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Chapter 19: Problem 44

What is Faraday's contribution to quantitative electrolysis?

Short Answer

Expert verified
Faraday's contribution to quantitative electrolysis lies in his two laws of electrolysis. The first law states that the mass of a substance deposited at an electrode is directly proportional to the quantity of electricity that flows through the electrolyte. His second law states that the masses of substances deposited are proportional to their respective chemical equivalent weights when the same quantity of electricity passes through several electrolytic solutions.

Step by step solution

01

Understanding Faraday's laws

Faraday's laws of electrolysis are two scientific laws that Michael Faraday discovered during his work on electrolysis. These laws provide information on the relationship between the quantity of substance produced at an electrode during electrolysis and the quantity of electricity which passes through the solution. They are preservative of the concepts of electrochemistry.
02

Detailing Faraday's First Law

Faraday's First Law states that: The mass of a substance deposited or dissolved at any electrode is directly proportional to the quantity of electricity that flows through the electrolyte (the solution). The law can be mathematically expressed as \( m = ZIt \), where:\n\n- \( m \) is the mass of the substance,\n- \( Z \) is the electrochemical equivalent of the substance,\n- \( I \) is the current, and\n- \( t \) is the time.
03

Detailing Faraday's Second Law

Faraday's Second Law states that: When the same quantity of electricity passes through several electrolytic solutions connected in series, the masses of the substances deposited (or dissolved) are proportional to their respective chemical equivalent weights. Mathematically, this law can be expressed as \( m1/m2 = E1/E2 \), where:\n\n- \( m1 \) and \( m2 \) are the masses of substances deposited (or dissolved), and\n- \( E1 \) and \( E2 \) are the chemical equivalent weights of the substances.

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Most popular questions from this chapter

Consider a galvanic cell consisting of a magnesium electrode in contact with \(1.0 \mathrm{MMg}\left(\mathrm{NO}_{3}\right)_{2}\) and a cadmium electrode in contact with \(1.0 \mathrm{M} \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) Calculate \(E^{\circ}\) for the cell, and draw a diagram showing the cathode, anode, and direction of electron flow.

Calculate the amounts of \(\mathrm{Cu}\) and \(\mathrm{Br}_{2}\) produced in \(1.0 \mathrm{~h}\) at inert electrodes in a solution of \(\mathrm{CuBr}_{2}\) by a current of 4.50 A.

Describe the basic features of a galvanic cell. Why are the two components of the cell separated from each other?

Predict whether the following reactions would occur spontaneously in aqueous solution at \(25^{\circ} \mathrm{C}\). Assume that the initial concentrations of dissolved species are all \(1.0 M\). (a) \(\mathrm{Ca}(s)+\mathrm{Cd}^{2+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{Cd}(s)\) (b) \(2 \mathrm{Br}^{-}(a q)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Br}_{2}(l)+\operatorname{Sn}(s)\) (c) \(2 \mathrm{Ag}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow 2 \mathrm{Ag}^{+}(a q)+\mathrm{Ni}(s)\) (d) \(\mathrm{Cu}^{+}(a q)+\mathrm{Fe}^{3+}(a q) \longrightarrow$$\mathrm{Cu}^{2+}(a q)+\mathrm{Fe}^{2+}(a q)\)

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is present in many plants and vegetables. (a) Balance the following equation in acid solution: $$ \mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2} $$ (b) If a 1.00 -g sample of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) requires \(24.0 \mathrm{~mL}\) of \(0.0100 M \mathrm{KMnO}_{4}\) solution to reach the equivalence point, what is the percent by mass of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the sample?
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