(s) + 2Ce4+(aq) → Ni2+(aq) + 2Ce3+(aq)
A. Ni(s) + 2Ce4+(aq) → Ni2+(aq) + 2Ce3+(aq)
B. Cr2O72-(aq) + 14H+(aq) + 6I-(aq) → 2 Cr3+(aq) + 3I2(s) + 7H2O(l)
C. Fe2+(aq) + Cu(s) → Fe(s) + Cu2+(aq)
D. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Half-reaction
|
Standard
potential
|
Ce3+(aq) → Ce4+(aq) + e-
|
-1.61V
|
Fe2+(aq) + 2e- → Fe(s)
|
-0.44V
|
Cu(s) → Cu2+(aq) + 2e-
|
-0.34V
|
Ni2+(aq)
+ 2e- → Ni(s)
|
-0.28V
|
I2(s)
+ 2e- → 2I-(aq)
|
+0.54V
|
Zn(s) → Zn2+(aq) + 2e-
|
+0.76V
|
Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H20(l)
|
+1.33V
|
Answer: C.
SHORT-AND-SWEET:
The idea behind the electrolytic cell is using energy from an external voltage source to force a non-spontaneous redox reaction. So, in order to answer the question we need to figure out which of the four offered reactions is non-spontaneous. Because it is non-spontaneous, this reaction will have a negative electromotive force (emf) or potential E0. Cell potential E0 we can calculate from the half-cell potentials:
Let's look at our answer choices. In A, the reduction half-reaction is Ce4+ reduction to Ce3+, and the oxidation half-reaction is Ni oxidation to Ni2+. The table suggests that the standard oxidation potential of Ce4+ is -1.61V. Simply change the sign to get the E0red: +1.61V. The E0red of Ni2+ is -0.28V, so the cell potential E0 = 1.61V - (-0.28V) = +1.89V, which means a spontaneous redox reaction. Ideal for a galvanic cell, but not for electrolytic cell (eliminate A). We can use the same process with the reaction in the answer choice B, whose cell potential is +0.79V, and answer choice D, whose cell potential is +1.10V, which eliminate both. In the answer choice C, Fe2+ is reduced, and its E0red is -0.44V. Cu is oxidized, and the E0red of this half-reaction is +0.34V, giving E0 = -0.44V - 0.34V = -0.78V. Negative E0 corresponds to a non-spontaneous reaction, which could be used to build an electrolytic cell (answer C).
Things to keep in mind:
E0 = E0red (reduction half-reaction) - E0red (oxidation half-reaction)
Let's look at our answer choices. In A, the reduction half-reaction is Ce4+ reduction to Ce3+, and the oxidation half-reaction is Ni oxidation to Ni2+. The table suggests that the standard oxidation potential of Ce4+ is -1.61V. Simply change the sign to get the E0red: +1.61V. The E0red of Ni2+ is -0.28V, so the cell potential E0 = 1.61V - (-0.28V) = +1.89V, which means a spontaneous redox reaction. Ideal for a galvanic cell, but not for electrolytic cell (eliminate A). We can use the same process with the reaction in the answer choice B, whose cell potential is +0.79V, and answer choice D, whose cell potential is +1.10V, which eliminate both. In the answer choice C, Fe2+ is reduced, and its E0red is -0.44V. Cu is oxidized, and the E0red of this half-reaction is +0.34V, giving E0 = -0.44V - 0.34V = -0.78V. Negative E0 corresponds to a non-spontaneous reaction, which could be used to build an electrolytic cell (answer C).
Things to keep in mind:
- When calculating the standard cell potential, always use standard REDUCTION potential for each half-reactions! If you are given a table, make sure you note whether you are given reduction or oxidation potentials (or a mix of both). If you are given the E0ox, just flip the sign to get the E0red!
- E0red is an intensive property, which means that the stoichiometric coefficient in a half-reaction does not affect its value!
THE WHOLE STORY:
Electrochemistry deals with the reduction-oxidation (redox) reactions, which involve transfer of electrons.
Oxidation occurs when a compound gives away an electron. The compound that receives this electron is reduced (reduction).
REMEMBER: Oxidation and reduction cannot occur independently of one another!!! An electron has to come off of one compound (oxidation), and it has to be picked up by another (reduction).
Oxidation occurs when a compound gives away an electron. The compound that receives this electron is reduced (reduction).
REMEMBER: Oxidation and reduction cannot occur independently of one another!!! An electron has to come off of one compound (oxidation), and it has to be picked up by another (reduction).
The chemical compounds that participate in this reaction have the appropriate names, based on what they do to the substance they are reacting with.
- Oxidizing agent (oxidant) → it oxidizes the other substance by taking electrons away from it. By taking these electrons, the oxidizing agent gets reduced.
- Reducing agent (reductant) → it reduces the other substance by giving away its electrons. By doing this, the reducing agent gets oxidized.
How can we tell whether a particular substance will be an oxidizing or reducing agent, whether it will be reduced or oxidizes? By knowing what other substance it has to react with!
The following analogy can help illustrate this. Let's say that you have two neighboring countries (our chemical compounds), and one of them wants to occupy a part of the other. The part of the land that might get occupied is our electron, and the country that wants to occupy it is -- you guessed it -- the oxidizing agent! Now, what will determine whether the electron (or the land) will get occupied? Two things: how strongly the attacking country wants it, and how willing the defending country is to defend it.
If the attacker really wants this land, it acts like a good oxidizing agent, which really wants to be reduced!
If the defending country doesn't particularly love this piece of land (and they want to get rid of it -- we know, unlikely to happen in the real world, but let's pretend it's possible!), it acts like a good reducing agent, which really wants to get rid of its electron and be oxidized!
If the attacker really wants this land, it acts like a good oxidizing agent, which really wants to be reduced!
If the defending country doesn't particularly love this piece of land (and they want to get rid of it -- we know, unlikely to happen in the real world, but let's pretend it's possible!), it acts like a good reducing agent, which really wants to get rid of its electron and be oxidized!
Standard reduction potential E0red tells us how badly a substance wants to be reduced -- compared to the standard hydrogen electrode. The more positive the E0red, more the substance wants to be reduced. Conversely, the more negative the E0red, less interested the substance is in being reduced.
Standard reduction potential is the one that you will most commonly hear about. However, there is also its counterpart -- the standard oxidation potential (E0ox), which tells you how badly a substance wants to be oxidized. A substance that really wants to be oxidized will have a very positive E0ox. Be very careful about noting whether the standard potential refers to reduction or oxidation. Though this measure is usually expressed as the standard reduction potential (e.g. Hg2+ + 2e- → Hg(l) E0 = +0.85V), sometimes MCAT will try to trick you on this (like we did in our question!), by expressing it as Hg(l) → Hg2+ + 2e- E0 = -0.85V. Mercury is clearly being oxidized here, so the standard potential is that of oxidation.
The tables of standard potentials can tell you
how particular substances will react with one another, if left to their own
devices. It's like the analogy above -- the land (or electron) in question will end up in the hands (or orbitals) of the country (or compound) wanting it more! Let’s look at an example of a reaction between platinum and copper. E0red of Pt2+ + 2e- → Pt is +1.19V, whereas E0red of Cu2+ + 2e- → Cu is +0.34V. Clearly, the
platinum cation wants to be reduced more, which is what will happen if platinum and copper react together: Pt2+ + Cu(s) → Pt(s) + Cu2+.
Now that we understand the redox reactions, let's move on to the galvanic (voltaic) and electrolytic cells.
Make sure you understand the fundamental difference between the two:
- Galvanic cell uses a spontaneous redox reaction, and transforms the chemical energy of the reaction into electrical energy, which enables it to do work. Think: battery!
- Electrolytic cell requires energy to be put into the electrolytic cell to cause otherwise nonspontaneous redox reaction to occur.
- Regardless of the type of cell, oxidation always occurs on the anode, and reduction always occurs on the cathode (remember: consonants (R-C) go to one side, vowels (O-A) to the other!).
- In the galvanic cell the electrons are released during oxidation on the anode and they flow through the external circuit to the cathode, which is why the anode is labeled as negative, and the cathode as positive. We can envision electrons being attracted to the positive cathode from the negative anode! (This, however, should not be interpreted as the charges on the electrodes -- the actual charges on the electrodes are essentially zero!)
- In the electrolytic cell the voltage source's negative terminal is connected to the cathode, which from there receives electrons to reduce a substance. The electrons that are removed during oxidation at the anode travel to the positive terminal of the voltage source, completing the circuit. Therefore, cathode is labeled as negative, and anode as positive.
Now, back to our question! You are asked to identify a reaction which could be used to build an electrolytic cell. As we said above, for electrolytic cell one needs a nonspontaneous reaction. Non-spontaneous reaction corresponds to a cell whose cell potential E0 is negative. Cell potential refers to the difference between the potentials of the half-cells:
E0 = E0red (reduction half-reaction) - E0red (oxidation half-reaction)
What does this mean in practice? Let's look at the answer choice A. The reduction
half-reaction is Ce4+ reduction to Ce3+, and the oxidation half-reaction is Ni oxidation to Ni2+. From the table we see that the standard oxidation potential of Ce4+
is -1.61V. In order to get the E0red, we just
flip the sign, making it +1.61V. The E0red of Ni2+
is -0.28V, giving the E0 of : +1.61V - (-0.28V) = +1.89V,
which means a spontaneous redox reaction. Ideal for a galvanic cell,
but not for electrolytic cell (eliminate A)!
One thing to note here. In this reaction the stoichiometric coefficient in front of Ce4+ and Ce3+ is 2. Commonly folks make the mistake of multiplying the E0red of a half-reaction by the corresponding stoichiometric coefficient. E0red is an intensive property, which means that the stoichiometric coefficient in a half-reaction does not affect its value.
We can use the same
process with the reaction in the answer choice B, whose cell potential
is +0.79V, and answer choice D, whose cell potential is +1.10V
(eliminate B and D). Now, let's look at the answer choice C. Fe2+
is reduced, and its E0red is -0.44V. Cu is
oxidized, and the E0red of this half-reaction is
+0.34V. Cell potential E0 will be -0.44V - 0.34V = -0.78V, which means that this is a
non-spontaneous reaction, which could be used to build an electrolytic
cell (answer C).
BIG PICTURE:
1. Galvanic cell = spontaneous reaction = does work! Electrolytic cell = non-spontaneous reaction = requires energy from another source!
2. When using standard potential to calculate cell potential E0, always use standard REDUCTION potential! If you are given standard oxidation potential, just flip the sign to get the reduction potential.
3. Stoichiometric coefficients in a half-reaction will not change the value of E0red!
~The MCATPOD Team ~
BIG PICTURE:
1. Galvanic cell = spontaneous reaction = does work! Electrolytic cell = non-spontaneous reaction = requires energy from another source!
2. When using standard potential to calculate cell potential E0, always use standard REDUCTION potential! If you are given standard oxidation potential, just flip the sign to get the reduction potential.
3. Stoichiometric coefficients in a half-reaction will not change the value of E0red!
~The MCATPOD Team ~
