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Wize High School Grade 12 Chemistry Textbook > Electrochemistry

Galvanic Cells

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Galvanic/Voltaic Cells

  • A galvanic cell constructed of two half cells is shown below.
  • Before we take a deeper look, let's review what we already know about galvanic cells:
  • Do galvanic cells have spontaneous or non-spontaneous electron flow?
    spontaneous
  • Is the Go positive or negative?
    negative
  • According to the equation deltaGo=-nFEoredox then is the Eoredox positive or negative?
    positive
  • Let's take a look at our 2 half reactions from the table again:
Zn2+ (aq) + 2e- -> Zn(s) Eo= -0.76V
Cu2+(aq) + 2e- -> Cu(s) Eo= +0.34V
  • In order to get an Eoredox that is positive, we needed to flip the Zn(s) equation:
Zn(s) -> Zn2+ (aq) + 2e- Eoox=+0.76
Cu2+(aq) + 2e- -> Cu(s) Eored=+0.34
Eoredox=1.1V (you might see the Eocell be referred to as the electromotive force or emf, they mean the same thing!)

Now let's label the galvanic cell...

1) Which is the reduction half equation and which is the oxidation half reaction?

2) Using the following "Wize Concept,"label the anode and cathode in the above diagram.


Wize Concept
The anode is always where oxidation happens.
The cathode is always where reduction happens.

There are different ways to remember this. One way is to remember that anode and oxidation both start with vowels, while cathode and reduction both start with consonants. The other way to remember this is: An Ox Red Cat! Pick your favourite way :)

3) Show where the ions are moving (into solution/onto electrode)

4) Which electrode is shrinking? Which is growing?

5) Consider the salt bridge. Let's say it's made of KCl. Which half cell would the K+ ions travel to? Which half cell would the Cl- ions travel to?


Wize Concept
Anions of the salt bridge move to the anodic solution
Cations of the salt bridge move to the cathodic solution.
This helps to complete the circuit and prevent charges from building up by keeping the solution neutral.

6) Draw the direction of electron flow.

  • Analogy: Running out of your last exam! Everyone is running in the same direction and to where they want to go (outside of the school!)
Wize Tip
Since the anode is always where oxidation happens and the cathode is always where reduction happens, then electrons will always flow from the anode to the cathode!

7) Write in the charges (+ or -) for each electrode.


Wize Concept
Since galvanic cells involve spontaneous electron flow, electrons will travel to where they want to go!
Electrons will always go from the anode to the cathode.
But here, since it's spontaneous flow it means that the cathode will have a positive charge (electrons want to go here) and the anode will have a negative charge (electrons want to get away from here!)
We will see how this is different for electrolytic cells...

8) Write in the spectator ions (in this case it is SO42-, you might also see NO3- in other cases).

  • These ions are called "spectator" ions because that's all they are doing...spectating! They don't participate in the actual reaction, but help to balance charges!


In a galvanic cell, oxidation occurs at the:
Which of the following is true about a galvanic cell?
checklist
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Construct a galvanic cell with the highest voltage given the following half reactions. Determine the voltage produced and express it in standard cell notation.


Na(aq)++e−→Na(s)Eo=−2.71VNa^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
S2O8(aq)2−+2e−→2SO4(aq)2−Eo=2.01VS_2O_{8(aq)}^{2-} +2e^- \to 2SO_{4(aq)}^{2-} \hspace{20pt} E^o=2.01V
I2(s)+2e−→2I(aq)−Eo=0.54VI_{2(s)}+2e^- \to 2I^-_{(aq)} \hspace{43pt} E^o=0.54V
Ni(aq)2++2e−→Ni(s)Eo=−0.23VNi_{(aq)}^{2+} +2e^- \to Ni_{(s)} \hspace{35pt} E^o=-0.23V

For a galvanic cell with Na/Na+ and Cu/Cu2+ answer the following questions.


Which element is being oxidized?
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Spontaneity Rule

  • We learned that we can calculate Eoredox (Eoredox=Eooxidation+ Eoreduction) and use the equation ΔG=-nFE to figure out spontaneity
  • In grade 12 there is a quicker way you may be more familiar with called the "spontaneity rule"
Before we take a look at this new rule, let's review what we already learned...


Question: Can solid aluminum and aqueous copper (II) sulphate react?
  • Find the Eored values in the table:
  • Al3+(aq) + 3e- --> Al(s) Eored=-1.66V
  • Cu2+(aq) + 2e- --> Cu(s) Eored=0.34V
Flip the equation with aluminum since Al(s) is supposed to be a reactant:
  • Al(s) --> Al3+(aq) + 3e- Eoox=+1.66V
  • Cu2+(aq) + 2e- --> Eored=0.34V
Calculate Eoredox=Eoox + Eored
Eoredox=1.66 + 0.34V
Eoredox=2V
Since Eoredox is positive, we know the reaction is spontaneous and so these will react!

Now, let's use the spontaneity rule to arrive at the same answer:
According to the spontaneity rule, any strong reducing agent + strong oxidizing agent will be able to react spontaneously (so a reaction will occur!)


Determine if the following reactions will occur or not.


Will the reaction occur between aqueous calcium nitrate and solid nickel?
Extra Practice
Practice: Galvanic Cells
Galvanic Cell Practice
Construct a galvanic cell with the highest voltage given the following half reactions. Determine the voltage produced and express it in standard cell notation.
Na(aq)++e−→Na(s)Eo=−2.71VNa^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
Practice: Galvanic Cells
Construct a galvanic cell with the highest voltage given the following half reactions. Determine the voltage produced and express it in standard cell notation.
Na(aq)++e−→Na(s)Eo=−2.71VNa^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
S2O8(aq)2−+2e−→2SO4(aq)2−Eo=2.01VS_2O_{8(aq)}^{2-} +2e^- \to 2SO_{4(aq)}^{2-} \hspace{20pt} E^o=2.01V
Practice: Galvanic Cells
Galvanic Cell Practice
Construct a galvanic cell with the highest voltage given the following half reactions. Determine the voltage produced and express it in standard cell notation.
Na(aq)++e−→Na(s)Eo=−2.71VNa^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
Electrochemistry: Galvanic Cells
In a galvanic cell, oxidation occurs at the:
Electrochemistry: Galvanic Cell
In a zinc-lead cell the reaction is:
Pb2+(aq)+Zn(s)→Pb(s)+Zn2+(aq)       E°=0.637VPb^{2+}\left(aq\right)+Zn\left(s\right)\to Pb\left(s\right)+Zn^{2+}\left(aq\right)\ \ \ \ \ \ \ E\degree=0.637V
Which of the following statements about this cell is FALSE?
Electrochemistry: Galvanic Cells
Construct a galvanic cell with the highest voltage given the following half reactions. Determine the voltage produced and express it in standard cell notation.
Na(aq)++e−→Na(s)Eo=−2.71VNa^+_{(aq)} +e^- \to Na_{(s)} \hspace{40pt} E^o=-2.71V
S2O8(aq)2−+2e−→2SO4(aq)2−Eo=2.01VS_2O_{8(aq)}^{2-} +2e^- \to 2SO_{4(aq)}^{2-} \hspace{20pt} E^o=2.01V
A voltaic cell is constructed with an Sn/Sn2+ half-cell and a Zn/Zn2+ half-cell. The zinc electrode is negative.
Write a balanced overall reaction for this cell
Draw a diagram of this cell, labeling electrodes with their charges and showing the direction of electron flow in the wire and anion/cation flow in the salt bridge
Choose the INCORRECT statement
Electrochemistry: Reduction potentials in a Galvanic Cell
Use the standard reduction potentials below to answer the following question: which metal could be used to create a sacrificial anode for an aluminum pipe?
Reduction half−reactionE°(V)𝐶𝑎2+(𝑎𝑞)+2𝑒−⇌𝐶𝑎(𝑠)−2.87𝑁𝑎+(𝑎𝑞)+𝑒−⇌𝑁𝑎(𝑠)−2.71𝐴𝑙3+(𝑎𝑞)+3𝑒−⇌𝐴𝑙(𝑠)−1.66𝑀𝑛2+(𝑎𝑞)+2𝑒−⇌𝑀𝑛(𝑠)−1.18𝐶𝑢2+(𝑎𝑞)+2𝑒−⇌𝐶𝑢(𝑠)+0.15𝐴𝑔+(𝑎𝑞)+𝑒−⇌𝐴𝑔(𝑠)+0.80\begin{array}{c:c}\rm Reduction\ half-reaction& E° (V)\\ \hline 𝐶𝑎^{2+} (𝑎𝑞)+2 𝑒^−⇌𝐶𝑎 (𝑠)& -2.87\\ \hline 𝑁𝑎^+ (𝑎𝑞)+𝑒^−⇌𝑁𝑎 (𝑠)&-2.71\\ \hline 𝐴𝑙^{3+} (𝑎𝑞)+3 𝑒^−⇌𝐴𝑙(𝑠) &-1.66\\ \hline 𝑀𝑛^{2+} (𝑎𝑞)+2 𝑒^−⇌𝑀𝑛 (𝑠)&-1.18\\ \hline 𝐶𝑢^{2+} (𝑎𝑞)+2 𝑒^− ⇌𝐶𝑢 (𝑠) &+0.15\\ \hline 𝐴𝑔^+ (𝑎𝑞)+𝑒^− ⇌𝐴𝑔 (𝑠)&+0.80\end{array}
Electrochemistry: Standard Reduction Potentials
A galvanic cell is constructed from chromium and copper electrodes immersed in aqueous solutions of Cr3+and Cu2+ respectively (half-reactions shown below).
Cr3+(aq)+3e−⇌Cr(s)Cr^{3+}(aq)+3e^-⇌Cr(s) E°red=−0.74VE°_{red}=-0.74V
Cu2+(aq)+2e−⇌Cu(s)Cu^{2+}(aq)+2e^-⇌Cu(s) E°red=+0.34VE°_{red}=+0.34V
A voltaic cell is constructed with an Sn/Sn2+ half-cell and a Zn/Zn2+ half-cell. The zinc electrode is negative.
Draw a diagram of this cell, labeling electrodes with their charges and showing the direction of electron flow in the wire and anion/cation flow in the salt bridge

Electrochemistry: Battery
Given the following half cells, what is the maximum voltage of a battery that could be constructed?
Li(aq)++e−→Li(s)Eo=−3.04 VCe(aq)4++e−→Ce(s)3+Eo=1.44 VCo(aq)3++e−→Co(aq)2+Eo=1.82 V\begin{array}{llll}&Li^+_{(aq)}&+&e^{-}&\rightarrow&Li_{(s)}&&E^o=-3.04\ V\\ \\&Ce^{4+}_{(aq)}&+&e^{-}&\rightarrow&Ce^{3+}_{(s)}&&E^o=1.44 \ V \\ \\&Co^{3+}_{(aq)}&+&e^{-}&\rightarrow&Co^{2+}_{(aq)}&&E^o=1.82 \ V \end{array}
Electrochemistry: Ksp calculations
Use the following reduction potentials to calculate the Ksp of La(OH)3(s)
La(OH)3(s)+3e−→La(s)+3 OH(aq)−Eo=−2.90VLa(aq)3++3e−→La(s)Eo=−2.379 V\begin{array}{llllll}&La(OH)_{3(s)}&+&3e-&\rightarrow& La_{(s)}&+&3\ OH^-_{(aq)}& &&E^o=-2.90 V \\ &La^{3+}_{(aq)}&+&3e^-&\rightarrow& La_{(s)}&&&&& E^o=-2.379\ V \end{array}