Electrochemistry #2

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)

B. Cr₂O₇^2-(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:

E⁰ = E⁰_{red (reduction half-reaction)} - E⁰_{red (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 E⁰_red: +1.61V.  The E⁰_red 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 E⁰_red is -0.44V.  Cu is oxidized, and the E⁰_red 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 E⁰_ox, just flip the sign to get the E⁰_red!
• E⁰_red 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).

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!      

Standard reduction potential E⁰_red tells us how badly a substance wants to be reduced -- compared to the standard hydrogen electrode.  The more positive the E⁰_red, more the substance wants to be reduced.  Conversely, the more negative the E⁰_red, 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 (E⁰_ox), which tells you how badly a substance wants to be oxidized.  A substance that really wants to be oxidized will have a very positive E⁰_ox. 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.  E⁰_red of Pt²⁺ + 2e^- → Pt is +1.19V, whereas E⁰_red 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.  

Here are some other useful things to know about these cells. 

• 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 = E⁰_{red (reduction half-reaction)} - E⁰_{red (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 E⁰_red, we just flip the sign, making it +1.61V.  The E⁰_red 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 E⁰_red of a half-reaction by the corresponding stoichiometric coefficient.  E⁰_red 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 E⁰_red is -0.44V.  Cu is oxidized, and the E⁰_red 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 E⁰_red!
 
~The MCATPOD Team ~

Electrochemistry #1

In a galvanic cell the voltage (emf) will change with all of the following, EXCEPT:


    A. the chemical reactions occurring in the half-cells
    B. the length of the wire connecting the half-cells
    C. the concentration of the solutions in the half-cells
    D. the temperature of the solutions in the half-cells


ANSWER:  B


SHORT-AND-SWEET:


Galvanic cells convert the chemical energy of a spontaneous redox reaction into electrical energy, which enables the cell to do work.  You can imagine the voltage of a galvanic cell as a measure of galvanic cell "strength".  What makes a galvanic cell "strong"? 


1. The redox reaction which is occurring in the cell:  different redox reactions will have different intrinsic "desires" to occur.  Those that want to occur spontaneously will have a lot of chemical energy that can be converted into electric energy.  The more chemical energy a reaction has, the higher the voltage produced in the galvanic cell.
2. How concentrated the solutions in the half-cells are:  This tells you how many charged units there are in the galvanic cell;  the more charge that there is, the stronger the galvanic cell!?
3. The temperature of the half-cell solutions:  in order to understand this one, we have to recall that the spontaneity of a reaction is partially determined by the temperature in the system (ΔG = ΔH - TΔS; G-free Gibbs free energy, H-enthalpy, T-temperature, S-entropy).  If we look back to 1., the more spontaneous reactions have more free chemical energy, which corresponds to higher voltage in the galvanic cell.


This leaves us with the length of the wire connecting the half-cells as the correct answer (answer choice B).  This variable will not affect the voltage of the galvanic cell.


THE WHOLE STORY:


"Ugh, I wish I could remember that stupid Nernst equation!"  If this is your first response after reading this question, do not despair.  A good understanding of galvanic cells will enable you to answer this question without memorizing that dreaded equation.  

In fact, the MCAT will frequently test your understanding of topics that are represented by complicated equations -- which does not mean that you have to memorize these equations.  This question is a perfect example.  Instead of forcing yourself to memorize this and a bunch of other complicated equations, what you should do is spend this time understanding what these equations mean.  

Let's see how we can apply this principle to our question.  Before thinking about the voltage of a galvanic cell, it is crucial to understand the galvanic cell itself.  

Galvanic cell uses chemical energy intrinsic to a redox reaction to produce electrical energy, which enables it to do useful electrical work.  The key is that galvanic cell requires a spontaneous redox reaction, because the spontaneity of the redox reaction is where the chemical energy comes from.  An example of a spontaneous chemical reaction used to make a galvanic cell is between zinc and copper:  


Cu²⁺ + Zn    Cu + Zn²⁺


Galvanic cell will transform the chemical energy of a spontaneous redox reaction into electric energy, which manifests as the electric current.

Now, if the chemical energy comes from the spontaneity of a redox reaction, you can guess that the redox reactions that are most spontaneous will produce the most energy, therefore generating the greatest voltage (emf) in the galvanic cell.  This means that if we change which chemical redox reactions are occurring in the galvanic cell, the voltage will change (which eliminates answer A).


The voltage of a particular reaction can be calculated from standard potentials (E°) of each of the oxidation and reduction half-reactions.  What does "standard" mean?  Remember that "standard" always refers to some "standard" conditions.  In this particular case, the standard potential of a reaction is measured at standard CONCENTRATION, which is 1M, and at the standard TEMPERATURE, which is 25°C.  


Obviously, if any of these conditions were to change, the potential of the reaction would change (which eliminates answers C and D).  


This leaves answer B.  The length of the wire between the two half-cells has nothing to do with the emf of the galvanic cell.

..........

Like we mentioned in the beginning, the other way of answering this question is to memorize the equation for voltage of a galvanic cell.  This is the dreaded Nernst equation: 

E = E °  - (RT / nF)  ln (Q)

    E = voltage of a redox reaction (under non-standard conditions)

    E°  = standard voltage of a redox reaction (under standard conditions)

    R =  universal gas constant: R = 8.314 J/Kmol

    T = absolute temperature

    n = number of moles of electrons transferred in the cell reaction or half-reaction

    F = Faraday constant, which is the number of coulombs per mole of electrons: F = 96,500 C/mol

    Q = reaction quotient, determined by the concentration of reactants and products in the chemical reaction

Nernst equation looks intimidating because it has so many components, but what it shows, in the simplest terms, is that E (emf) of the galvanic cell depends on:
    -standard potential E°, which is specific to the chemical species reacting in the galvanic cell,
    -temperature T in the system, and 
    -reaction quotient Q, which depends on the concentration of chemicals in the galvanic cell.




BIG PICTURE: 


1.  When you run into a complicated-looking equation (e.g. Nernst equation), don't waste time trying to memorize it.  Look at the variables in the equation, and how they relate to each other.  Finally, practice by explaining these relations in words.


2.  Galvanic cells require a spontaneous redox reaction, whose intrinsic chemical energy the galvanic cell transforms into electric energy and electric work, i.e. current.  


3.  What redox reaction is occurring, and at which reactant concentration, as well as the temperature of the system (one of the determinants of the spontaneity of the reaction) determine what voltage (emf) the galvanic cell will produce. 


~The MCAT POD Team~