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The electrochemical equivalents of two substances are \(E_{1}\) and \(E_{2}\). The current that must pass to deposit the same amount at the cathodes in the same time must be in the ratio of (a) \(E_{1}: E_{2}\) (b) \(E_{2}: E_{1}\) (c) \(\left(E_{1}-E_{2}\right): E_{2}\) (d) \(E_{1}:\left(E_{2}-E_{1}\right)\)

Short Answer

Expert verified
The ratio of the current that must pass to deposit the same amount at the cathodes in the same time is \(E_{2} : E_{1}\).

Step by step solution

01

Understanding Electrochemical Equivalents

Understand that the electrochemical equivalent (E) of a substance is the mass of that substance deposited by a certain quantity of electricity (usually one coulomb). According to Faraday's laws of electrolysis, the mass (m) of a substance deposited at an electrode is directly proportional to the quantity of electricity (Q) that passes through the solution and to the electrochemical equivalent (E) of the substance. The relationship can be expressed as: \( m = E \times Q \).
02

Applying Faraday's Law to Both Substances

For two substances with electrochemical equivalents of \(E_{1}\) and \(E_{2}\), the same amount of mass \(m\) would be deposited by different amounts of charge \(Q_{1}\) and \(Q_{2}\) respectively, if the amounts are to be the same in the same time. Using the proportionality, the relations would be: \(m = E_{1} \times Q_{1}\) and \(m = E_{2} \times Q_{2}\).
03

Establishing the Ratio of Mass Deposit

Since the same mass \(m\) is deposited by both substances in the same time, the equations derived from Faraday's law are equal to each other: \(E_{1} \times Q_{1} = E_{2} \times Q_{2}\).
04

Deriving the Current Ratio from the Established Relationship

The current (I) is the rate of flow of charge (Q) over time (t), so \(I = Q/t\). We can assume that time \(t\) is the same for both cases, thus giving us \(Q_{1} = I_{1} \times t\) and \(Q_{2} = I_{2} \times t\). Plugging this back into the equation from Step 3, we get \(E_{1} \times I_{1} \times t = E_{2} \times I_{2} \times t\).
05

Solving for the Current Ratio

Canceling out the common factor of time \(t\) from both sides of the equation, we have \(E_{1} \times I_{1} = E_{2} \times I_{2}\). Hence, the required ratio of current that must pass is \(I_{1} : I_{2} = E_{2} : E_{1}\). So the ratio of currents is the inverse of the ratio of electrochemical equivalents.

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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
The foundation of electrochemistry, Faraday's laws of electrolysis, tell us about the relationship between the amount of substance deposited at an electrode during electrolysis and the quantity of electric charge 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 amount of electricity used. In simple terms, the more electrons you push through the circuit, the more atoms will be deposited or dissolved at the electrodes.

Michael Faraday's second law provides a further insight: when the same quantity of electricity is passed through different substances, the mass of substances altered is proportional to their chemical equivalent weights. This means that different materials will have different 'electrochemical equivalents' based on their properties, such as atomic weight and valency, which affect how they react during electrolysis.

To apply these concepts practically, consider an experiment where you're depositing copper and silver out of solution. By understanding Faraday's laws, you can predict how much copper versus silver will be deposited by a given amount of electricity. These laws are not just theoretical; they're essential for industries such as metal plating and the manufacturing of batteries.
Mass Deposit in Electrolysis
Now, it's important to thoroughly understand what 'mass deposit in electrolysis' means. As mentioned in the solution steps provided earlier, the mass of the substance deposited is proportional to its electrochemical equivalent and the quantity of electricity passed through the electrolyte. But what does that really mean for our understanding and applications?

In practical scenarios like gold plating, for instance, a jeweler would need to know how much gold can be deposited on a piece of jewelry within a certain time frame and using a specific electric current. This is where the concept of mass deposit shines – it allows the jeweler to calculate the exact amount of gold needed for the plating procedure. This concept isn't just limited to jewelry making; it's crucial in various electrochemical processes such as electrorefining, where the purity of a metal is increased by electrolysis.

To make this even clearer, imagine you have two metals, A and B, with different electrochemical equivalents. If you pass the same current through solutions of both metals for the same amount of time, more mass of metal A will be deposited if it has a larger electrochemical equivalent. Understanding the nuances of these amounts is key to controlling the thickness and quality of the deposited layer in electroplating applications.
Current Ratio in Electrolysis
When we look at 'current ratio in electrolysis', we are basically examining how the currents needed to deposit equal masses of two substances compare. This is tightly connected to the electrochemical equivalents of these substances. In the context of our earlier example, if you want to deposit the same mass of two different metals, you'll need to adjust the current passing through each solution based on their electrochemical equivalents.

The concept is quite straightforward when you think about it. Substances with higher electrochemical equivalents will need a smaller current to deposit the same mass as a substance with a lower equivalent, given the same amount of time. This understanding comes into play in precision manufacturing and quality control, where consistent and accurate material deposition is the name of the game.

Real-World Example

Consider two metals, let's say silver and lead. If you want to plate an object with equal masses of these metals, the current ratio will depend on the electrochemical equivalents of silver and lead. Using the formula from Faraday's laws, by ensuring you have the correct current ratio, you can achieve the desired mass deposition effectively and efficiently. This principle is pivotal in the field of electroplating, where the aesthetics and functionality often depend on precise and uniform coatings.

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Most popular questions from this chapter

The EMF of the cell: \(\mathrm{Hg}(1) \mid \mathrm{Hg}_{2} \mathrm{Cl}_{2}(\mathrm{~s})\), \(\mathrm{KCl}\) sol. \((1.0 \mathrm{~N}) \mid\) Quinohydrone \(\mid \mathrm{Pt}\), is \(0.210 \mathrm{~V}\) at \(298 \mathrm{~K}\). What is the \(\mathrm{pH}\) of the quinohydrone solution, the potential of the normal calomel electrode is \(0.279 \mathrm{~V}\) and \(E^{\circ}\) for the quinohydrone electrode is \(0.699 \mathrm{~V}\), both at the same temperature. \([2.303 R T / F=0.06]\) (a) \(3.5\) (b) \(7.0\) (c) \(1.85\) (d) \(-3.5\)

What is the equilibrium constant of the reaction: \(2 \mathrm{Fe}^{3+}+\mathrm{Au}^{+} \rightarrow 2 \mathrm{Fe}^{2+}+\mathrm{Au}^{3+} ?\) Given \(E_{\mathrm{Au}^{+} \mid \mathrm{Au}}^{\circ}=1.68 \mathrm{~V}, E_{\mathrm{Au}^{3 \cdot} \mid \mathrm{Au}}^{\circ}=1.50 \mathrm{~V}\), \(E_{\mathrm{Fe}^{3 *} \mid \mathrm{Fe}^{\mathrm{e}}^{\circ}}=0.75 \mathrm{~V}\) and \(2.303 R T / F=0.06\) (a) \(1 \times 10^{22}\) (b) \(1 \times 10^{-22}\) (c) \(1 \times 10^{-11}\) (d) \(1 \times 10^{-72}\)

In an experimental set-up for the measurement of EMF of a half-cell using a reference electrode and a salt bridge, when the salt bridge is removed, the voltage (a) remains the same (b) increases to maximum (c) decreases half the value (d) drops to zero

A Tl'|Tlcouple was prepared by saturating \(0.10 \mathrm{M}-\mathrm{KBr}\) with TIBr and allowing \(\mathrm{Tl}^{+}\) ions form the insoluble bromide to equilibrate. This couple was observed to have a potential \(-0.444 \mathrm{~V}\) with respect to \(\mathrm{Pb}^{2+} \mid \mathrm{Pb}\) couple in which \(\mathrm{Pb}^{2+}\) was \(0.10 \mathrm{M} .\) What is the \(K_{\mathrm{sp}}\) of TIBr. [Given: \(E_{\mathrm{Pb}^{2+} \mid \mathrm{Pb}}^{\circ}=-0.126 \mathrm{~V}, E_{\mathrm{T}^{+} \mid \mathrm{T}}^{o}=-0.336 \mathrm{~V}\) \(\log 2.5=0.4,2.303 R T / F=0.06]\) (a) \(4.0 \times 10^{-6}\) (b) \(2.5 \times 10^{-4}\) (c) \(4.0 \times 10^{-5}\) (d) \(6.3 \times 10^{-3}\)

The EMF of a galvanic cell composed of two hydrogen electrodes is \(272 \mathrm{mV}\). What is the \(\mathrm{pH}\) of the solution in which the anode is immersed if the cathode is in contact with a solution of \(\mathrm{pH}=3\) ? (a) 3 (b) \(6.7\)

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