Problem 13 - Chemistry and Chemical Reactivity

Question

For each of the following electrochemical cells, write equations for the oxidation and reduction half-reactions and for the overall reaction. (a) \(\mathrm{Pb}(\mathrm{s})\left|\mathrm{Pb}^{2+}(\mathrm{aq}) \| \mathrm{Sn}^{4+}(\mathrm{aq}), \mathrm{Sn}^{2+}(\mathrm{aq})\right| \mathrm{C}(\mathrm{s})\) (b) \(\mathrm{Hg}(\ell)\left|\mathrm{Hg}_{2} \mathrm{Cl}_{2}(\mathrm{s})\right| \mathrm{Cl}^{-}(\mathrm{aq}) \| \mathrm{Ag}^{+}(\mathrm{aq}) | \mathrm{Ag}(\mathrm{s})\)

Step-by-Step Solution

Answer

(a) Pb + Sn⁴⁺ → Pb²⁺ + Sn²⁺ ; (b) Hg₂Cl₂ + 2Ag⁺ → 2Hg + 2Ag + 2Cl⁻.

Step 1: Identify the components of the cell and their roles for part (a)

For the Pb|Pb²⁺ and Sn⁴⁺|Sn²⁺ cell, we need to identify which species will undergo oxidation and which will undergo reduction. In an electrochemical cell, the anode is where oxidation occurs and the cathode is where reduction occurs. - Anode reaction for Pb/Pb²⁺: Pb(s) → Pb²⁺(aq) + 2e⁻ - Cathode reaction for Sn⁴⁺/Sn²⁺: Sn⁴⁺(aq) + 2e⁻ → Sn²⁺(aq)

Step 2: Write the overall cell reaction for part (a)

To find the overall cell reaction, we combine the oxidation and reduction half-reactions such that the number of electrons in each half-reaction cancels out. - Overall reaction: Pb(s) + Sn⁴⁺(aq) → Pb²⁺(aq) + Sn²⁺(aq)

Step 3: Identify the components of the cell and their roles for part (b)

For the Hg|Hg₂Cl₂|Cl⁻ and Ag⁺|Ag cell, identify the reactions occurring at each electrode. - Anode reaction for Hg/Hg₂Cl₂: Hg₂Cl₂(s) + 2e⁻ → 2Hg(l) + 2Cl⁻(aq) - Cathode reaction for Ag⁺/Ag: Ag⁺(aq) + e⁻ → Ag(s)

Step 4: Write the overall cell reaction for part (b)

Combine the half-reactions for part (b) following the electron transfer. Since the anode reaction involves 2 electrons, and the cathode only 1 electron, multiply the cathode reaction by 2 to balance the electrons. - Overall reaction: Hg₂Cl₂(s) + 2Ag⁺(aq) → 2Hg(l) + 2Ag(s) + 2Cl⁻(aq)

Key Concepts

Oxidation-Reduction Reactions

Electrochemical cells operate based on the principle of oxidation-reduction reactions, commonly known as redox reactions. These reactions involve the transfer of electrons between chemical species, leading to a change in the oxidation state of the substances involved.
In these reactions:

• Oxidation describes the process where a species loses electrons. For example, in the reaction Pb(s) → Pb²⁺(aq) + 2e⁻, lead is oxidized by losing electrons to form lead ions.
• Reduction is the opposite process where a species gains electrons. For instance, in the reaction Sn⁴⁺(aq) + 2e⁻ → Sn²⁺(aq), tin ions are reduced by gaining electrons.

This transfer of electrons is what drives the operation of electrochemical cells, which are essential in generating electric currents.

Half-Reactions

Half-reactions are an essential part of understanding electrochemical processes. Every redox reaction can be separated into two half-reactions: an oxidation half-reaction and a reduction half-reaction. Each of these reactions shows either the loss or gain of electrons.

For example, in our electrochemical cell examples:

• The oxidation half-reaction for lead, Pb(s) → Pb²⁺(aq) + 2e⁻, shows lead losing electrons and thus being oxidized.
• The reduction half-reaction for tin, Sn⁴⁺(aq) + 2e⁻ → Sn²⁺(aq), indicates the gaining of electrons by tin.

By writing equations for each half-reaction, it becomes easier to see the specific changes occurring at each electrode and balance the overall cell reaction. Balancing ensures that the number of electrons lost in oxidation equals the number of electrons gained in reduction, ensuring the reaction is complete.

Cell Reactions

A cell reaction is the combination of both the oxidation and reduction half-reactions to form the overall balanced reaction for the electrochemical cell. This reaction should ensure that the number of electrons released during oxidation equals the number of electrons consumed during reduction.

To achieve this electron balance:

• In example (a), the reactions are combined as Pb(s) + Sn⁴⁺(aq) → Pb²⁺(aq) + Sn²⁺(aq), ensuring that electrons from the lead oxidation are consumed in the tin reduction.
• In example (b), notice that the oxidation half-reaction involves two electrons while the reduction only one. Multiplying the reduction reaction by 2 balances it with the oxidation, leading to: Hg₂Cl₂(s) + 2Ag⁺(aq) → 2Hg(l) + 2Ag(s) + 2Cl⁻(aq).

When combining these reactions, it's crucial to ensure that the stoichiometry matches, reflecting a balanced transfer of electrons between reactants and products.

Electrode Processes

Electrode processes are the reactions occurring at each electrode within an electrochemical cell. These processes are crucial for the conversion of chemical energy into electrical energy. Each electrode has its role in facilitating either oxidation or reduction:

**At the Anode:**

• Oxidation takes place. Electrons are produced here in a reaction such as at the Pb/Pb²⁺ interface where Pb is oxidized, releasing electrons.
• For the Hg/Hg₂Cl₂ cell, the anode involves the oxidation of mercury in Hg₂Cl₂, producing Hg and Cl⁻ ions.

**At the Cathode:**

• Reduction occurs. This electrode serves as the site where electrons are consumed. In the Sn⁴⁺/Sn²⁺ cell, electrons are absorbed by tin ions to reduce Sn⁴⁺ to Sn²⁺.
• Similarly, in the Ag⁺/Ag cell, Ag⁺ ions gain electrons to form solid silver.

The movement of electrons between these electrodes via an external circuit allows for the generation of electrical current, with the ionic movement in the solution maintaining charge balance.