/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 119 A student designs an ammeter (a ... [FREE SOLUTION] | 91影视

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Chapter 20: Problem 119

A student designs an ammeter (a device that measures electrical current) that is based on the electrolysis of water into hydrogen and oxygen gases. When electrical current of unknown magnitude is run through the device for \(2.00 \mathrm{~min}, 12.3 \mathrm{~mL}\) of water-saturated \(\mathrm{H}_{2}(g)\) is collected. The temperature of the system is \(25.5^{\circ} \mathrm{C}\), and the atmospheric pressure is 768 torr. What is the magnitude of the current in amperes?

Short Answer

Expert verified
The magnitude of the current used in the ammeter is approximately \(1.60 \mathrm{~A}\).

Step by step solution

01

The Ideal Gas Law states that \(PV = nRT\), where: - \(P\) is the pressure, - \(V\) is the volume, - \(n\) is the number of moles, - \(R\) is the gas constant, and - \(T\) is the temperature in Kelvin. Let's first convert the given temperature to Kelvin and adjust pressure to exclude water vapor pressure to get the partial pressure of \(\mathrm{H}_{2}(g)\). Given temperature \(25.5^{\circ} \mathrm{C}\), we convert to Kelvin as: \(T = 25.5 + 273.15 = 298.65\mathrm{K}\) Water vapor pressure at \(25.5^{\circ} \mathrm{C}\) (available in standard tables) is approximately \(23.8 \mathrm{~torr}\). Hence, partial pressure of \(\mathrm{H}_{2}(g)\): \(P_\mathrm{H_2} = 768 - 23.8 = 744.2 \mathrm{~torr}\). We convert the pressure to atmospheres: \(P_\mathrm{H_2} = \frac{744.2}{760} = 0.9792 \mathrm{~atm}\) Now we have temperature, pressure, and volume of hydrogen gas \((25.5^{\circ} \mathrm{C}, 0.9792 \mathrm{~atm}, 12.3 \mathrm{~mL} )\). We can calculate the number of moles of hydrogen gas using the Ideal Gas Law with \(R = 0.0821 \mathrm{\frac{L\cdot atm}{mol\cdot K}}\): #Step 2: Calculate number of moles of hydrogen gas#

Using the Ideal Gas Law, we have: \(PV = nRT\). Solve for \(n\): \[n = \frac{PV}{RT} = \frac{(0.9792\mathrm{~atm})(12.3\mathrm{~mL})(1\mathrm{~L}/1000\mathrm{~mL})}{(0.0821\mathrm{~\frac{L\cdot atm}{mol\cdot K}})(298.65\mathrm{~K})} = 4.977 \times 10^{-4} \mathrm{~mol}\] #Step 3: Use Faraday's Law to compute total charge during the electrolysis#
02

Faraday's Law of electrolysis states that the amount of substance produced at an electrode is proportional to the total charge passed through the system. In this case, we are interested in the total charge that causes the production of hydrogen gas through the electrolysis of water. The stoichiometry of the reaction is \(\mathrm{2H_{2}O \rightarrow 2H_{2} + O_{2}}\), which means not only 4 moles of electrons are required to produce 2 moles of hydrogen gas. For hydrogen gas production, we apply Faraday's Law as: \[\text{Total charge} = n_\mathrm{H_2} \times F \times 4 e^{-}\] where \(n_\mathrm{H_2}\) is the number of moles of hydrogen gas and \(F\) is Faraday's constant (\(96,485 \mathrm{~ C/mol}\)). Now we can calculate the total charge passed through the system: #Step 4: Calculate total charge passed through the system#

Using the Faraday's law, we get: \[\text{Total charge} = (4.977\times 10^{-4} \mathrm{~mol})(96485 \mathrm{~ \frac{C}{mol}})(4e^{-}) = 191.99\mathrm{~C}\] #Step 5: Determine the electric current using total charge and time taken#
03

To get the magnitude of the current, we use the formula: \[\mathrm{Current} = \frac{\text{Total charge}}{\text{Time}}\] Given the time during which the current flows is \(2.00\mathrm{~min}\), we can convert it to seconds and calculate the current: #Step 6: Calculate current magnitude in amperes#

Using the current formula, we have: \[\mathrm{Current} = \frac{191.99\mathrm{~C}}{2.00\mathrm{~min} \times 60\mathrm{~s/min}} = 1.60\mathrm{~A}\] The magnitude of the current used in the ammeter is approximately \(1.60 \mathrm{~A}\).

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