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Chapter 16: Problem 565

In an electrolytic cell, a liter of a \(1 \mathrm{M}\) aqueous solution of \(\mathrm{MnO}_{4}^{-}\) is reduced at the cathode. Determine the number of faradays required for each of the following to be made a) a solution that is \(.01 \mathrm{M} \mathrm{MnO}_{4}^{2-}\); b) 1 gram of \(\mathrm{MnO}_{2}\); c) 1 gram-equivalent of Min metal.

Short Answer

Expert verified
The number of faradays required for each case is: a) Approximately \(1.04 \times 10^{-7}\) faradays b) Approximately \(4.76 \times 10^{-6}\) faradays c) Approximately \(1.88 \times 10^{-5}\) faradays

Step by step solution

01

Balance the Reduction Reactions

We first need to find the balanced reduction reactions for each case. In each case, permanganate ions (\(\mathrm{MnO}_{4}^-\)) are reduced to the desired product. a) \(\mathrm{MnO}_{4}^-\) is reduced to \(\mathrm{MnO}_{4}^{2-}\): $$\mathrm{MnO}_{4}^{-} + 2H_2O \rightarrow \mathrm{MnO}_{4}^{2-} + 4H^{+} + e^{-}$$ b) \(\mathrm{MnO}_{4}^-\) is reduced to \(\mathrm{MnO}_{2}\): $$2\mathrm{MnO}_{4}^{-} + 4H_2O + 4e^{-} \rightarrow 2\mathrm{MnO}_{2} + 8H^{+} + 5O^{-}$$ c) \(\mathrm{MnO}_{4}^-\) is reduced to Mn metal: $$2\mathrm{MnO}_{4}^{-} + 16H^{+} + 10e^{-} \rightarrow 2\mathrm{Mn} + 8H_2O$$
02

Calculate Moles of Product

Next, we calculate the moles of the desired product in each case: a) We have 1 liter \(.01 \mathrm{M}\) solution of \(\mathrm{MnO}_{4}^{2-}\), so the moles of \(\mathrm{MnO}_{4}^{2-}\) are: \(moles = M \times volume = 0.01 \times 1 = 0.01 \hspace{2mm} moles\) b) We have 1 gram of \(\mathrm{MnO}_{2}\), so the moles of \(\mathrm{MnO}_{2}\) are: \(moles = \frac{mass}{Molar \hspace{2mm} mass} =\frac{1}{\approx 87} \approx 0.0115 \hspace{2mm} moles\) c) We have 1 gram-equivalent of Mn metal which means it's equal to its molar mass, so the moles of Mn metal are: \(moles = \frac{mass}{Molar \hspace{2mm} mass} =\frac{1}{\approx 55} \approx 0.0182 \hspace{2mm} moles\)
03

Use Faraday's Law of Electrolysis to Calculate Required Faradays

Now, we can use Faraday's Law of Electrolysis to determine the required number of faradays for each case: a) $$required \hspace{1mm} faradays = moles \times \frac{electrons \hspace{2mm} transferred}{Faraday \hspace{1mm} constant} = 0.01 \times \frac{1}{96485} \approx 1.04 \times 10^{-7} \hspace{1mm} faradays$$ b) $$required \hspace{1mm} faradays = moles \times \frac{electrons \hspace{2mm} transferred}{Faraday \hspace{1mm} constant} = 0.0115 \times \frac{4}{96485} \approx 4.76 \times 10^{-6} \hspace{1mm} faradays$$ c) $$required \hspace{1mm} faradays = moles \times \frac{electrons \hspace{2mm} transferred}{Faraday \hspace{1mm} constant} = 0.0182 \times \frac{10}{96485} \approx 1.88 \times 10^{-5} \hspace{1mm} faradays$$ In conclusion, the number of faradays required for each case is: a) Approximately \(1.04 \times 10^{-7}\) faradays b) Approximately \(4.76 \times 10^{-6}\) faradays c) Approximately \(1.88 \times 10^{-5}\) faradays

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Key Concepts

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

Reduction Reactions
In the world of electrochemistry, reduction reactions are key players. Reduction refers to the gain of electrons by a substance. In an electrolytic cell, we have a cathode where reduction occurs. For instance, the reduction of permanganate ions \(\mathrm{MnO}_{4}^{-}\) in an aqueous solution involves accepting electrons and converting into different manganese compounds based on the context and reaction conditions.
For example:
  • In the first reaction, \(\mathrm{MnO}_{4}^{-}\) is reduced to \(\mathrm{MnO}_{4}^{2-}\) by gaining an electron.
  • The second reaction sees \(\mathrm{MnO}_{4}^{-}\) reduced to manganese dioxide \(\mathrm{MnO}_{2}\), involving multiple electron transfers and interactions with water.
  • Lastly, \(\mathrm{MnO}_{4}^{-}\) can be reduced all the way to manganese metal after a series of electron gains.
Each of these processes requires different amounts of electrons and therefore different amounts of electricity or faradays. Thus, understanding these mechanisms provides clarity on the nature and outcome of reduction reactions.
Faraday's Law of Electrolysis
Faraday's law of electrolysis is fundamental in calculating the electric charge needed to drive a chemical reaction within an electrolytic cell. This law states that the amount of chemical change is directly proportional to the quantity of electricity used.
This means you can calculate how many faradays are needed based on the moles of electrons transferred in a reaction.
For example:
  • When forming \(\mathrm{MnO}_{4}^{2-}\), knowing the moles of electrons involved helps determine the faradays required, as was shown in the solution \(0.01 \times \frac{1}{96485} \approx 1.04 \times 10^{-7} \hspace{1mm} \text{faradays}\).
  • Similarly, producing \(\mathrm{MnO}_{2}\) and manganese metal requires calculations for electron transfer, as seen in the solution.
Understanding and applying Faraday's law is essential in predicting the outcome of electrolysis reactions.
Manganese Compounds
Manganese is a versatile element, forming multiple oxidation states and thus various compounds. In this electrolytic cell process, permanganate ion \(\mathrm{MnO}_{4}^{-}\) is a starting material. Upon reduction, it forms intermediate and end products such as \(\mathrm{MnO}_{4}^{2-}\), \(\mathrm{MnO}_{2}\), and even manganese metal.
These transformations are interesting:
  • \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{MnO}_{4}^{2-}\) is an example of mild reduction.
  • \(\mathrm{MnO}_{4}^{-}\) to \(\mathrm{MnO}_{2}\) involves more significant electron transfer compared to the first case.
  • When reduced to manganese metal, it exemplifies complete reduction with maximum electron transfer involved.
The ability of manganese to form such diverse compounds makes it a special focus in redox chemistry, particularly in electrolytic cells.
Molar Mass
Molar mass plays an essential role when dealing with chemical reactions, especially in converting between grams and moles. Molar mass is the weight of one mole of a substance and is usually expressed in grams per mole (g/mol). It is crucial in determining how much material is reacting or required in a chemical process.
In the problem, molar mass is used to calculate moles needed:
  • For 1 gram of \(\mathrm{MnO}_{2}\), the molar mass approximately 87 g/mol is used to find the moles \(\approx 0.0115 \text{ moles}\).
  • When dealing with manganese metal, its molar mass around 55 g/mol helped determine the moles of manganese metal equivalent to 1 gram \(\approx 0.0182 \text{ moles}\).
Understanding molar mass and being able to perform these calculations accurately is imperative for solving such chemical problems efficiently.

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Most popular questions from this chapter

For the following oxidation-reduction reaction, (a) write out the two half- reactions and balance the equation, (b) calculate \(\Delta E^{\circ}\), and (c) determine whether the reaction will proceed spontaneously as written; \(\mathrm{Fe}^{2+}+\mathrm{MnO}^{-}{ }_{4}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (1) \(\mathrm{Fe}^{3+}+\mathrm{e}^{-} \leftrightarrows \mathrm{Fe}^{2+}, \mathrm{E}^{\circ}=0.77 \mathrm{eV}\) (2) \(\mathrm{MnO}^{-}{ }_{4}+8 \mathrm{H}^{+}+6 \mathrm{e}^{-} \leftrightarrows \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}, \mathrm{E}^{\circ}=1.51 \mathrm{eV}\)

Given \(\mathrm{Zn} \rightarrow \mathrm{Zn}^{+2}+2 \mathrm{e}^{-}\) with \(\mathrm{E}^{\circ}=+.763\), calculate \(\mathrm{E}\) for a Zn electrode in which \(\mathrm{Zn}^{+2}=.025 \mathrm{M}\)

For 1000 seconds, a current of \(0.0965\) amp is passed through a \(50 \mathrm{ml}\) of \(0.1 \mathrm{M} \mathrm{NaCl}\). You have only the reduction of \(\mathrm{H}_{2} \mathrm{O}\) to \(\mathrm{H}_{2}\) at the cathode and oxidation of \(\mathrm{Cl}^{-}\) to \(\mathrm{Cl}_{2}\) at the anode. Determine the average concentration of \(\mathrm{OH}^{-}\) in the final solution.

Knowing that the \(\mathrm{K}_{3 p}\) for AgCl is \(1.8 \times 10^{-10}\), calculate \(\mathrm{E}\), the electrode potential, for a silver-silver chloride electrode immersed in \(1 \mathrm{M} \mathrm{KCI}\). The standard oxidation potential for the \(\left(\mathrm{Ag}, \mathrm{Ag}^{+}\right)\) half reaction is \(-0.799\) volts.

The same quantity of electricity was passed through two separate electrolytic cells. The first of these contained a solution of copper sulfate \(\left(\mathrm{CuSO}_{4}\right)\) and exhibited the following cathode reaction (reduction): \(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})\) The second of these contained a solution of silver nitrate \(\left(\mathrm{AgNO}_{3}\right)\) and exhibited the following cathode reaction: \(\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s})\) If \(3.18 \mathrm{~g}\) of Cu were deposited in the first cell, how much (Ag) was deposited in the second cell?
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