Question

Predict whether the following reactions will be spontaneous in acidic solution under standard conditions: (a) oxidation of $Sn$ to $S{n}^{2+}$ by $I_2($ to form I $),$ (b) reduction (a) oxidation of $\mathrm{Sn}$ to ${\mathrm{Sn}}^{2+}$ by ${\mathrm{I}}_2$ $(\text{ to form }\mathrm{I})$; (b) reduction of ${\mathrm{Ni}}^{2+}$ to $\mathrm{Ni}$ by ${\mathrm{I}}^{-}($ to form ${\mathrm{I}}_2),(\mathbf{c})$ reduction of ${\mathrm{Ce}}^{4+}$ to ${\mathrm{Ce}}^{3+}$ by ${\mathrm{H}}_2{\mathrm{O}}_2$ (d) reduction of ${\mathrm{Cu}}^{2+}$ to Cu by ${\mathrm{Sn}}^{2+}($ to form ${\mathrm{Sn}}^{4+})$.

Short answer

In summary, based on the standard cell potential (E°) calculations, we can predict the following reactions in acidic solution under standard conditions: (a) The oxidation of Sn to Sn²⁺ by I₂ to form I is spontaneous (E° = 0.68 V, positive). (b) The reduction of Ni²⁺ to Ni by I⁻ to form I₂ is non-spontaneous (E° = -0.79 V, negative).

Step-by-step solution

Write the half-reactions of the reaction

We can write the half-reactions as follows: Oxidation of Sn: Sn -> Sn²⁺ + 2e⁻ Reduction of I2: I₂ + 2e⁻ -> 2I⁻

Look up the standard reduction potentials for each half-reaction

We can use the standard reduction potentials table to find the values for each half-reaction: E°(Sn²⁺/Sn) = -0.14 V E°(I₂/I⁻) = +0.54 V

Calculate the standard cell potential (E°) for the reaction

To calculate E° for the reaction, we subtract E° of the reduction (oxidation) half-reaction from E° of the oxidation (reduction) half-reaction: E° = E°(I₂/I⁻) - E°(Sn²⁺/Sn) = 0.54 - (-0.14) = 0.68 V Since E° is positive, the reaction is spontaneous in acidic solution under standard conditions. (b) Reduction of Ni²⁺ to Ni by I⁻ to form I₂

Write the half-reactions of the reaction

We can write the half-reactions as follows: Oxidation of I⁻: 2I⁻ -> I₂ + 2e⁻ Reduction of Ni²⁺: Ni²⁺ + 2e⁻ -> Ni

Look up the standard reduction potentials for each half-reaction

We can use the standard reduction potentials table to find the values for each half-reaction: E°(Ni²⁺/Ni) = -0.25 V E°(I₂/I⁻) = +0.54 V

Calculate the standard cell potential (E°) for the reaction

To calculate E° for the reaction, we subtract E° of the reduction (oxidation) half-reaction from E° of the oxidation (reduction) half-reaction: E° = E°(Ni²⁺/Ni) - E°(I₂/I⁻) = -0.25 - 0.54 = -0.79 V Since E° is negative, the reaction is non-spontaneous in acidic solution under standard conditions. We will follow the same process for reactions (c) and (d).

Key concepts

Oxidation-Reduction Reactions

In electrochemistry, oxidation-reduction reactions, also known as redox reactions, involve the transfer of electrons between chemical species. Understanding these reactions is crucial as they are the foundation for many processes, including metabolism and energy production in batteries.

- **Oxidation** refers to the loss of electrons. When a species is oxidized, it becomes more positive. - **Reduction** involves the gain of electrons. When a species is reduced, it becomes more negative.
These reactions always occur in pairs, as electrons lost by one species must be gained by another. Identifying what gets oxidized and what gets reduced is key in solving problems related to electrochemistry.

Standard Reduction Potentials

Standard reduction potentials (E°) are essential for predicting the direction of redox reactions. They provide the tendency of a substance to gain electrons, measured in volts. Tables listing these potentials help in determining whether a reaction will occur spontaneously.

- Higher E° values indicate a greater likelihood of being reduced. - Lower E° values show a tendency to be oxidized.
These potentials are always measured under standard conditions, which are 1 M concentration, 1 atm pressure, and 25°C temperature. By comparing E° values for reactants in a half-reaction, we can predict how electrons will flow.

Spontaneity of Reactions

The spontaneity of a reaction can be determined by calculating the standard cell potential (E° cell) for the redox reaction. A positive E° cell indicates a spontaneous process, while a negative E° cell signifies non-spontaneity.

To find the E° cell:

• Identify the relevant half-reactions and their standard potentials.
• Use the equation: $$E{°}_{\text{cell}}=E{°}_{\text{cathode}}-E{°}_{\text{anode}}$$
• A positive result means the reaction is favorable under standard conditions.

Understanding this concept is crucial for predicting how and where reactions will proceed without external input.

Half-Reactions

Half-reactions are an effective way to represent redox reactions because they separately show the oxidation and reduction processes. This separation helps clarify electron flow and balance both mass and charge.

Each half-reaction includes:

• Electron transfer, with oxidation involving electron loss and reduction involving electron gain.
• A balance of atoms and charges on both sides.

Writing out half-reactions is a foundational skill in electrochemistry, making complex redox reactions easier to comprehend and solve.

Acidic Solutions

Acidic environments influence redox reactions significantly. In acidic solutions, hydrogen ions (H+) are plentiful, which can affect the balance and direction of the reaction.

- **Balancing:** Additional H+ may be required to balance both sides of a redox equation in an acidic medium. - **Influence on E° values:** Some reduction potentials are measured specifically under acidic conditions.
Understanding how to adjust equations for acidic solutions is important when determining the feasibility and spontaneity of a chemical reaction.

Electrochemical Cells

Electrochemical cells are systems that convert chemical energy into electrical energy through redox reactions. They have a wide range of applications from batteries to electrolysis.

There are two main types:

• **Galvanic Cells:** These produce electrical energy naturally through spontaneous reactions. They have a positive E° cell.
• **Electrolytic Cells:** These require external electrical energy to drive non-spontaneous reactions, indicating a negative E° cell.

By understanding how these cells function, you can grasp how we harness and utilize energy in our daily lives.