/*! 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 64 Hydrogen and oxygen can be produ... [FREE SOLUTION] | 91Ó°ÊÓ

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Hydrogen and oxygen can be produced by the electrolysis of water according to $$ 2 \mathrm{H}_{2} \mathrm{O}(l) \stackrel{\text { electrolysis }}{\longrightarrow} 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ Calculate the volume of \(\mathrm{O}_{2}(g)\) produced at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) when a current of \(30.35\) amperes is passed through a \(\mathrm{K}_{2} \mathrm{SO}_{4}(a q)\) solution for \(2.00\) hours.

Short Answer

Expert verified
The volume of \( \mathrm{O}_2(g) \) produced is calculated using the ideal gas law after determining moles from the charge passed.

Step by step solution

01

Calculate Total Charge

Use the formula \( Q = I \times t \) to find the total charge \( Q \) in coulombs, where \( I \) is the current in amperes and \( t \) is the time in seconds. Convert 2.00 hours to seconds (\( 2.00 \times 60 \times 60 \)). Multiply by the current \( 30.35 \) amperes to find \( Q \).
02

Determine Moles of Electrons

Based on Faraday's constant (\( 96485 \) C/mol), calculate the moles of electrons using \( \, n = \frac{Q}{96485} \). Use the charge found in Step 1 to find the number of moles of electrons passed during the reaction.
03

Relate Moles of Electrons to Moles of Oxygen Gas

From the balanced equation \( 2 \mathrm{H}_2\mathrm{O} \rightarrow 2 \mathrm{H}_2 + \mathrm{O}_2 \), it requires 4 moles of electrons to produce 1 mole of \( \mathrm{O}_2 \). Calculate the moles of \( \mathrm{O}_2 \) using the ratio \( \text{{moles of }} \mathrm{O}_2 = \frac{n}{4} \).
04

Calculate Volume of Oxygen at STP

Use the ideal gas law \( PV = nRT \) to calculate the volume \( V \) of \( \mathrm{O}_2(g) \). Use \( P = 1.00 \space \text{atm}, \space T = 298 \space \text{K} \), \( R = 0.0821 \space \text{L}\cdot\text{atm}/\text{mol}\cdot \text{K} \), and the moles of \( \mathrm{O}_2 \) found in Step 3.
05

Solve for Volume

Rearrange the ideal gas law to \( V = \frac{nRT}{P} \) and plug in the known values to solve for the volume \( V \) of \( \mathrm{O}_2 \) produced.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

ideal gas law
The Ideal Gas Law is a fundamental equation in chemistry and physics, which describes the behavior of an ideal gas under various conditions. The equation is given as:\[ PV = nRT \]Here's what each symbol represents:
  • P: Pressure of the gas (in atmospheres, atm).
  • V: Volume of the gas (in liters, L).
  • n: Number of moles of gas.
  • R: The ideal gas constant (approximately 0.0821 L·atm/mol·K).
  • T: Temperature of the gas (in Kelvin, K).
In our exercise, we use the Ideal Gas Law to calculate the volume of oxygen gas produced during electrolysis. First, we calculate the moles of oxygen gas (found from an earlier step in the process using different calculations) and then apply the formula. We rearrange the formula to solve for the volume \( V \):\[ V = \frac{nRT}{P} \]This allows us to determine how much space the oxygen gas will occupy under the stated conditions of temperature and pressure. It's like figuring out how big of a container you would need to hold the gas at a specific pressure and temperature. Remember, temperature needs to be in Kelvin, so always convert Celsius to Kelvin by adding 273.15 to the Celsius temperature.
Faraday's constant
Faraday's Constant is an essential number in electrochemistry, especially when dealing with electrolysis. It represents the charge of one mole of electrons. The value is approximately:\[ 96485 \text{ C/mol} \]This constant allows us to relate electrical charge (which we can measure) to chemical reactions (which involve moles of reactants). In our problem, we use Faraday's Constant to determine how many moles of electrons are involved in the reaction.When you know the total amount of charge in coulombs (calculated as current multiplied by time), you can find out how many moles of electrons were transferred:\[ n = \frac{Q}{F} \]where \(n\) is the number of moles of electrons, \(Q\) is the total charge in coulombs, and \(F\) is Faraday's Constant.This step helps us bridge the gap between the electric current used in electrolysis and the chemical processes occurring in the solution. This particular application is pivotal for predicting the amount of a substance generated at electrodes.
current and charge calculation
Calculating the total charge in a system is crucial for analyzing chemical processes happening during electrolysis. To find total charge, we use the equation:\[ Q = I \times t \]where:
  • Q: Total charge in coulombs (C).
  • I: Current in amperes (A).
  • t: Time in seconds (s).
To ensure accuracy, convert all measurements to proper units. For example, converting hours to seconds involves multiplying by 3600 (since there are 3600 seconds in an hour).In our exercise, we calculated charge by first converting the electrolysis time from hours to seconds — \(2 \times 3600 = 7200 \) seconds. We then multiply by the current, given as 30.35 amperes.This calculation gives us the total amount of charge that has flowed through the system, directly impacting the amount of material formed at the electrodes in an electrolysis process. Understanding this concept helps predict product yield and manage resources efficiently in scientific and industrial applications.

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Most popular questions from this chapter

Suppose the leads of an electrochemical cell are connected together external to the cell and the cell is allowed to come to equilibrium. What will be the value of the cell voltage at equilibrium?

Given that \(E_{\text {cll }}^{\circ}=0.728 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\) for the cell $$ \mathrm{Ag}(s)\left|\mathrm{Br}^{-}(a q)\right| \mathrm{AgBr}(s) \| \mathrm{Ag}^{+}(a q) \mid \mathrm{Ag}(s) $$ write the cell equation and determine the solubilityproduct constant of \(\mathrm{AgBr}(s)\) in water at \(25^{\circ} \mathrm{C}\).

The reaction described by the equation $$ \mathrm{Zn}(s)+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) \leftrightharpoons 2 \mathrm{Hg}(l)+\mathrm{Zn}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ is run in an electrochemical cell. The measured voltage of the cell at \(25^{\circ} \mathrm{C}\) is \(1.03 \mathrm{~V}\) when \(Q=1.00\). Suppose that the measured voltage of the cell at \(25^{\circ} \mathrm{C}\) is \(1.21 \mathrm{~V}\) when \(\left[\mathrm{Cl}^{-}\right]=0.10 \mathrm{M}\) in the cell solution and \(\left[\mathrm{Zn}^{2+}\right]\) is unknown. Calculate the value of \(\left[\mathrm{Zn}^{2 *}\right]\) in the cell solution.

A battery that operates at \(-50^{\circ} \mathrm{C}\) was developed for the exploration of the moon and Mars. The electrodes are magnesium metal-magnesium chloride and silver chloride-silver. The electrolyte is potassium thiocyanate, KSCN, in liquid ammonia. Write the equation for the cell reaction and the cell diagram for the cell.

The deteriorating iron framework inside the Statue of Liberty was replaced with stainless steel as part of a major restoration project. The work was finished in 1986, exactly one hundred years after the statue was first completed. To avoid any electrochemical contact between the metals, the new stainless steel frame and the external copper plates covering the statue were separated using Teflon spacers. The original statue was constructed using asbestos pads as insulating spacers. Apparently, the pads were still able to act as a conductor (in conjunction with moisture and gases from the atmosphere). Why was the iron framework on the interior of the statue most in need of repair and not the copper plating exposed to the atmosphere on the exterior of the statue?

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