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Chapter 8: Problem 115

When a solution of dithionite ions \(\left(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\right)\) is added to a solution of chromate ions \(\left(\mathrm{CrO}_{4}^{2-}\right),\) the products of the ensuing chemical reaction that occurs under basic conditions include soluble sulfite ions and solid chromium(III) hydroxide. This reaction is used to remove \(\mathrm{Cr}^{6+}\) from wastewater generated by factories that make chrome- plated metals. a. Write the net ionic equation for this redox reaction. b. Which element is oxidized and which is reduced? c. Identify the oxidizing and reducing agents in the reaction. d. How many grams of sodium dithionite would be needed to remove the \(\mathrm{Cr}^{6+}\) in \(100.0 \mathrm{L}\) of wastewater that contains \(0.00148 M\) chromate ion?

Short Answer

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#tag_content#Answer The net ionic equation for the reaction is: 2 S鈧侽鈧劼测伝 + CrO鈧劼测伝 + 12 H鈧侽 鉄� 2 SO鈧兟测伝 + 3 Cr(OH)鈧� + 12 H鈧侽. The oxidizing agent is the chromate ion (CrO鈧劼测伝) and the reducing agent is the dithionite ion (S鈧侽鈧劼测伝). To remove Cr鈦垛伜 from 100 L of wastewater, 51.54 grams of sodium dithionite are required.

Step by step solution

01

Write the balanced redox reaction equation

First, we need to write the balanced redox reaction equation between dithionite ions (S鈧侽鈧劼测伝) and chromate ions (CrO鈧劼测伝), which involves identifying the oxidation states of sulfur and chromium, and their respective changes during the reaction. The balanced redox reaction equation is as follows: 2 S鈧侽鈧劼测伝 + CrO鈧劼测伝 + 6 H鈧侽 鉄� 2 SO鈧兟测伝 + 3 Cr(OH)鈧�
02

Write the net ionic equation

Now, we can write the net ionic equation for the reaction, which includes only the ions that change during the reaction. 2 S鈧侽鈧劼测伝 + CrO鈧劼测伝 + 12 H鈧侽 鉄� 2 SO鈧兟测伝 + 3 Cr(OH)鈧� + 12 H鈧侽
03

Identify which element is oxidized and which is reduced

By examining the oxidation states of sulfur and chromium in the initial reactants and final products, we can determine which element is oxidized and which is reduced: Sulfur in dithionite ion: S鈧侽鈧劼测伝 鈬� +4 oxidation state. Sulfur in sulfite ion: SO鈧兟测伝 鈬� +4 oxidation state. Since the oxidation state of sulfur does not change during the reaction, it is not involved in the redox process. Chromium in chromate ion: CrO鈧劼测伝 鈬� +6 oxidation state. Chromium in chromium(III) hydroxide: Cr(OH)鈧� 鈬� +3 oxidation state. Chromium is reduced from +6 to +3 during the reaction. Thus, chromium is reduced (gains electrons) and oxygen is oxidized (loses electrons).
04

Identify the oxidizing and reducing agents in the reaction

The oxidizing agent is the species that is reduced (gains electrons), and the reducing agent is the species that is oxidized (loses electrons). In this case, the oxidizing agent is the chromate ion (CrO鈧劼测伝), and the reducing agent is the dithionite ion (S鈧侽鈧劼测伝).
05

Calculate the grams of sodium dithionite needed to remove Cr鈦垛伜 from waste-water

To calculate the amount of sodium dithionite needed, we need to know the stoichiometry of the redox reaction. From the balanced redox equation, we can see that 2 moles of dithionite ions react with 1 mole of chromate ions. Given the volume (100 L) and concentration (0.00148 M) of chromate ions in wastewater: Moles of chromate ions = volume 脳 concentration Moles of chromate ions = 100 L 脳 0.00148 mol/L = 0.148 mol Now, use the stoichiometry to find the moles of sodium dithionite required: Moles of sodium dithionite = 2 脳 moles of chromate ions = 2 脳 0.148 mol = 0.296 mol Finally, convert moles of sodium dithionite to grams: Mass of sodium dithionite = moles 脳 molar mass Mass of sodium dithionite = 0.296 mol 脳 174.11 g/mol = 51.54 g Therefore, 51.54 grams of sodium dithionite are required to remove Cr鈦垛伜 from 100 L of wastewater with a 0.00148 M chromate ion concentration.

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Key Concepts

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

Oxidizing Agents
In a redox reaction, the oxidizing agent plays a key role by accepting electrons from another substance. Recognizing this agent is crucial because it undergoes reduction itself while causing the oxidation of another species. In our example reaction between dithionite ions \(\left(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\right)\) and chromate ions \(\left(\mathrm{CrO}_{4}^{2-}\right)\), the chromate ion acts as the oxidizing agent. While it starts with a +6 oxidation state, it is reduced to a +3 state when ending as chromium(III) hydroxide.
  • Oxidizing agents always gain electrons as a part of a redox process.
  • They cause other substances to lose electrons, thereby undergoing oxidation.
Understanding the role of oxidizing agents helps in designing processes for industrial applications such as removing harmful ions from wastewater.
Reducing Agents
The reducing agent, on the other hand, donates electrons in a redox reaction, becoming oxidized in the process. In the previous example involving dithionite ions and chromate ions, the dithionite ion \(\left(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\right)\) serves as the reducing agent. This is because it helps in reducing the chromate ion while undergoing a chemical change itself.
  • Reducing agents lose electrons, becoming oxidized themselves.
  • They enable oxidizing agents to gain electrons and thus get reduced.
Recognizing the reducing agent is integral for calculating quantities needed for reactions, particularly in wastewater treatment processes.
Balanced Chemical Equations
For every chemical reaction, especially redox reactions, writing a balanced chemical equation is essential. It ensures that both mass and charge are conserved throughout the reaction. The balanced equation for the reaction of dithionite ions and chromate ions under basic conditions can be written as:\[2 \mathrm{S}_{2} \mathrm{O}_{4}^{2-} + \mathrm{CrO}_{4}^{2-} + 6 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{SO}_{3}^{2-} + 3 \mathrm{Cr(OH)}_{3}\]
  • Each side of the equation must have the same number of each type of atom.
  • The charges must also balance, which can often involve adding water \(\mathrm{H}_{2} \mathrm{O}\), protons \(\mathrm{H}^{+}\), or hydroxide ions \(\mathrm{OH}^{-}\) depending on the reaction conditions.
Balancing chemical equations is a foundational skill that aids in correctly predicting the progression and impact of the reactions.
Stoichiometry
Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. By interpreting the balanced chemical equation, it guides calculations of reactant and product amounts. In our method of extracting chromium from wastewater, stoichiometry allows us to precisely calculate how much reducing agent is necessary:The balanced equation shows that 2 moles of dithionite ions react with 1 mole of chromate ions. To find out how many grams of sodium dithionite are needed for treating 100.0 L of wastewater with a 0.00148 M concentration of chromate ions, use the stoichiometry derived from the balanced equation.
  • Calculate moles from concentration: \( \text{Moles of } \mathrm{CrO}_{4}^{2-} = \text{Volume} \times \text{Concentration} = 0.148 \text{ mol} \)
  • Use stoichiometric coefficients to find moles of dithionite required: \(2 \times 0.148 = 0.296 \text{ mol} \)
  • Convert to grams using molar mass: \(0.296 \text{ mol} \times 174.11 \text{ g/mol} = 51.54 \text{ g} \)
With stoichiometry, industrial applications can be scaled up with precision, ensuring effective treatment of each batch of wastewater.

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Most popular questions from this chapter

Sodium fluoride is added to drinking water in many municipalities to protect teeth against cavities. The target of the fluoridation is hydroxyapatite, \(\mathrm{Ca}_{10}\left(\mathrm{PO}_{4}\right)_{6}(\mathrm{OH})_{2},\) a compound in tooth enamel. There is concern, however, that fluoride ions in water may contribute to skeletal fluorosis, an arthritis-like disease. a. Write a net ionic equation for the reaction between hydroxyapatite and sodium fluoride that produces fluorapatite, \(\mathrm{Ca}_{10}\left(\mathrm{PO}_{4}\right)_{6} \mathrm{F}_{2}\) b. The U.S. EPA currently restricts the concentration of \(\mathrm{F}^{-}\) in drinking water to \(4 \mathrm{mg} / \mathrm{L}\). Express this concentration of \(F^{-}\) in molarity. c. One study of skeletal fluorosis suggests that drinking water with a fluoride concentration of \(4 \mathrm{mg} / \mathrm{L}\) for 20 years raises the fluoride content in bone to \(6 \mathrm{mg} / \mathrm{g}\), a level at which a patient may experience stiff joints and other symptoms. How much fluoride (in milligrams) is present in a 100 mg sample of bone with this fluoride concentration?

A sample of crude oil contains \(3.13 \mathrm{m} M\) naphthalene, \(12.0 \mathrm{mM}\) methylnaphthalene, \(23.8 \mathrm{mM}\) dimethylnaphthalene, and \(14.1 \mathrm{m} M\) trimethylnaphthalene. What is the total number of moles of all the naphthalene compounds in \(100.0 \mathrm{mL}\) of the oil?

Determine the oxidation numbers of each of the elements in the following reactions, and identify which of them, if any, are oxidized or reduced: a. \(\mathrm{SiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{4} \mathrm{SiO}_{4}(a q)\) b. \(2 \mathrm{MnCO}_{3}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{MnO}_{2}(s)+2 \mathrm{CO}_{2}(g)\) c. \(3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g)+2 \mathrm{H}^{+}(a q)\)

Iron is oxidized in a number of chemical weathering processes. How many moles of \(\mathrm{O}_{2}\) are consumed when one mole of magnetite (Fe \(_{3} \mathrm{O}_{4}\) ) is converted into hematite \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right) ?\)

Give the formulas of two strong bases and two weak bases.
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