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For the cell \(\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\), if the concentration of both, \(\mathrm{Zn}^{2+}\) and \(\mathrm{Cu}^{2+}\) ions are doubled, the EMF of the cell (a) doubles (b) reduces to half (c) remains same (d) becomes zero

Short Answer

Expert verified
The EMF of the cell remains the same when the concentration of both Zn^2+ and Cu^2+ ions are doubled.

Step by step solution

01

Understanding the Cell and Nernst Equation

Firstly, it's important to understand that the electromotive force (EMF) for the cell is given by the Nernst equation, which relates the EMF of a cell to the concentrations of the ions involved in the cell. The Nernst equation for a cell at temperature T is given by: \[ E = E^0 - \frac{0.0591}{n} \log\frac{[\text{Products}]}{[\text{Reactants}]} \]where: \( E \) is the cell potential, \( E^0 \) is the standard cell potential, n is the number of moles of electrons transferred in the redox reaction, and [Reactants] and [Products] are the concentrations of the reactants and products, respectively.
02

Applying the Nernst Equation to the Given Cell

For the given cell (Zn|Zn^2+||Cu^2+|Cu), the Nernst equation can be applied separately to each half-cell and then combined to find the overall cell potential. Since we are doubling the concentrations of both Zn^2+ and Cu^2+, according to the Nernst equation, any change in concentration would affect the term involving the logarithm. However, since the concentrations of both the reactants and the products are being doubled, the ratio \( \frac{[\text{Products}]}{[\text{Reactants}]} \) remains unchanged.
03

Analyzing the Effect of Doubled Concentrations on EMF

Given that the ratio of concentrations in the logarithmic term of the Nernst equation does not change when both concentrations are doubled, the log term remains the same, and thus the EMF of the cell does not change. Thus, the change in concentration has no effect on the EMF of the cell as long as the ratio remains constant.

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Key Concepts

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

Electromotive Force (EMF)
When dealing with the concept of electromotive force (EMF), we're talking about the energy provided by a cell or battery that drives the flow of electrons through an external circuit. This isn't just any force; it's the heart of what makes any electrochemical cell tick, a measure of the cell's ability to do work electrically.

In essence, think of EMF as a pressure from behind, nudging electrons forward through the circuit. It's measured in volts (V), which could be seen as similar to water pressure measured in terms of how effectively water is pushed through pipes. However, unlike the name suggests, EMF isn't actually a force but a potential difference between the two terminals of a cell when no current is flowing.

Understanding EMF is crucial because it gives us a glimpse into how much oomph a cell can offer and subsequently how it might power various electrical devices. Imagine you have two batteries; the one with a higher EMF can be thought of as the stronger one, capable of doing more work or powering a device for longer.
Redox Reaction
At the heart of an electrochemical cell lies a redox reaction, an elegant interchange of electrons that gives life to the cell's electromotive force. A redox reaction involves two complementary processes: reduction and oxidation. During reduction, a substance gains electrons, while during oxidation, another substance loses electrons.

A simple way to remember this is the acronym 'OIL RIG': Oxidation Is Loss, Reduction Is Gain. These two half-reactions occur in different parts of the cell, and it's the flow of electrons from one side to the other that generates current.

In the Zn-Cu cell from the exercise, zinc atoms lose electrons (oxidation) and copper ions gain electrons (reduction). This electron handoff is the crux of the process that creates the EMF, with electrons being the currency that facilitates this electrical trade. Understanding redox reactions not only helps in grasping how batteries work but also provides insight into countless processes in both industry and nature, from the rusting of iron to photosynthesis in plants.
Chemical Concentration
Chemical concentration is a measure of how much of a given substance is present in a mixture or solution. In the realm of electrochemistry, the concentrations of reactants and products in a cell's half-reactions are key players determining the cell's potential and EMF.

Imagine a crowd at a concert. If the concentration of fans is high, there's a lot of energy and movement. In similar terms, high concentrations of chemical reactants or products can increase activity within an electrochemical cell. However, it is the ratio of these concentrations that affects the EMF as noted in the Nernst equation.

If you double the number of fans but also double the size of the venue, the density, or ratio, of fans to space remains the same, akin to doubling both the reactant and product concentrations in our cell— the ratio stays constant, and thus, the EMF remains unaffected.
Cell Potential
Cell potential, simply put, is the potential of an electrochemical cell to generate an EMF given a set of conditions, including the chemical concentrations of the reactants and products. It's kind of like a rating of how well the cell performs under those specific circumstances.

Measured in volts, cell potential is influenced by factors such as temperature, pressure, and the nature of the materials in the cell. This concept is crucial because it determines how much power you can get from the cell. It's the difference in potential between two half-cells where the action happens; if they're out of sync, you get less power.

In our example with the Zn-Cu cell, the Nernst equation tells us that even after the concentration of ions is doubled, the potential difference between the cells' halves doesn't get affected, much like increasing the volume of two speakers to the same degree won't change the difference in their loudness—it remains the same. Understanding cell potential helps us engineer better batteries and optimize them for all kinds of uses, from powering smartphones to electric cars.

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Most popular questions from this chapter

Electrolytic cell is used to convert (a) Chemical energy to electrical energy (b) Electrical energy to chemical energy (c) Chemical energy to mechanical energy (d) Electrical energy to mechanical energy

From the following \(E^{\circ}\) values for the half-cells: (i) \(\mathrm{D} \rightarrow \mathrm{D}^{2+}+2 \mathrm{e}^{-} ; E^{\circ}=-1.5 \mathrm{~V}\) (ii) \(\mathrm{B}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{B} ; E^{\circ}=-0.5 \mathrm{~V}\) (iii) \(\mathrm{A}^{3-} \rightarrow \mathrm{A}^{2-}+\mathrm{e}^{-} ; E^{\circ}=1.5 \mathrm{~V}\) (iv) \(\mathrm{C}^{2+}+\mathrm{e}^{-} \rightarrow \mathrm{C}^{+} ; E^{\circ}=+0.5 \mathrm{~V}\) Which combination of two half-cells would result in a cell with largest potential? (a) \(\mathrm{i}\) and iii (b) \(\mathrm{i}\) and iv (c) iii and iv (d) ii and iv

The conductivity of saturated solution of \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(1.2 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1} .\) The limiting equivalent conductivities of \(\mathrm{BaCl}_{2}, \mathrm{~K}_{3} \mathrm{PO}_{4}\) and \(\mathrm{KCl}\) are 160,140 and \(100 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\), respectively. The solubility product of \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\), is (a) \(10^{-5}\) (b) \(1.08 \times 10^{-23}\) (c) \(1.08 \times 10^{-25}\) (d) \(1.08 \times 10^{-27}\)

How does the electrical conductivity of \(20 \mathrm{ml}\) of \(0.2 \mathrm{M}-\mathrm{MgSO}_{4}\) change when \(0.5 \mathrm{M}-\mathrm{Ba}(\mathrm{OH})_{2}\) solution is gradually added in it, to excess? (a) decreases continuously (b) increases continuously (c) increases and then decreases (d) decreases am

How many electrons flow when a current of \(5 \mathrm{~A}\) is passed through a solution for \(200 \mathrm{~s}\) ? (a) \(6.022 \times 10^{23}\) (b) \(6.24 \times 10^{21}\) (c) \(6.024 \times 10^{21}\) (d) \(6.022 \times 10^{20}\)

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