# Galvanic Cell

A galvanic cell produces electrical energy (electricity) from spontaneous chemical energy (redox reactions). In the figure below (from OpenStax), a copper wire is dipped in an aqueous solution containing Ag+.

A redox reaction occurs due to the spontaneous transfer of electrons from the copper to the silver solution and silver precipitate forms. The half-reactions (oxidation and reduction) as well as the overall reaction, are listed below, respectively.

\begin{align*} \mathrm{Cu}(s) &\longrightarrow \mathrm{Cu^{2+}}(aq) + 2e^-\\ \mathrm{2Ag^+}(aq) + 2e^- &\longrightarrow \mathrm{2Ag}(s)\\[2ex] \mathrm{2Ag^+}(aq) + \mathrm{Cu}(s) &\longrightarrow \mathrm{Cu^{2+}}(aq) + \mathrm{2Ag}(s) \end{align*}

One can separate the copper and the silver solution and still carry out a redox reaction. This allows us to take advantage of the movement of electrons from the copper to the silver solution by shuttling them through a wire (electricity) and turn it into work. This leads us to a galvanic cell.

## Anatomy of a Galvanic Cell

Below is a cartoon of a galvanic cell for the redox reaction introduced above.

Anatomy

1. Oxidation takes place at the anode (–). Cu(s) dissolves into the Cu(NO3)2 solution. The anode loses mass.

$\mathrm{Cu}(s) \longrightarrow \mathrm{Cu^{2+}}(aq) + 2e^-$

1. Electrons produced by the oxidation reaction (in 1) flows through the wire and produces a voltage (here shown as +0.46 V) to the cathode. Electrons spontaneously flows from low to high potential, generating a positive voltage.

2. Reduction takes place at the cathode (+). Ag+(aq) precipitates out of the AgNO3 solution and deposits as solid Ag on the cathode. The cathode gains mass.

$\mathrm{2Ag^+}(aq) + 2e^- \longrightarrow \mathrm{2Ag}(s)$

1. The NaNO3 salt bridge contains a salt and is necessary for charge balance.

2. The cathode half-cell experiences a build-up of negative charge as it receives e from the anode. The salt bridge provides positive charge (here, Na+) to keep the cathode charge-neutral.

3. The anode half-cell experiences a build-up of positive charge as it loses e to the cathode. The salt bridge provides negative charge (here, NO3) to keep the anode charge-neutral.

## Cell Diagram

A cell diagram is a shorthand scheme to represent a galvanic cell. For example, the following cell notation represents the galvanic cell pictured above.

$\mathrm{Cu}(s) ~ \lvert ~\mathrm{Cu^{2+}}(aq) ~\rvert\lvert ~\mathrm{Ag^+}(aq) ~\rvert ~\mathrm{Ag}(s)$

Looking closely, the anode is represented by the left-half of the cell diagram and demonstrates the oxidation half-reaction

$\mathrm{Cu}(s) \longrightarrow \mathrm{Cu^{2+}}(aq) + 2e^-$

whereas the cathode is represented by the right-half of the cell diagram and demonstrates the reduction half-reaction

$\mathrm{Ag^+}(aq) + e^- \longrightarrow \mathrm{Ag}(s)$

The double vertical bars in the middle of the cell diagram represents the salt bridge. Electrons flow from left-to-right (anode-to-cathode) in the cell diagram.

### Concentration

Cell diagrams can include additional information such as the concentration of the solution at the anode and cathode such as

$\mathrm{Cu}(s) ~ \lvert ~\mathrm{Cu^{2+}}(aq, 1~M) ~\rvert\lvert ~\mathrm{Ag^+}(aq, 1 M) ~\rvert ~\mathrm{Ag}(s)$ or, in a more detailed way,

$\mathrm{Cu}(s) ~ \lvert ~\mathrm{Cu(NO_3)}(aq, 1~M) ~\rvert\lvert ~\mathrm{AgNO_3}(aq, 1 M) ~\rvert ~\mathrm{Ag}(s)$

Concentrations can be non-standard and vary between half-cells.

$\mathrm{Cu}(s) ~ \lvert ~\mathrm{Cu(NO_3)}(aq, 0.1~M) ~\rvert\lvert ~\mathrm{AgNO_3}(aq, 0.25 M) ~\rvert ~\mathrm{Ag}(s)$

### Electrodes

Cell diagrams can also contain information about inert electrodes. An inert electrode is one that does not take part in a redox reaction (it is inert). It is simply there to help facilitate the transfer of electrons to the solution. The following image shows an inert platinum electrode at the cathode. The reduction taking place does not involve the platinum metal in the reaction. The reduction reaction takes place in the acidic solution where protons (or hydronium ions) react to form hydrogen gas.

This can be represented in a cell diagram such that

$\mathrm{Mg}(s) ~ \lvert ~\mathrm{Mg^{2+}}(aq) ~\rvert\lvert ~\mathrm{Fe^{3+}}(aq) ~\rvert ~\mathrm{Fe^{2+}}(aq) ~\vert~ \mathrm{Pt}(s)$

## Cell Potential

Let us consider, again, the following redox reaction.

$\mathrm{Cu}(s) ~ \lvert ~\mathrm{Cu^{2+}}(aq) ~\rvert\lvert ~\mathrm{Ag^+}(aq) ~\rvert ~\mathrm{Ag}(s)$

Why is it that electrons flow from copper to silver? Why do the electrons not flow in the opposite direction from silver to copper? It is because electrons spontaneously flow from low to high potential, generating a positive voltage. That is, copper has a lower potential than silver. The electron flow in this redox reaction generates +0.46 V, the cell potential. The cell potential, E°cell, is the difference between the potential, E°, for copper and the potential for silver. The equation for determining the cell potential of a redox reaction is given as follows

$E_{\mathrm{cell}}^{\circ} = E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ}$

or, in terms of oxidation and reduction, we have

$E_{\mathrm{cell}}^{\circ} = E_{\mathrm{reduction}}^{\circ} - E_{\mathrm{oxidation}}^{\circ}$

since reduction takes place at the cathode and oxidation takes place at the anode.

For our Cu/Ag redox reaction,

\begin{align*} E_{\mathrm{cell}}^{\circ} &= E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ}\\ &= E_{\mathrm{Ag^+/Ag}}^{\circ} - E_{\mathrm{Cu^{2+}/Cu}}^{\circ}\\ &= (+0.80~\mathrm{V}) - (+0.34~\mathrm{V})\\ &= 0.46~\mathrm{V} \end{align*}

### How potentials are measured

Potentials for substances are relative and a reference half-cell is needed. This half-cell has been chosen to be the standard hydrogen electrode (SHE) and is based on the following half-reaction

$2\mathrm{H^+}(aq) + 2e^- \longrightarrow \mathrm{H_2}(g)$

Create two half-cells of the SHE and hook them together and you get

$\mathrm{H_2}(g) ~\lvert ~ \mathrm{2H^+}(aq) ~\rvert\lvert ~\mathrm{2H^+}(aq) ~\rvert ~\mathrm{H_2}(g)$

Both half-cells are identical and, therefore, have the same potential. The cell potential for this is therefore, 0 V, and each half-reaction is defined to have a potential of 0 V. This is our reference.

Now, hook up any other half-cell to the SHE and measure the cell potential. This becomes the potential for half-reaction in question. For example, we can determine the potential for the following half-reactions

$\textcolor{magenta}{\mathrm{X}(s) ~ \lvert ~\mathrm{X}^{n+}(aq)} ~\rvert\lvert ~\mathrm{2H^+}(aq) ~\rvert ~\mathrm{H_2}(g)$ if the unknown undergoes oxidation, or,

$\mathrm{H_2}(g) ~\rvert ~\mathrm{2H^+}(aq) ~\rvert\lvert~ \textcolor{magenta}{\mathrm{X}^{n+}(aq) ~ \lvert ~\mathrm{X}(s)}$ if the unknown undergoes reduction.

The half-cell colored in magenta reprsents the unknown species, X, we plan on determining the cell potential for.

• If X undergoes oxidation (loses electrons), then X is the anode (left side of the cell diagram) and has a lower potential than the SHE
• If X undergoes reduction (gains electrons), then X is the cahotde (right side of the cell diagram) and has a higher potential than the SHE
• If the cell potential is zero, then both half-cells are the same

Example:

Let’s say you want to know the potential for the following half-reaction

$\mathrm{Cu}(s) \longrightarrow \mathrm{Cu^{2+}}(aq) + \mathrm{2e^-}$

Create this half-cell and hook it up to the SHE. A redox reaction occurs and a cell potential of Ecell = +0.34 V. We can determine the potential for the copper reduction reaction such that

\begin{align*} E_{\mathrm{cell}} &= E_{\mathrm{Cu}} - E_{\mathrm{SHE}}\\ 0.34~\mathrm{V} &= E_{\mathrm{Cu}} - 0~\mathrm{V}\\ 0.34~\mathrm{V} &= E_{\mathrm{Cu}} \end{align*}

Therefore,

• ESHE = 0 V (by definition)
• ECu/Cu2+ = +0.34 V

Since electrons flow from low-to-high potential, the copper is undergoing reduction and is therefore the cathode in this redox process. The following cell diagram represents this setup.

$\mathrm{H_2}(g) ~\rvert ~\mathrm{2H^+}(aq) ~\rvert\lvert~ \textcolor{magenta}{\mathrm{Cu}^{2+}(aq) ~ \lvert ~\mathrm{Cu}(s)}$

### Standard Reduction Potentials

Potentials have been carefully measured (using the process outlined above) under standard conditions and 25 °C and have been tabulated. Below is a list of standard reduction potentials for select substances and listed in descending order. For more cell potentials, see Appendix L.

E° (V)
F2(g) + 2e 2F(aq) +2.87
Au3+(aq) + 3e Au(s) +1.50
Cl2(g) + 2e 2Cl(aq) +1.36
Br2(l) + 2e 2Br(aq) +1.09
Ag+(aq) + e Ag(s) +0.80
Fe3+(aq) + e Fe2+(aq) +0.77
I2(s) + 2e 2I(aq) +0.54
Cu2+(aq) + 2e Cu(s) +0.34
2H+(aq) + 2e H2(g) 0
Pb2+(aq) + 2e Pb(s) –0.13
Ni2+(aq) + 2e Ni(s) –0.26
Co2+(aq) + 2e Co(s) –0.28
Fe2+(aq) + 2e Fe(s) –0.45
Zn2+(aq) + 2e Zn(s) –0.76
Cr2+(aq) + 2e Cr(s) –0.913
Al3+(aq) + 3e Al(s) –1.66
Mg2+(aq) + 2e Mg(s) –2.37
Na+(aq) + e Na(s) –2.71
K+(aq) + e K(s) –2.93
Li+(aq) + e Li(s) –3.04

Remember that electrons spontaneously flow from low to high potential. That means, if you have two half-reactions and you want to determine the cell potential, the half-cell with the lower potential shows up lower in the table while the half-cell with the higher potential shows up higher in the table. From this you can quickly determine the anode (oxidation) and cathode (reduction).

For example, a galvanic cell is created with the following two half-reactions

1. $$\mathrm{Fe^{3+}}(aq) + e^- \longrightarrow \mathrm{Fe^{2+}}(aq)$$
2. $$\mathrm{Zn^{2}}(aq) + 2e^- \longrightarrow \mathrm{Zn}(s)$$

Now we want to determine what the standard cell potential, E°cell, is. First, we determine what will undergo oxidation and what will undergo reduction. To do this, simply find these reactions in the table. Whichever is lower will undergo oxidation and whichever is higher will undergo reduction. We easily conclude that the Zn half-reaction will undergo oxidation whereas the Fe3+ half-reaction will undergo reduction. Our cell diagram can now be written as

$\mathrm{Zn}(s) ~\rvert ~\mathrm{Zn^{2+}}(aq) ~\rvert\lvert~ \mathrm{Fe}^{3+}(aq) ~ \lvert ~\mathrm{Fe^{2+}}(aq)$

We now solve the cell potential equation using the reduction potentials given in the table above.

\begin{align*} E_{\mathrm{cell}}^{\circ} &= E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ}\\ &= E_{\mathrm{Fe^{3+}/Fe^{2+}}}^{\circ} - E_{\mathrm{Zn^{2+}/Zn}}^{\circ}\\ &= (0.77~\mathrm{V}) - (-0.76~\mathrm{V})\\ &= 1.53~\mathrm{V} \end{align*}

This galvanic cell produces very little voltage when using these two half-reactions. For reference, a standard AA battery produces 1.5 V, a standard AAA battery produces 1.5 V, and a standard 9-volt battery produces, well, 9 V.

### Example

What is the standard cell potential for the following galvanic cell?

$\mathrm{Cr}(s) ~\rvert ~\mathrm{Cr^{2+}}(aq) ~\rvert\lvert~ \mathrm{Pb}^{2+}(aq) ~ \lvert ~\mathrm{Pb}(s)$

\begin{align*} E_{\mathrm{cell}}^{\circ} &= E_{\mathrm{cathode}}^{\circ} - E_{\mathrm{anode}}^{\circ}\\ &= E_{\mathrm{Pb^{2+}/Pb}}^{\circ} - E_{\mathrm{Cr^{2+}/Cr}}^{\circ}\\ &= (-0.13~\mathrm{V}) - (-0.913~\mathrm{V})\\ &= 0.78~\mathrm{V} \end{align*}

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