Electrolytic Cells and Electrolysis

Electrolytic Cells And Electrolysis

Electrolytic Cell: An electrolytic cell is an electrochemical cell that uses electrical energy to drive a non-spontaneous redox reaction. This process is called electrolysis.

Components:

Electrodes: Anode (positive, where oxidation occurs) and Cathode (negative, where reduction occurs).
Electrolyte: A molten ionic compound or an aqueous solution containing ions.
External Power Source: A DC power supply (like a battery) that forces electrons to flow and drive the non-spontaneous reaction.

Electrolysis: The process of using an electric current to drive a non-spontaneous chemical reaction. In an electrolytic cell, the external power supply forces electrons onto the cathode, causing reduction, and pulls electrons from the anode, causing oxidation.

Anode and Cathode Sign Convention in Electrolytic Cells:

Anode: The electrode where oxidation occurs. Electrons are pulled away from it by the positive terminal of the power supply, making it the positive electrode.
Cathode: The electrode where reduction occurs. Electrons are pushed onto it by the negative terminal of the power supply, making it the negative electrode.

Contrast with Galvanic Cells: In galvanic cells, the anode is negative (site of oxidation), and the cathode is positive (site of reduction), with the reaction occurring spontaneously to produce electricity. In electrolytic cells, electricity is supplied to force a non-spontaneous reaction.

Quantitative Aspects Of Electrolysis

Faraday's Laws of Electrolysis: These laws describe the quantitative relationship between the amount of electricity passed through an electrolyte and the amount of chemical change produced.

First Law of Electrolysis: The mass of a substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte.

$$m \propto Q$$

Where:

$m$ is the mass of the substance.
$Q$ is the quantity of electricity passed (in Coulombs, C).

Second Law of Electrolysis: The masses of substances deposited or liberated by the same quantity of electricity passing through the electrolyte are proportional to their chemical equivalents.

Combining the Laws (Faraday's Constant):

The quantity of electricity required to deposit one mole of a substance is related to its molar mass and the number of electrons ($n$) involved in the reduction process.

1 Faraday ($F$) = $6.022 \times 10^{23}$ electrons $\times$ (charge of one electron) = $N_A \times e = 96485$ Coulombs (approximately 96500 C).

The mass ($m$) of a substance deposited by passing a charge $Q$ is given by:

$$m = \frac{M \times Q}{n \times F}$$

Where:

$m$ = mass deposited (in grams)
$M$ = molar mass of the substance (in g/mol)
$Q$ = quantity of electricity passed (in Coulombs, $Q = I \times t$, where $I$ is current in Amperes and $t$ is time in seconds)
$n$ = number of electrons transferred per mole of substance in the reduction process (equivalent weight factor)
$F$ = Faraday constant (96485 C/mol)

Equivalent Weight: The equivalent weight of a substance in electrolysis is its molar mass divided by the number of electrons transferred per mole in the relevant half-reaction ($EW = M/n$).

Thus, $m = EW \times \frac{Q}{F}$.

Example: Calculate the mass of copper deposited when 2 Amperes of current are passed through a $CuSO_4$ solution for 30 minutes.

Example 1. Calculate the mass of copper deposited when 2 Amperes of current are passed through a $CuSO_4$ solution for 30 minutes.

Answer:

Step 1: Identify the relevant information.

Current ($I$) = 2 A
Time ($t$) = 30 minutes = $30 \times 60$ seconds = 1800 s
Molar mass of Copper ($Cu$) ($M$) $\approx 63.5$ g/mol
Half-reaction for copper deposition: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$. So, $n=2$.
Faraday constant ($F$) $\approx 96500$ C/mol

Step 2: Calculate the quantity of electricity passed ($Q$).

$Q = I \times t = 2 \text{ A} \times 1800 \text{ s} = 3600$ C

Step 3: Use the formula $m = \frac{M \times Q}{n \times F}$.

$m = \frac{63.5 \text{ g/mol} \times 3600 \text{ C}}{2 \text{ mol}^{-1} \times 96500 \text{ C/mol}}$

$m = \frac{228600}{193000}$

$m \approx 1.18$ g

Approximately 1.18 grams of copper will be deposited.

Products Of Electrolysis

Predicting Products: When an electrolytic cell is set up, the products formed at the electrodes depend on the relative ease of oxidation and reduction of the species present in the electrolyte and at the electrodes.

At the Cathode (Reduction):

The species that is most easily reduced will be reduced. This means the species with the most positive (or least negative) reduction potential will gain electrons.

Consider the possible reductions:

Cations of alkali metals, alkaline earth metals, $Al^{3+}$: These are very difficult to reduce. If present, water is more likely to be reduced.
Reduction of Water: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$ ($E^\circ = -0.83$ V at standard conditions, but depends on pH).
Cations of less reactive metals (e.g., $Zn^{2+}$, $Fe^{2+}$, $Cu^{2+}$, $Ag^+$, $Au^{3+}$): These are more easily reduced than water (if the solution is neutral or acidic).

Example: Electrolysis of aqueous $NaCl$ solution.

Possible cations: $Na^+$ and $H_2O$.
Reduction potentials: $Na^+ + e^- \rightarrow Na$ ($E^\circ = -2.71$ V); $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ ($E^\circ = -0.83$ V).
Since water has a less negative reduction potential, water is reduced at the cathode, producing hydrogen gas and hydroxide ions.

At the Anode (Oxidation):

The species that is most easily oxidized will be oxidized. This means the species with the most negative (or least positive) oxidation potential will lose electrons.

Consider the possible oxidations:

Oxidation of anions: If anions like $Cl^-$, $Br^-$, $I^-$, $S^{2-}$, $SO_4^{2-}$ are present.
Oxidation of Water: $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$ ($E^\circ = +1.23$ V).
Oxidation of Electrode Material: If the anode is made of a reactive metal (e.g., copper), it might oxidize.

Order of Ease of Oxidation:

Generally: $I^- > Br^- > Cl^- > H_2O > SO_4^{2-}$ (Ease of oxidation decreases in this order).
Oxidation of $Cl^-$ to $Cl_2$ ($E^\circ = +1.36$ V) vs. Oxidation of water to $O_2$ ($E^\circ = +1.23$ V). In dilute $NaCl$ solutions, water oxidation is favored. However, in concentrated $Cl^-$ solutions, the overpotential for $O_2$ evolution can make $Cl_2$ formation more favorable.

Example: Electrolysis of molten $NaCl$.

Only $Na^+$ and $Cl^-$ ions are present.
Cathode (Reduction): $Na^+ + e^- \rightarrow Na(l)$
Anode (Oxidation): $2Cl^- \rightarrow Cl_2(g) + 2e^-$

Example: Electrolysis of aqueous $AgNO_3$ solution.

Cathode: $Ag^+(aq) + e^- \rightarrow Ag(s)$ ($Ag^+$ is more easily reduced than water).
Anode: Possible oxidations are $NO_3^-$ and $H_2O$. Water is more easily oxidized than nitrate ions. So, $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$. However, if the anode is made of silver, silver itself will oxidize: $Ag(s) \rightarrow Ag^+(aq) + e^-$. This is often the case in electroplating with silver.

Electrolytic Refining: The process of purifying metals using electrolysis, where impure metal acts as the anode and pure metal acts as the cathode.