AP Chemistry 9.9-9.10: Nonstandard Cell Potential and Electrolysis

Standard cell potential, $E^\circ_{\text{cell}}$

The cell potential when all species are in their standard states. For a galvanic cell, $E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$, using tabulated reduction potentials for both half-reactions.

Faraday's constant

Faraday's constant is the charge of one mole of electrons: $F=96{,}485\ \text{C mol}^{-1}e^-$. It connects moles of electrons to electrical charge.

Why must the number of transferred electrons, $n$, be determined from the balanced redox equation?

The value of $n$ is needed in electrochemical calculations and represents the number of moles of electrons transferred per mole of the overall reaction as written.

What does the sign of $E^\circ_{\text{cell}}$ indicate about a redox reaction under standard conditions?

If $E^\circ_{\text{cell}}>0$, the forward reaction is spontaneous under standard conditions. If $E^\circ_{\text{cell}}<0$, the forward reaction is nonspontaneous.

Reaction quotient, $Q$

The reaction quotient has the same form as the equilibrium-constant expression but uses the current, nonequilibrium activities or concentrations. For a reaction such as $\text{Co}+\text{Fe}^{2+}\rightarrow\text{Co}^{2+}+\text{Fe}$, $Q=\frac{[\text{Co}^{2+}]}{[\text{Fe}^{2+}]}$ because pure solids are omitted.

How are pure solids and pure liquids treated in an electrochemical reaction quotient?

Pure solids and pure liquids have activity approximately equal to 1, so they are omitted from $Q$. Only dissolved species, gases, and other species whose activity changes are included.

Nernst equation

The Nernst equation gives the cell potential under nonstandard conditions: $E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{RT}{nF}\ln Q$. At $25\ ^\circ\text{C}$, use $E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.0592\ \text{V}}{n}\log Q$.

How does changing $Q$ affect the cell potential under nonstandard conditions?

If $Q>1$, the logarithmic term is positive and $E_{\text{cell}}$ decreases relative to $E^\circ_{\text{cell}}$. If $Q<1$, $E_{\text{cell}}$ increases. At equilibrium, $Q=K$ and $E_{\text{cell}}=0$.

How can the Nernst equation be used to predict spontaneity under nonstandard conditions?

Calculate $Q$ using the specified composition and then find $E_{\text{cell}}$. A positive calculated potential indicates a spontaneous forward reaction, while a negative potential indicates that the reverse reaction is spontaneous.

For the cell $\text{Al}|\text{Al}^{3+}(0.15\ \text{M})||\text{Cu}^{2+}(0.025\ \text{M})|\text{Cu}$, what are $n$ and $Q$?

Al is oxidized and copper(II) is reduced. Balancing electron transfer gives $n=6$, and $Q=\frac{[\text{Al}^{3+}]^2}{[\text{Cu}^{2+}]^3}=\frac{(0.15)^2}{(0.025)^3}=1440$.

For $\text{Co}+\text{Fe}^{2+}\rightarrow\text{Co}^{2+}+\text{Fe}$ with $[\text{Co}^{2+}]=0.15\ \text{M}$ and $[\text{Fe}^{2+}]=1.94\ \text{M}$, what are $Q$ and $E_{\text{cell}}$?

$Q=0.15/1.94=0.077$ and $n=2$. With $E^\circ_{\text{cell}}=-0.17\ \text{V}$, the Nernst equation gives $E_{\text{cell}}\approx-0.14\ \text{V}$, so the reaction remains nonspontaneous.

Concentration cell

A concentration cell contains two half-cells based on the same redox couple but with different concentrations. Since the standard potentials cancel, its voltage arises entirely from the concentration difference.

Why does a concentration cell spontaneously transfer ions from the more concentrated half-cell toward equal concentrations?

The more concentrated side has a greater driving force for reduction of the dissolved species, so it acts as the cathode. Operation lowers the concentration at the cathode and raises it at the anode until the concentrations become equal and $E_{\text{cell}}=0$.

What is the potential of the concentration cell $\text{Zn}|\text{Zn}^{2+}(0.10\ \text{M})||\text{Zn}^{2+}(0.50\ \text{M})|\text{Zn}$ at $25\ ^\circ\text{C}$?

The concentrated side is the cathode, and $Q=0.10/0.50$. With $n=2$ and $E^\circ_{\text{cell}}=0$, $E_{\text{cell}}=-\frac{0.0592}{2}\log(0.10/0.50)=+0.021\ \text{V}$.

What happens to a concentration cell when it reaches equilibrium?

The concentrations of the redox species become equal in the two half-cells, the net driving force disappears, and $E_{\text{cell}}=0$. In the zinc concentration-cell example, both concentrations become $0.30\ \text{M}$.

Galvanic cell

An electrochemical cell in which a spontaneous redox reaction produces electrical energy. Oxidation occurs at the anode, reduction occurs at the cathode, and the cell has a positive potential for the spontaneous direction.

Electrolysis

Electrolysis uses an external power source to drive a redox reaction that is otherwise nonspontaneous. It is used in processes such as metal production, electroplating, ore refinement, and battery recharging.

How do galvanic and electrolytic cells differ in energy conversion and cell potential?

A galvanic cell converts chemical energy into electrical energy through a spontaneous reaction and has $E_{\text{cell}}>0$ for that direction. An electrolytic cell uses electrical energy to force a nonspontaneous reaction, whose written reaction has $E_{\text{cell}}<0$.

Which electrode is the anode and which is the cathode in an electrolytic cell?

Oxidation always occurs at the anode and reduction always occurs at the cathode, regardless of cell type. The electrode signs differ between cell types, but the definitions based on oxidation and reduction do not.

What are the electrode signs in an electrolytic cell?

The anode is positive because it is connected to the positive terminal of the external power source, and the cathode is negative because it is connected to the negative terminal.

How does recharging a rechargeable battery relate to electrolysis?

Discharge is a spontaneous galvanic reaction. Recharging applies external electrical energy to force the reverse, nonspontaneous reaction, partially restoring the original compositions and voltage.

What half-reactions occur during electrolysis of molten sodium chloride?

At the anode, $2\text{Cl}^-\rightarrow\text{Cl}_2+2e^-$. At the cathode, $\text{Na}^++e^-\rightarrow\text{Na}$, giving the net reaction $2\text{Na}^++2\text{Cl}^-\rightarrow2\text{Na}+\text{Cl}_2$.

Why must an external voltage be applied during the electrolysis of molten sodium chloride?

The decomposition of molten sodium chloride is nonspontaneous and has a negative cell potential as written. An applied potential with sufficient magnitude must overcome this thermodynamic barrier and drive the reaction.

What are the electrode reactions and standard cell potential for the electrolysis of water?

The anode reaction is $2\text{H}_2\text{O}\rightarrow\text{O}_2+4\text{H}^++4e^-$, and the cathode reaction is $2\text{H}^++2e^-\rightarrow\text{H}_2$. The net reaction is $2\text{H}_2\text{O}\rightarrow2\text{H}_2+\text{O}_2$ with $E^\circ_{\text{cell}}=-1.229\ \text{V}$.

What minimum thermodynamic voltage is suggested for water electrolysis under standard conditions?

The applied voltage must exceed $1.229\ \text{V}$ in the direction that drives water decomposition. Actual operating voltages may be higher because the standard-potential calculation is only an estimate when conditions are not standard and practical losses occur.

Why does electrolysis of aqueous sodium chloride not produce sodium metal at the cathode?

Although sodium-ion reduction is possible in principle, aqueous conditions favor reduction of water instead. The observed cathode reaction is $2\text{H}_2\text{O}+2e^-\rightarrow\text{H}_2+2\text{OH}^-$ rather than $\text{Na}^++e^-\rightarrow\text{Na}$.

What products can form during electrolysis of aqueous sodium chloride?

At the cathode, water is reduced to produce hydrogen gas and hydroxide ion. At the anode, chloride ion and water can compete for oxidation; under typical conditions both chlorine and oxygen may be produced, and the industrial chlor-alkali process produces chlorine and sodium hydroxide.

Electroplating

Electroplating is the electrolytic deposition of a thin layer of one metal onto a conducting object. The coating can improve corrosion resistance, strength, appearance, or purity.

In silver electroplating, which electrode is the object being coated and what reactions occur?

The object being coated is the cathode, where $\text{Ag}^++e^-\rightarrow\text{Ag}(s)$. A silver anode is oxidized by $\text{Ag}(s)\rightarrow\text{Ag}^++e^-$, transferring silver from the anode to the cathode.

Electric current and charge

Current is the rate of charge flow, with $1\ \text{A}=1\ \text{C s}^{-1}$. For constant current, the charge transferred is $Q=It$.

How can electrical charge be converted into moles of electrons?

Use $n_{e^-}=Q/F=It/F$, where $Q$ is in coulombs, $I$ is in amperes, $t$ is in seconds, and $F=96{,}485\ \text{C mol}^{-1}e^-$. This is the starting point for quantitative electrolysis problems.

What stoichiometric information is essential when solving an electrolysis deposition problem?

The cathode half-reaction determines how many moles of electrons are needed per mole of deposited metal. For $\text{M}^{z+}+ze^-\rightarrow\text{M}$, one mole of metal requires $z$ moles of electrons, so $Q=z(n_{\text{metal}})F$.

How is the mass of metal deposited during electrolysis calculated?

First find moles of electrons using $n_{e^-}=It/F$. Then use the cathode half-reaction to convert moles of electrons to moles of metal, followed by $m=nM$, where $M$ is the metal's molar mass.

What cathode reaction produces aluminum metal from aluminum(III) ions, and why are three moles of electrons required per mole of aluminum?

The reaction is $\text{Al}^{3+}+3e^-\rightarrow\text{Al}(s)$. Three electrons are required because aluminum changes from oxidation state $+3$ to 0.

A current of $25.0\ \text{A}$ passes for $15.0$ minutes through a solution containing $\text{Al}^{3+}$. What mass of aluminum is produced ideally?

The charge is $Q=It=(25.0)(900)=22{,}500\ \text{C}$, giving $0.233\ \text{mol }e^-$. Dividing by 3 and multiplying by aluminum's molar mass gives approximately $2.10\ \text{g Al}$.

How can the time required to electroplate a metal layer be determined from surface area, thickness, density, and current?

Calculate volume from $V=\text{area}\times\text{thickness}$, mass from $m=\rho V$, and moles of metal from $n=m/M$. Convert to required charge using the electron stoichiometry, then calculate time with $t=Q/I$.