/*! 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 43 In aqueous solution, hydrogen su... [FREE SOLUTION] | 91影视

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In aqueous solution, hydrogen sulfide reduces (a) \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+},(\mathbf{b}) \mathrm{Br}_{2}\) to \(\mathrm{Br}^{-},(\mathbf{c}) \mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+},(\mathbf{d}) \mathrm{HNO}_{3}\) to \(\mathrm{NO}_{2}\) . In all cases, under appropriate conditions, the product is elemental sulfur. Write a balanced net ionic equation for each reaction.

Short Answer

Expert verified
The balanced net ionic equations for the given reactions are: Reaction a: \(2 H_2S(aq) + 2 Fe^{3+}(aq) \to 2 Fe^{2+}(aq) + 3 S(s) + 4 H^+(aq)\) Reaction b: \(H_2S(aq) + 2 Br_2(aq) \to 3 S(s) + 4 Br^-(aq) + 2 H^+(aq)\) Reaction c: \(5 H_2S(aq) + 2 MnO_4^-(aq) + 6 H^+(aq) \to 2 Mn^{2+}(aq) + 8 H_2O(l) + 5 S(s)\) Reaction d: \(3 H_2S(aq) + 4 H^+(aq) \to 3 NO_2(g) + 4 H_2O(l) + 5 S(s)\)

Step by step solution

01

Reaction a - Reduction of Fe鲁鈦 to Fe虏鈦

Step 1 - Unbalanced chemical equation: H鈧係(aq) + Fe鲁鈦(aq) -> Fe虏鈦(aq) + S(s) Step 2 - Balancing the chemical equation: 2 H鈧係(aq) + 2 Fe鲁鈦(aq) -> 2 Fe虏鈦(aq) + 3 S(s) + 4 H鈦(aq) Step 3 - Total ionic equation: 2 H鈧係(aq) + 2 Fe鲁鈦(aq) -> 2 Fe虏鈦(aq) + 3 S(s) + 4 H鈦(aq) Step 4 - Net ionic equation (No spectator ions found in this reaction): 2 H鈧係(aq) + 2 Fe鲁鈦(aq) -> 2 Fe虏鈦(aq) + 3 S(s) + 4 H鈦(aq)
02

Reaction b - Reduction of Br鈧 to Br鈦

Step 1 - Unbalanced chemical equation: H鈧係(aq) + Br鈧(aq) -> S(s) + Br鈦(aq) Step 2 - Balancing the chemical equation: H鈧係(aq) + 2 Br鈧(aq) -> 3 S(s) + 4 Br鈦(aq) + 2 H鈦(aq) Step 3 - Total ionic equation: H鈧係(aq) + 2 Br鈧(aq) -> 3 S(s) + 4 Br鈦(aq) + 2 H鈦(aq) Step 4 - Net ionic equation (No spectator ions found in this reaction): H鈧係(aq) + 2 Br鈧(aq) -> 3 S(s) + 4 Br鈦(aq) + 2 H鈦(aq)
03

Reaction c - Reduction of MnO鈧勨伝 to Mn虏鈦

Step 1 - Unbalanced chemical equation: H鈧係(aq) + MnO鈧勨伝(aq) -> Mn虏鈦(aq) + S(s) Step 2 - Balancing the chemical equation: 5 H鈧係(aq) + 2 MnO鈧勨伝(aq) + 6 H鈦(aq) -> 2 Mn虏鈦(aq) + 8 H鈧侽(l) + 5 S(s) Step 3 - Total ionic equation: 5 H鈧係(aq) + 2 MnO鈧勨伝(aq) + 6 H鈦(aq) -> 2 Mn虏鈦(aq) + 8 H鈧侽(l) + 5 S(s) Step 4 - Net ionic equation (No spectator ions found in this reaction): 5 H鈧係(aq) + 2 MnO鈧勨伝(aq) + 6 H鈦(aq) -> 2 Mn虏鈦(aq) + 8 H鈧侽(l) + 5 S(s)
04

Reaction d - Reduction of HNO鈧 to NO鈧

Step 1 - Unbalanced chemical equation: H鈧係(aq) + HNO鈧(aq) -> NO鈧(g) + S(s) Step 2 - Balancing the chemical equation: 3 H鈧係(aq) + 4 HNO鈧(aq) -> 3 NO鈧(g) + 4 H鈧侽(l) + 5 S(s) Step 3 - Total ionic equation: 3 H鈧係(aq) + 4 H鈦(aq) + 4 NO鈧冣伝(aq) -> 3 NO鈧(g) + 4 H鈧侽(l) + 5 S(s) Step 4 - Net ionic equation (Canceling out spectator ions - 4 NO鈧冣伝): 3 H鈧係(aq) + 4 H鈦(aq) -> 3 NO鈧(g) + 4 H鈧侽(l) + 5 S(s) We have now written balanced net ionic equations for all 4 reactions involving hydrogen sulfide as the reducing agent and producing elemental sulfur.

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Key Concepts

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

Balancing Chemical Equations
Understanding how to balance chemical equations is fundamental in chemistry. It ensures that the law of conservation of mass is obeyed, meaning that atoms are neither created nor destroyed in a chemical reaction. For instance, in the reaction provided, when hydrogen sulfide reduces Fe鲁鈦 to Fe虏鈦, the initial unbalanced equation is H鈧係(aq) + Fe鲁鈦(aq) 鈫 Fe虏鈦(aq) + S(s). To balance it, every atom on the left side of the equation must be accounted for on the right side.

Start by counting the number of each type of atom in the reactants and products. For the reaction mentioned, sulfur and iron must be balanced by ensuring that the number of iron and sulfur atoms in both the reactants and products are equal. Hydrogen atoms are balanced last since they are often found in multiple compounds. After applying stoichiometry, we arrive at a balanced equation of 2 H鈧係(aq) + 2 Fe鲁鈦(aq) 鈫 2 Fe虏鈦(aq) + 3 S(s) + 4 H鈦(aq). This reflects that the mass and the charge are both conserved.

Remember to adjust the coefficients of the compounds accordingly and not to alter the chemical formulas of the reactants or products. Balancing equations allows us to predict the ratios in which chemicals will react and the amounts of substances produced or consumed in a reaction.
Net Ionic Equations
Net ionic equations provide a more straightforward depiction of a chemical reaction by focusing only on the entities that change in the course of the reaction. They exclude spectator ions, which are ions that do not participate in the reaction. The process of writing a net ionic equation involves three steps: writing the full ionic equation, identifying the spectator ions, and removing them to write the net ionic equation.

For instance, when writing the net ionic equation for the reduction of Br鈧 to Br鈦 by hydrogen sulfide, you'll transform the balanced molecular equation, H鈧係(aq) + 2 Br鈧(aq) 鈫 3 S(s) + 4 Br鈦(aq) + 2 H鈦(aq), into a total ionic equation. After that, examine this equation for spectator ions; however, in this particular reaction, there are no spectator ions. Therefore, the net ionic equation is the same as the total ionic equation.

A well-written net ionic equation makes understanding the changes that occur during the reaction easier and helps avoid the complexity that spectator ions can introduce.
Oxidation States
The concept of oxidation states is central to understanding redox reactions, where oxidation and reduction occur simultaneously. An oxidation state is an indicator that suggests how many electrons an atom has lost, gained, or shared when bonding with other atoms. During the process of balancing equations, changes in oxidation states can signal which atoms are oxidized and which are reduced.

For example, in the reaction where MnO鈧勨伝 is reduced to Mn虏鈦, MnO鈧勨伝 has a higher oxidation state which decreases during the reaction, indicating that manganese is being reduced. To find the oxidation state, consider the known oxidation states of other atoms in the compound and the overall charge of the species. Oxygen typically has an oxidation state of -2, which helps in deducing the oxidation state of manganese in MnO鈧勨伝 as +7. This state changes to +2 in Mn虏鈦 upon reduction.

Determining changes in oxidation state is pivotal in writing balanced chemical and net ionic equations for redox processes. Always assign oxidation numbers to all atoms in a reaction and track their changes to identify the oxidized and reduced species properly.

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Most popular questions from this chapter

Indicate whether each of the following statements is true or false (a) \(\mathrm{H}_{2}(g)\) and \(\mathrm{D}_{2}(g)\) are allotropic forms of hydrogen. (b) \(\mathrm{ClF}_{3}\) is an interhalogen compound. (c) MgO(s) is an acidic anhydride. (d) \(\mathrm{SO}_{2}(g)\) is an acidic anhydride. (e) \(2 \mathrm{H}_{3} \mathrm{PO}_{4}(l) \rightarrow \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}(l)+\mathrm{H}_{2} \mathrm{O}(g)\) is an example of a condensation reaction. (f) Tritium is an isotope of the element hydrogen. (g) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\) is an example of a disproportionation reaction.

Write balanced equations for each of the following reactions.(a) When mercury (II) oxide is heated, it decomposes to form \(\mathrm{O}_{2}\) and mercury metal. (b) When copper(II) nitrate is heated strongly, it decomposes to form copper(II) oxide, nitrogen dioxide, and oxygen. (c) Lead(II) sulfide, PbS(s) reacts with ozone to form PbSO \(_{4}(s)\) and \(\mathrm{O}_{2}(g) .\) (d) When heated in air, \(Z n S(s)\) is converted to ZnO. (e) Potassium peroxide reacts with \(\mathrm{CO}_{2}(g)\) to give potassium carboate and \(\mathrm{O}_{2} .(\mathbf{f})\) Oxygen is converted to ozone in the upper atmosphere.

Explain each of the following observations: (a) At room temperature I \(_{2}\) is a solid, Br \(_{2}\) is a liquid, and \(C l_{2}\) and \(F_{2}\) are both gases. (b) \(F_{2}\) cannot be prepared by electrolytic oxidation of aqueous \(\mathrm{F}^{-}\) solutions. (c) The boiling point of HF is much higher than those of the other hydrogen halides. (d) The halogens decrease in oxidizing power in the order \(\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2} .\)

Write the Lewis structure for each of the following species, describe its geometry, and indicate the oxidation state of the nitrogen: (a) HNO \(_{2},(\mathbf{b}) \mathrm{N}_{3},(\mathbf{c}) \mathrm{N}_{2} \mathrm{H}_{5}^{+},(\mathbf{d}) \mathrm{NO}_{3}^{-} .\)

What is the anhydride for each of the following acids: (a) \(\mathrm{H}_{2} \mathrm{SO}_{4},(\mathbf{b}) \mathrm{HClO}_{3},(\mathbf{c}) \mathrm{HNO}_{2},(\mathbf{d}) \mathrm{H}_{2} \mathrm{CO}_{3},(\mathbf{e}) \mathrm{H}_{3} \mathrm{PO}_{4} ?\)

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