Non-standard Cells

So far we have only covered the galvanic cell under standard conditions. A galvanic cell is actually a subset of electrochemical cells. An electrochemical cell is any combination of two half-cells. A galvanic cell has the requirement that it create a voltage difference that results from a redox reaction.

The other type of electrochemical cell, the electrolytic cell, doesn’t produce electricity. Instead, this type of cell utilized electricity to drive a reaction. In essence, it is the opposite of the galvanic cell.

An electrolytic cell typically does not have the anode and cathode directly connected by a conductive medium. Instead, a voltage is applied by hooking each side up to a battery or other voltage source. The current that battery puts out is used to fuel the redox reaction of the cell.

Typically, these cells are used to decompose chemicals, commonly water. In some cases, galvanic cells can be reversed by applying a voltage opposing the one created, a concept used in rechargeable batteries, as will be explored later.

For the electrolysis of water, the cells are:

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

$O_2+4H^++4e^-\rightarrow 2H_2O$

For an overall reaction of:

$2H_2O\rightarrow 2H_2+O_2$

Now that reaction has a $E^\circ_{cell}=-1.23V$. This means that the Gibb’s free energy will be positive, indicating a non-favorable reaction. For this reason, the application of a voltage across the anode and cathode is needed to produce a reaction.

With the ability to determine the potential of electrochemical cells under standard conditions, we need a way of converting them to non-standard conditions. In order to do so, we will use the Nernst Equation, given as:

$E = E^\circ-\frac{RT}{nF}\ln{Q}$

If we assume that the temperature will remain at $25^\circ C$, we can simplify that equation down to:

$E=E^\circ-\frac{0.02569}n\ln Q$

This approach can apply to both a half-reaction and a complete cell. The only difference between the two is that in a half-reaction, disregard the ‘concentration’ of electrons whilst calculating $Q$. For example, in the half reaction for copper, $Cu^{2+}+2e^-\rightarrow Cu_{(s)}$, which has a standard potential of $E^\circ=0.34V$, we would determine $E$ as:

$E = E^\circ-\frac{RT}{nF}\ln{Q}$

Assuming $T=25^\circ C$.

$E=E^\circ-\frac{0.02569}n\ln Q$

$E=0.34-\frac{0.02569}n\ln Q$

$Q = \frac 1{[Cu^{2+}]}$

$E=0.34-\frac {0.02569}n\ln\frac1{[Cu^{2+}]}$

This means that the effectiveness of the cell relies on the concentration of the copper ion in solution. In fact, the concentration can greatly affect the performance of an electrolytic cell.

Example Problem 8.3.1

What is the potential of a galvanic cell made up of a copper electrode in a $4M$ solution of copper sulfate, and a zinc electrode in $0.5M$ zinc sulfate which is kept at $10^\circ C$?

Example Problem 8.3.1

Reviewed replacement: use an explicitly ideal aqueous Zn/Cu galvanic cell at 298.15 K, with a(Zn²⁺)=0.100, a(Cu²⁺)=0.0100, E°_cell=1.100 V, R=8.31446 J mol⁻¹ K⁻¹ and F=96485.3 C mol⁻¹. Calculate E and ΔG for Zn(s)+Cu²⁺(aq)→Zn²⁺(aq)+Cu(s). This replaces the incomplete calcium/chloride electrolysis calculation.