The Galvanic Cell

So redox reactions transfer electrons. Big whoop. They just move an electron off of one atom directly to another. How is that useful to us?

Well yes, most redox reactions occur with out being able to utilize the moving electrons as they move directly from the donor atom to the acceptor atom.

However, if we can somehow get a redox reaction going with some sort of way to separate the reducing agent and the oxidizing agent, we can utilize the electrons as they move from one place to the next. How might we do that?

The easiest way is by creating a pathway between the two reactants along which electrons can travel. When we have a system that allows for the electrons to move along one path, and the reacting ions to move along another, we can capture the flow of electrons as the ions react. This set-up is called a Galvanic Cell.

The galvanic cell has four main components, a pair of unlike metals, an electrolyte, a salt bridge, and a metal wire connecting the two metals. We’ll cover each component individually.

Zn(s)+Cu²⁺(aq)→Zn²⁺(aq)+Cu(s) is the usual standard Zn/Cu galvanic discharge reaction.

If we eliminate the spectator sulfate ion, we have:

In this standard galvanic direction, zinc loses electrons and copper ions gain them. Zinc is the reducing agent; Cu²⁺ is the oxidizing agent.

Zinc metal is oxidized at the anode: Zn→Zn²⁺+2 e⁻. Copper ions are reduced at the cathode: Cu²⁺+2 e⁻→Cu. Copper metal does not spontaneously plate zinc under these standard conditions. Actual cell direction depends on E under its current composition.

The two half-cells are then connected via a conductive wire strewn between the solid copper and zinc. Then the solutions are connected via a salt bridge. The salt bridge allows the two solution to exchange ions from one solution to the other.

Unlike metals tend to readily form redox reactions. However, a redox reaction requires both a metal and another metal’s ion. Therefore, simply having two metals in electrical contact is not sufficient. There needs to be some medium for the metal to be able to dissolve into.

This solution is the electrolyte we introduced before. The electrolyte allows the metal to produce ions which can then be used in the redox reaction.

If we didn’t have any electrolyte, and just connected two metals, there would be no reaction, as both reactants aren’t present. In the copper-zinc cell, the reaction that occurs was:

$Zn_{(s)}+Cu^{2+}_{(aq)}\rightarrow Zn^{2+}_{(aq)}+Cu_{(s)}$

If you only had zinc and copper, there would be no $Cu_{(aq)}^{2+}$ to react with the zinc. However, placing the two metals into a solution would permit the metal to dissolve somewhat, and form metal ions needed to permit the reaction.

With the two metal ions present, by connecting each half-cell with the other through a conductive wire, electrons flow from the zinc side to the copper side, causing the copper ions to reduce and form solid copper. Conversely, the solid zinc atoms, having given up their electrons, dissolve into solution as zinc ions.

A galvanic cell without a salt bridge will indeed generate a flow of electrons, however the life of such a setup is exceedingly shorter than the typical galvanic cell, which includes a salt bridge to transport ions from one half-cell to the other, equalizing the two.

Each galvanic cell is composed of two different half-cells, and the overall performance of the galvanic cells is reliant upon the two half-cells involved. Some half-cells combine to give great voltage potential (electron movement) across the galvanic cell. Others aren’t so great. The actual performance of the cell is determined by the Gibb’s free energy , $\Delta G^\circ$, of the cell reaction. The more negative the Gibb’s free energy, the more favorable the redox reaction, and the more powerful the cell.

Anode always means oxidation and cathode always means reduction. During galvanic discharge, the anode is negative and cathode positive. In an externally driven electrolytic cell, the anode is positive and cathode negative. A discharging battery is a galvanic cell; for a rechargeable battery, specify charging or discharging before assigning electrode roles.

As we covered before, the two half-cells that comprise the anode and the cathode are both needed for a redox reaction to occur. If we want to successfully measure the potential of the cell, we have to calculate it using the Gibb’s free energy of the reaction involved in the galvanic cell in question.

For any given cell, the Gibb’s free energy can be given as:

$\Delta G^\circ =-nFE^\circ_{cell}$

You already know $\Delta G^\circ$ to be the Gibb’s free energy, but you most likely do not know the rest.

In this equation, $n$ is the number of moles of electrons involved in each mole of reaction. For the reaction:

$Zn_{(s)}+Cu^{2+}_{(aq)}\rightarrow Zn^{2+}_{(aq)}+Cu_{(s)}$

$n$ is 2, as there are 2 electrons that pass from each zinc to each copper.

In the Gibb’s free energy equation, $F$ refers to Faraday’s constant, the total charge carried by one mole of electrons, given as $F=96\,485\frac{C}{mol}$.

The final new concept is $E^\circ_{cell}$. This is the Standard Electrode Potential (SEP) of the cell, which determines the voltage of a galvanic cell.

The SEP is measured for half-cells, so the galvanic cell has an overall $E^\circ_{cell}$ given as:

$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$

So what is the SEP, and how do we find it for each half-cell?

First off, the SEP is the voltage associated with one half-cell at standard conditions. For the purposes of this use, standard conditions are given to be the cell in an electrolyte containing is ions at $1M$ concentration and under $1atm$ of pressure, while at $25^\circ C$.

One problem that arises with using these half-cell voltages is that a voltage can only be measured for a whole galvanic cell. For a half-cell to provide a voltage, it needs to be hooked up to another half-cell, so how can we standardize this?

Well, electrochemists decided that by using a standard half-cell as the anode in all SEP measurements, we can achieve useful data for our values.

The chosen half-cell was the Standard Hydrogen Cell, SHC. Not all redox half-cells are based on metals. The hydrogen cell relies on the half reaction:

$2H^++2e^-\rightarrow H_2$

This is a unique cell, as neither part is a useful electrode (a conductive material for the transfer of electrons), so a secondary electrode must be used if you wish to have a working cell.

To get around that, we use a nonreactive electrode, typically platinum. This allows us to feed the half-cell with electrons, permitting the redox reaction to take place.

Notice that this reaction is given as a reduction reaction. Any half-cell will be shown based on its reduction half-reaction. You only reverse the reaction when you are describing a full galvanic cell, and in that case you have to decide which half-cell reaction to reverse. More on that shortly.

So how does the SHC allow us to calculate the SEP for other cells if we don’t even know the SEP for the SHC?

Well, as we are basing every half-cell off of the same SHC, we can simply set a value for the SHC’s SEP, and then use the following equation to solve for the SEP of the accompanying half-cell:

What would be an easy value to set as our reference potential? 10? -5? 42.44? The easiest value to use, and what is in fact used is 0. This way, we can determine the SEP for any half-cell by pairing that cell with the SHC, and then measuring the potential of the overall cell.

$E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$

Under convention, the SHC always represents the anode. Practically, this is not always true. By assuming the SHC to be the anode, we may sometimes get negative voltages in our measurements of SEP. This just means that the reaction actually proceeds with the SHC as the cathode. By retaining the negative convention, we eliminate confusion when a half-cell’s SEP is lower than hydrogen’s.

When listing out the SEPs of various cells, the standard form of each half-cell (as a reduction reaction) is used. The SEP is also given relative to the SHC potential. The lower the SEP of the half-cell, the better that half-cell is at reducing other half-cells. This means that if two cells are combined into a single galvanic cell, a spontaneous redox reaction would occur, with the half-cell of lower SEP reducing the half-cell of higher SEP. In our copper-zinc galvanic cell, the zinc reduces the copper, so therefore the zinc half-cell must have a lower SEP. This is true, as the half-cell for copper has an SEP of $0.34V$, while zinc has $-0.762V$.

Example Problem 8.2.1

What is the anode in a Mercury-Aluminum galvanic cell?

$\begin{aligned} \bold{Half\ }-&\bold{Reaction}&&\boldsymbol{E^\circ(V)}\\ Li^++e^-&\rightleftharpoons Li_{(s)} &&-3.04\\ Cs^++e^-&\rightleftharpoons Cs_{(s)} &&-3.026\\ K^++e^-&\rightleftharpoons K_{(s)} &&-2.931\\ Ba^{2+}+2e^-&\rightleftharpoons Ba_{(s)} &&-2.912\\ Ca^{2+}+2e^-&\rightleftharpoons Ca_{(s)} &&-2.868\\ Mg^{2+}+2e^-&\rightleftharpoons Mg_{(s)} &&-2.372\\ Al^{3+}+3e^-&\rightleftharpoons Al_{(s)} &&-1.662\\ Ti^{3+}+3e^-&\rightleftharpoons Ti_{(s)} &&-1.37\\ Mg^{2+}+2e^-&\rightleftharpoons Mg_{(s)} &&-2.372\\ Al^{3+}+3e^-&\rightleftharpoons Al_{(s)} &&-1.662\\ Ti^{3+}+3e^-&\rightleftharpoons Ti_{(s)} &&-1.37\\ 2H_2O+2e^-&\rightleftharpoons H_{2(g)}+2OH^- &&-0.8277\\ Zn^{2+}+2e^-&\rightleftharpoons Zn_{(s)} &&-0.7618\\ Ga^{3+}+3e^-&\rightleftharpoons Ga_{(s)} &&-0.53\\ Pb^{2+}+2e^-&\rightleftharpoons Pb_{(s)} &&-0.13\\ 2H^{+}+2e^-&\rightleftharpoons H_{2(g)} &&\pm0.000\\ Cu^{2+}+2e^-&\rightleftharpoons Cu_{(s)} &&+0.34\\ Hg^{2+}+2e^-&\rightleftharpoons Hg_{(l)} &&+0.85\\ Cl_2+2e^-&\rightleftharpoons 2Cl^- &&+1.36\\ \end{aligned}$

Now that table is not comprehensive, it just covers many of the more common half-cells you might encounter. If we were to give you a problem regarding a half-cell not on that table, the $E^\circ$ for that half-cell will be provided if needed.

When you determine the $E^\circ_{cell}$ of a galvanic cell, sometimes the redox equation for the cell is not balanced simply by using the half-reaction as it appears on the table.

If that is the case, you multiply the coefficients of the half-reaction by some number to balance the overall redox reaction, and then you can still use the same $E^\circ$ value for that half-cell.

Recall that the Gibb’s free energy equation for a galvanic cell can be found by:

$\Delta G^\circ =-nFE^\circ_{cell}$

The sign of the $E^\circ_{cell}$ factor determines whether the reaction is spontaneous and will produce a positive voltage, or if the reaction requires energy to proceed.

Perhaps you remember from before, but for a reaction to be spontaneous, it needs to have a negative Gibb’s free energy. So how can that happen?

$\Delta G^\circ =-nFE^\circ_{cell}$

In this equation, $n$ is the stoichiometric number of electrons in the reaction. It can never be negative. $F$ is Faraday’s constant, always a positive $F=96\,485\frac{C}{mol}$.

That leaves $E^\circ_{cell}$. If this is negative, then the Gibb’s free energy will become positive. This means that in order for a galvanic cell to produce a useful voltage, the $E^\circ_{cell}$ has to be positive.

Now, the stoichiometry of the redox reaction does not affect $E^\circ_{cell}$, but it does determine $n$, so it is needed in determining the magnitude of Gibb’s free energy for the reaction.

The $n$ we use is the stoichiometric coefficient of the number of electrons passed in a fully balanced redox equation. This means that we have to find a balanced equation for a galvanic cell before we can know the cell’s Gibb’s free energy.

Example Problem 8.2.2

Reviewed standard-state model: a Ti/Ti³⁺ and Cu/Cu²⁺ galvanic cell uses E°(Cu²⁺/Cu)=+0.340 V and E°(Ti³⁺/Ti)=−1.370 V. With F=96485 C mol⁻¹, determine cathode and ΔG° per balanced reaction 3 Cu²⁺+2 Ti→3 Cu+2 Ti³⁺.