An electrochemical cell, in which the reaction Cu+Hg2 SO4 →2 Hg+CuSo4 takes place, is connected

An electrochemical cell, in which the reaction Cu+Hg2 SO4 →2...

An electrochemical cell, in which the reaction $$ \mathrm{Cu}+\mathrm{Hg}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Hg}+\mathrm{CuSo}_{4} $$ takes place, is connected to a motor having a back emf only slightly smaller than the $\mathrm{emf}$ of the cell. The emf of this cell is given by Eq. (2.14), with $\mathcal{S}_{20}=0.3497 \mathrm{~V}$, $\alpha=-6.35 \times 10^{-4} \mathrm{~V} / \mathrm{deg}, \beta=-2.4 \times 10^{-6} \mathrm{~V} / \mathrm{deg}^{2}$, and $\gamma=0 .$ If the cell is kept at a constant temperature of $25^{\circ} \mathrm{C}$ and $0.1 \mathrm{~mol}$ of copper reacts, then how much work is done on the motor?

An electrochemical cell, in which the reaction
$$
\mathrm{Cu}+\mathrm{Hg}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Hg}+\mathrm{CuSo}_{4}
$$
takes place, is connected to a motor having a back emf only slightly smaller than the $\mathrm{emf}$ of the cell. The emf of this cell is given by Eq. (2.14), with $\mathcal{S}_{20}=0.3497 \mathrm{~V}$, $\alpha=-6.35 \times 10^{-4} \mathrm{~V} / \mathrm{deg}, \beta=-2.4 \times 10^{-6} \mathrm{~V} / \mathrm{deg}^{2}$, and $\gamma=0 .$ If the cell is kept at a
constant temperature of $25^{\circ} \mathrm{C}$ and $0.1 \mathrm{~mol}$ of copper reacts, then how much work is done on the motor?

Key Concepts

Electrochemical Cell

An electrochemical cell is a device that converts chemical energy into electrical energy through redox reactions. It consists of electrodes where oxidation and reduction occur, and its ability to do work depends on the potential difference generated by these reactions. This concept is essential for understanding how chemical reactions can drive electrical current in devices such as batteries and fuel cells.

Electromotive Force (emf)

The electromotive force (emf) of an electrochemical cell is the voltage difference between its two electrodes when no current is flowing. It represents the driving force for electron motion through an external circuit. The emf depends on the chemical nature of the electrodes, the reaction conditions, and can be influenced by temperature and concentration through relationships like the Nernst equation.

Temperature Dependence in Electrochemical Cells

The potential of an electrochemical cell can vary with temperature due to the dependence of reaction kinetics and equilibrium constants on temperature. Parameters such as temperature coefficients (e.g., linear and quadratic terms) describe how the emf changes with temperature. Understanding this dependence is crucial for predicting cell performance under different environmental conditions while performing energy conversion.

Faraday's Laws of Electrolysis

Faraday's laws establish the quantitative relationship between the amount of substance transformed in an electrochemical reaction and the total electric charge passed through the cell. The first law states that the mass of a substance altered at an electrode during electrolysis is proportional to the total charge, while the second law relates mass changes between different substances to their equivalent weights. This concept is essential for calculating how much charge—and thus energy—is associated with a given amount of reactant.

Electrical Work and Energy Conversion

Electrical work in an electrochemical cell is the energy transferred when charge moves through an external circuit against a potential difference. The work done is calculated by multiplying the charge by the effective voltage across the device. When connected to a load, such as a motor, the cell’s chemical energy is converted into mechanical work. This process involves accounting for all potential differences in the circuit, including effects like back emf in motors, to determine the net work performed.

Transcript

00:01 So to answer the first question and we needed to know that's what is the formula to determine the cell potential.

00:13 So cell potential is the standard redux potential of cathode minus the reduction potential of anode.

00:28 So we need to figure out which one is cathode and which one is the anode.

00:33 So to do that, we're going to write down two half reactions right here.

00:40 So in fact, to solve this problem, you really need to start from the b, okay? so the copper with two electrons, you form two solid, an iron, a copper solid, and another half reaction is the standard reduction potential for this one.

01:23 Is 0 .34 volt and standard reduction potential for this one is 1 .06 volt.

01:34 Now if you look these two numbers, this one is more positive, so this will be the reaction in cathode and another one will be in the anode.

01:48 So now we figured out which one will be the cathode, which one will be anode, and we plug in into this react equations we will be able to solve this problem.

02:00 So that's be 0 .06 minus 0 .34 volt that will give you 0 .72 volt.

02:11 That will give you 0 .72 volt.

02:11 That will be the cell potentials.

02:15 And these are the standard in the form of redox potentials, but if you really want to write down this half react and you need to reverse one of those.

02:28 So as i say right here, this is a cathode, so which means this reaction must be the reduction, reaction, so this is a reduction form, and this one will be oxidation form, so you really wanted to write as oxidation reaction form, so it should be copper plus, well, it would be copper, become copper two electron so this is the oxidation reduction this is a reduction reaction okay so that is the two half reactions right here okay so that will be the correct answer for bees of course you can write this complete reaction that will be the copper plus bromine that you form copper two ion and bromine ions...

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