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Chapter 6: Problem 131

If \(\mathrm{E}_{\mathrm{MnO4}-\mathrm{Mn} 2+}^{\mathrm{o}}=1.51 \mathrm{~V}\) and \(\mathrm{E}^{\circ} \mathrm{MnO}_{2} / \mathrm{Mn}^{2+}=1.23\) \(\mathrm{V}\), the \(\mathrm{E}_{\mathrm{M}=04-\mathrm{Mn} \mathrm{O}_{2}}^{\mathrm{o}}\) is a. \(+1.69 \mathrm{~V}\) b. \(-1.69 \mathrm{~V}\) c. \(-3.38 \mathrm{~V}\) d. \(+0.845 \mathrm{~V}\)

Short Answer

Expert verified
The calculated standard potential is +0.28 V, but it does not match any available options.

Step by step solution

01

Identify the Redox Couples

Recognize the given standard electrode potentials: 1. \( ext{MnO}_4^- / ext{Mn}^{2+} \) with \( E^ ext{o} = 1.51 \, ext{V} \)2. \( ext{MnO}_2 / ext{Mn}^{2+} \) with \( E^ ext{o} = 1.23 \, ext{V} \)
02

Write the Half-Reactions

Write the half-reactions for the given electrode potentials:1. \( ext{MnO}_4^- + 8 ext{H}^+ + 5 ext{e}^- ightarrow ext{Mn}^{2+} + 4 ext{H}_2 ext{O} \)2. \( ext{MnO}_2 + 4 ext{H}^+ + 2 ext{e}^- ightarrow ext{Mn}^{2+} + 2 ext{H}_2 ext{O} \)
03

Determine the Direct Conversion Half-Reaction

We want to find the electrode potential for the direct conversion between the species \( ext{MnO}_4^- \) and \( ext{MnO}_2 \):\( ext{MnO}_4^- + 2 ext{H}^+ + ext{e}^- ightarrow ext{MnO}_2 + ext{H}_2 ext{O} \)
04

Calculate the Standard Potential

Subtract the electrode potential for \( ext{MnO}_2 + 4 ext{H}^+ + 2 ext{e}^- ightarrow ext{Mn}^{2+} \) from \( ext{MnO}_4^- + 8 ext{H}^+ + 5 ext{e}^- ightarrow ext{Mn}^{2+} \):\[E^ ext{o}_{ ext{MnO}_4^- ext{-} ext{MnO}_2} = E^ ext{o}_{ ext{MnO}_4^- ext{-} ext{Mn}^{2+}} - E^ ext{o}_{ ext{MnO}_2 ext{-} ext{Mn}^{2+}}\]Substitute the known values:\[E^ ext{o}_{ ext{MnO}_4^- ext{-} ext{MnO}_2} = 1.51 ext{ V} - 1.23 ext{ V} = 0.28 ext{ V}\]
05

Select the Correct Answer

None of the options exactly match our calculated value of \( +0.28 ext{ V} \). Thus, the given options may need revision or re-evaluation.(There may have been a mistake in the given answers, but based on standard procedure, the calculated answer should be checked against revised options if available.)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Redox Reactions
Redox reactions are fascinating chemical processes involving the transfer of electrons between two substances. It gets its name from the two main components involved: Reduction and Oxidation.
  • Reduction: This is the gain of electrons by a molecule, atom, or ion. It often results in a decrease in oxidation state.
  • Oxidation: This is the loss of electrons, leading to an increase in oxidation state.
Redox reactions are important in various chemical and biological processes, from rusting metals to cellular respiration. In a typical redox reaction, one reactant donates electrons (oxidized) while another accepts electrons (reduced). This electron exchange fundamentally alters the chemical properties of the substances involved, facilitating various processes critical to life and industry.
Standard Electrode Potentials
Standard electrode potentials, often denoted as \( E^o \), indicate the tendency of a chemical species to gain or lose electrons during a redox reaction. These potentials are measured under standard conditions, typically at 25掳C, 1 atm pressure, and 1 M concentration.Standard electrode potentials are crucial in predicting the direction of redox reactions. They serve as a measure of potential difference between species:
  • Positive \( E^o \): Indicates a greater tendency to gain electrons (reduction).
  • Negative \( E^o \): Suggests a propensity to lose electrons (oxidation).
Understanding these potentials helps in determining how easily a redox reaction can proceed. For example, species with higher positive standard electrode potentials will more readily undergo reduction than those with lower potentials.
Half-Reaction Method
The half-reaction method is a systematic approach used to balance redox reactions. By focusing on the separate oxidation and reduction processes, this method breaks down complex reactions into manageable parts.To apply the half-reaction method, follow these steps:
  • **Identify the substances being oxidized and reduced.** Determine which species is losing electrons and which is gaining.
  • **Write the half-reactions.** Separate the overall reaction into two half-reactions, one for oxidation and one for reduction.
  • **Balance each half-reaction individually.** Ensure atoms other than oxygen and hydrogen are balanced, then balance oxygen by adding \( H_2O \), and hydrogen by adding \( H^+ \).
  • **Balance the charge.** Add electrons to one side to equalize the charge, ensuring both half-reactions have the same number of electrons.
  • **Combine the half-reactions.** Sum the balanced half-reactions, ensuring electrons lost and gained cancel each other out.
This technique not only simplifies balancing but also provides deeper insights into the redox processes at play, especially in electrochemical cells.

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