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If two electrolytic cells are placed in series, the same number of electrons must pass through both cells. One student argues that you can get twice as much product if two cells are placed in series compared to a single cell and therefore the cost of production (i.e., the cost of electricity) will decrease greatly and profits will increase. Is the student correct? Explain your reasoning based on the principles of electrochemistry.

Short Answer

Expert verified
The student is incorrect; placing two cells in series won't double the product because the same charge passes through both. The electricity cost and the profits remain unchanged, hence there is no increase in profits from series arrangement of cells.

Step by step solution

01

Understanding Electrolytic Cells

Electrolytic cells use electricity to drive a non-spontaneous chemical reaction. The amount of product formed in an electrolytic cell is directly proportional to the amount of electric charge (number of electrons) that flows through the cell, as described by Faraday's laws of electrolysis. When two cells are placed in series, the same electric current (same number of electrons) flows through both cells.
02

Analyzing the Effect of Series Arrangement on Product Formation

When two electrolytic cells are placed in series, they do not produce twice as much product as a single cell because the same amount of charge passes through both cells. The reaction in each cell will consume exactly the amount of electrons provided, and hence each cell will produce its respective amount of product according to Faraday's laws. The total amount of product produced in both cells combined will be the same as the amount produced in a single cell, given the same amount of charge.
03

Considering the Cost of Electricity and Profit

The cost of electricity is a significant factor in the overall cost of production in electrolytic processes. Since the series arrangement of the cells does not produce more product but requires the same amount of electricity, the cost of production remains unchanged. Therefore, profits will not increase simply by placing cells in series.

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Key Concepts

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

Faraday's Laws of Electrolysis
Understanding Faraday's laws of electrolysis is crucial for anyone interested in the field of electrochemistry. These laws quantitatively relate the amount of substance produced at an electrode during electrolysis to the quantity of electricity passing through the electrolyte.

The first law states that the mass of a substance altered at an electrode during electrolysis is directly proportional to the quantity of electricity used. The mathematical representation of this law is given by the equation \( m = Q \times \frac{M}{nF} \), where \( m \) is the mass of the substance, \( Q \) is the total electric charge, \( M \) is the molar mass of the substance, \( n \) is the valence number of the substance, and \( F \) is Faraday's constant.

The second law states that when the same quantity of electricity passes through different electrolytes, the mass of substances produced at the respective electrodes is directly proportional to their equivalent weights (molar mass divided by valence).

These fundamental insights help us in predicting the amount of substance produced during the process of electrolysis. For instance, when looking at the scenario where two electrolytic cells are placed in series, Faraday's laws imply that despite having two cells, since the same charge passes through both cells, the total mass of product from both cells will not be double compared to a single cell.
Electrochemistry Principles
Electrochemistry principles form the theoretical framework behind the conversion of chemical energy into electrical energy and vice versa. In the context of electrolytic cells, we are dealing with the latter: using electrical energy to drive chemical reactions that are not spontaneous on their own.

The key principles include the understanding of oxidation and reduction reactions (redox), where electrons are transferred between species. Redox reactions are split between two different locations, called the anode and cathode. The anode is where oxidation occurs (loss of electrons), and the cathode is where reduction takes place (gain of electrons).

Electrolytic Cell Mechanism

In an electrolytic cell, an external voltage source applies an electric potential that forces electrons to move in the opposite direction to the spontaneous flow. This causes reduction reactions to occur at the negative electrode (cathode), and oxidation reactions to happen at the positive electrode (anode).

In practice, the cell must be designed considering the nature of the electrolyte, electrodes, and the electrochemical series which predicts the potential at which different elements will be oxidized or reduced. The electrochemistry principles are indispensable in predicting product formation and are directly applied in the exercise, where we evaluate the effect of series arrangement on product yield in electrolytic cells.
Cost of Electricity in Production
The cost of electricity in production processes, especially for those involving electrolysis, is a major component of the total operational expenses. It's important to consider the amount of electrical energy consumed in relation to the yield of the desired product.

Energy costs can be assessed by multiplying the power consumption in kilowatts by the operation time and the cost per kilowatt-hour. In electrolytic processes, the use of electricity is dictated by Faraday's laws: a certain amount of electric charge is needed to produce a certain quantity of product. Even if multiple electrolytic cells are used, if they are set up in series and hence the same charge passes through each, the energy consumed for the total production does not decrease.

There's a misconception that more cells in series might translate to less cost of electricity per product unit. However, as Faraday's laws suggest, increasing the number of cells in series does not increase the total mass of the product. The input energy requirement will remain the same, therefore, without improving the efficiency of the cell or reducing the cost of electricity, profits will not increase.

Accounting for electricity costs in electrolytic production is essential to evaluate the financial feasibility of a process. Engineers and chemists work on optimizing electrolytic cell designs and operational conditions to improve the energy efficiency, thereby reducing costs and potentially increasing profitability.

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Most popular questions from this chapter

Determine whether the reaction: $$ \begin{aligned} \mathrm{S}_{2} \mathrm{O}_{8}^{2-}+\mathrm{Ni}(\mathrm{OH})_{2}+2 \mathrm{OH}^{-} & \longrightarrow \\ 2 \mathrm{SO}_{4}^{2-}+\mathrm{NiO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \end{aligned} $$ will occur spontaneously under standard state conditions. Use \(E_{\text {cell }}^{\circ}\) calculated from the data below to answer the question. $$ \begin{aligned} \mathrm{NiO}_{2}+2 \mathrm{H}_{2} \mathrm{O}+2 e^{-} \rightleftharpoons \mathrm{Ni}(\mathrm{OH})_{2}+2 \mathrm{OH}^{-} \\ E_{\mathrm{NiO}_{2}}^{\circ}=0.49 \mathrm{~V} \\ \mathrm{~S}_{2} \mathrm{O}_{8}^{2-}+2 e^{-} \rightleftharpoons 2 \mathrm{SO}_{4}^{2-} & E_{\mathrm{SO}_{4}^{2-}}^{\circ}=2.01 \mathrm{~V} \end{aligned} $$

Most flashlights use two or more batteries in series. Use the concepts of galvanic cells in this chapter to explain why a flashlight with two new batteries and one "dead" battery will give only a dim light if any light is obtained at all.

Describe the hydrogen electrode. What is the value of its standard reduction potential?

Write the half-reactions and the balanced cell reaction for the following galvanic cells. (a) \(\operatorname{Cd}(s)\left|\mathrm{Cd}^{2+}(a q) \| \mathrm{Au}^{3+}(a q)\right| \mathrm{Au}(s)\) (b) \(\mathrm{Fe}(s)\left|\mathrm{Fe}^{2+}(a q) \| \operatorname{Br}_{2}(a q), \operatorname{Br}^{-}(a q)\right| \operatorname{Pt}(s)\) (c) \(\operatorname{Cr}(s)\left|\mathrm{Cr}^{3+}(a q) \| \mathrm{Cu}^{2+}(a q)\right| \mathrm{Cu}(s)\)

What is the standard cell potential and the net reaction in a galvanic cell that has the following half reactions? $$ \begin{aligned} \mathrm{MnO}_{2}+4 \mathrm{H}^{+}+2 e^{-} \rightleftharpoons \mathrm{Mn}^{2+} &+2 \mathrm{H}_{2} \mathrm{O} \\ & E_{\mathrm{MnO}_{2}}^{\circ}=1.23 \mathrm{~V} \\ \mathrm{PbCl}_{2}+2 e^{-} \rightleftharpoons \mathrm{Pb}+2 \mathrm{Cl}^{-} & E_{\mathrm{PbCl}}^{\mathrm{o}}=-0.27 \mathrm{~V} \end{aligned} $$

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