Question

Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations. Standard reduction potentials are found in Table 18.1. a. \(\mathrm{MnO}_{4}^{-(a q)}+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(a q)+\mathrm{Mn}^{2+}(a q)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{F}^{-}(a q) \longrightarrow \mathrm{F}_{2}(g)+\mathrm{Mn}^{2+}(a q)\)

Short answer

The balanced half-reactions are: a. \( \mathrm{MnO}_{4}^{-}+\mathrm{I}^{-}+8 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{I}_{2}+\mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \) with E° = 0.97 V (spontaneous) b. \( \mathrm{MnO}_{4}^{-}+\mathrm{F}^{-}+8 \mathrm{H}^{+}+9 \mathrm{e}^{-} \rightarrow \mathrm{F}_{2}+\mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \) with E° = -1.36 V (non-spontaneous)

Step-by-step solution

Balance the half-reactions

We will first balance the two given half-reactions in the acidic medium using the half-reaction method: a. MnO4- + I- → I2 + Mn2+ 1. Balance the atoms other than oxygen and hydrogen: As Mn and I are already balanced, move to the next step. 2. Balance the oxygen atoms: Add 4 H2O to the right side. MnO4- + I- → I2 + Mn2+ + 4H2O 3. Balance hydrogen atoms: Add 8 H+ to the left side. MnO4- + I- + 8H+ → I2 + Mn2+ + 4H2O 4. Balance the charge: 2 electrons are added to the left side. MnO4- + I- + 8H+ + 2e- → I2 + Mn2+ + 4H2O b. MnO4- + F- → F2 + Mn2+ 1. Balance the atoms other than oxygen and hydrogen: As Mn and F are already balanced, move to the next step. 2. Balance oxygen atoms: Add 4H2O to the right side. MnO4-+ F- → F2 + Mn2+ + 4H2O 3. Balance hydrogen atoms: Add 8 H+ on the left side. MnO4-+ F- + 8H+ → F2 + Mn2+ + 4H2O 4. Balance the charge: 9 electrons are added to the left side. MnO4-+ F- + 8H+ + 9e- → F2 + Mn2+ + 4H2O

Calculate the standard cell potential for each reaction

We can now calculate the E° using reduction potentials from Table 18.1. For both reactions, we have MnO4- (reduced) and Mn2+ (oxidized), so, the reduction potential for MnO4- is the same (1.51 V). We will calculate E° for both reactions: a. I2/I-: For this reaction, we have E°(I2) = 0.54 V. ∆E°(cell) = E°(cathode) - E°(anode) = E°(MnO4-/Mn2+) - E°(I2/I-) = 1.51 - 0.54 = 0.97 V b. F2/F-: For this reaction, we have E°(F2) = 2.87 V. ∆E°(cell) = E°(cathode) - E°(anode) = E°(MnO4-/Mn2+) - E°(F2/F-) = 1.51 - 2.87 = -1.36 V

Determine if reactions are spontaneous

To determine if a reaction is spontaneous, we will check the sign of E°. If E° is positive, the reaction is spontaneous (as written). a. ∆E°(cell) = 0.97 V; The reaction is spontaneous as E° > 0. b. ∆E°(cell) = -1.36 V; The reaction is not spontaneous as E° < 0.

Summary

The given half-reactions have been balanced, and their standard cell potentials have been calculated using the standard reduction potentials provided in Table 18.1. For reaction (a), the E° is 0.97 V, making it a spontaneous reaction. For reaction (b), the E° is -1.36 V, making it a non-spontaneous reaction.

Key concepts

Standard Reduction Potentials

Standard reduction potentials are essential for understanding electrochemical cells. These potentials, measured in volts, indicate how easily a chemical species will be reduced.
Each substance has a characteristic potential, listed in reduction tables where higher values suggest a greater tendency to gain electrons.
In our exercise, the standard reduction potential for MnO_4^- is 1.51 V, meaning it has a strong ability to gain electrons compared to other elements.
Conversely, the potential for I2 is 0.54 V, and for F2 it is 2.87 V.
This information lets us predict which reactions will occur or how effectively they can proceed.

• Find the potential in the tables
• Identify if higher or lower compared to others
• Apply to specific reactions

Using standard potentials critically evaluates the potential viability and spontaneity of an electrochemical reaction.

Half-Reaction Method

The half-reaction method is a handy tool for balancing redox reactions, which involve transfer of electrons between two species.
First, reactions are split into two parts: one showing oxidation and the other reduction.
For our exercise, MnO4- gains electrons and gets reduced to Mn2+, while I- loses electrons to form I2.

When balancing these:

• Equalize elements other than O and H
• Adjust O by adding H_2O
• Balance H by adding H^+
• Finally, equalize charges using electrons

Applying this method ensures that matter and charge remain consistent in both half-reactions.
This method is systematic and simplifies complex redox equations into manageable steps.

Spontaneity of Reactions

Determining a reaction's spontaneity is key in electrochemistry. It tells us if a reaction can proceed without external energy.
Positive standard cell potential (E^0 > 0) indicates a spontaneous reaction.
In our task, reaction (a) with ∆E°(cell) = 0.97 V shows spontaneity.
Contrarily, a negative potential (E^0 < 0) , like in reaction (b) where ∆E°(cell) = -1.36 V, suggests non-spontaneity.

To assess spontaneity:

• Calculate the cell potential
• Check the sign of E^0
• Positive value means spontaneous
• Negative value means not spontaneous

These calculations and interpretations help foresee the reaction's natural progression.

Balancing Redox Equations

Balancing redox equations accurately exhibits conservation of mass and charge.
In electrochemical processes, it's vital to reflect the equal exchange of electrons between reacting species.
Our exercise involves balancing equations in an acidic medium, which involves careful attention to H^+ and H_2O . Ensuring all aspects balance:

• Start with elements other than O and H
• Balance O with H_2O
• Adjust H with H^+
• Finalize by balancing charge with electrons

This orderly approach helps us create a coherent and accurate representation of the redox reaction.
The benefits of mastering redox balancing include enhanced understanding of the process, ensuring accuracy in calculations, and applying the results practically in various electrochemical applications.