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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鈥檚 first law, the amount of electricity required to discharge one mole of a substance is known as one Faraday. Hence, the number of Faraday鈥檚 (\(F\)) 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鈥檚 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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