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Chapter 19: Problem 54

A constant electric current flows for \(3.75 \mathrm{~h}\) through two electrolytic cells connected in series. One contains a solution of \(\mathrm{AgNO}_{3}\) and the second a solution of \(\mathrm{CuCl}_{2}\). During this time \(2.00 \mathrm{~g}\) of silver are deposited in the first cell. (a) How many grams of copper are deposited in the second cell? (b) What is the current flowing, in amperes?

Short Answer

Expert verified
a) The mass of copper deposited in the second cell is approximately \(1.18 \mathrm{~g}\). b) The current flowing through the cells is approximately \(0.45 \mathrm{~A}\).

Step by step solution

01

Calculate the number of faradays

According to Faraday’s first law, the amount of electricity required to discharge one mole of a substance is known as one Faraday. Hence, the number of Faraday’s (\(F\)) is equal to the mass of the Ag deposited divided by its atomic mass times the number of exchanged electrons, which equals 1 in this case. Thus, we can say that \(F = \frac{2}{107.8682}\)
02

Calculate the mass of copper deposited

The molar mass of copper (\(Cu\)) is approximately 63.5 g/mol. So we can calculate the mass of copper deposited in the second cell by multiplying the number of faradays by the molar mass of \(Cu\) and its charge (which is +2). So we have: \(Mass_{Cu} = 2 * F * 63.5 = 2 * \frac{2}{107.8682} * 63.5\)
03

Conversion of time

The time given is in hours, so this needs to be converted into seconds. As one hour equals 3600 seconds, multiply 3.75 h by 3600 to get the time in seconds.
04

Calculate the current

Finally, we can calculate the current through the cells. Faraday’s law says that the electric charge (\(Q\)) is equal to the current (\(I\)) times the time (\(t\)), which means \(I = \frac{Q}{t}\). The electric charge can be calculated by the formula \(Q = F * 96500\), where 96500 is Faraday's constant, the electric charge in a mole of electrons. Substituting \(Q\) into the formula, we get \(I = \frac{F * 96500}{t}\). Now substitute the values calculated previously into this formula to obtain the current.

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Most popular questions from this chapter

Describe an experiment that would enable you to determine which is the cathode and which is the anode in a galvanic cell using copper and zinc electrodes.

A piece of magnesium ribbon and a copper wire are partially immersed in a \(0.1 \mathrm{M} \mathrm{HCl}\) solution in a beaker. The metals are joined externally by another piece of metal wire. Bubbles are seen to evolve at both the \(\mathrm{Mg}\) and \(\mathrm{Cu}\) surfaces. (a) Write equations representing the reactions occurring at the metals. (b) What visual evidence would you seek to show that \(\mathrm{Cu}\) is not oxidized to \(\mathrm{Cu}^{2+} ?\) (c) At some stage, \(\mathrm{NaOH}\) solution is added to the beaker to neutralize the HCl acid. Upon further addition of \(\mathrm{NaOH},\) a white precipitate forms. What is it?

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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?

Explain why chlorine gas can be prepared by electrolyzing an aqueous solution of \(\mathrm{NaCl}\) but fluorine gas cannot be prepared by electrolyzing an aqueous solution of NaF.
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