Identify the atoms that are oxidized and reduced, the change in oxidation state for each, and the oxidizing and reducing agents in each of the following equations: (a) \(\mathrm{Mg}(s)+\mathrm{NiCl}_{2}(a q) \rightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{Ni}(s)\) (b) \(\mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(s)\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g)\) (e) \(2 \mathrm{K}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(s)+\mathrm{I}_{2}(s) \longrightarrow \mathrm{K}_{2} \mathrm{S}_{4} \mathrm{O}_{6}(s)+2 \mathrm{KI}(s)\) (f) \(3 \mathrm{Cu}(s)+8 \mathrm{HNO}_{3}(a q) \rightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) 21\. Complete and balance the following acid-base equations: (a) HCl gas reacts with solid Ca(OH) \(_{2}(s)\). (b) A solution of \(\operatorname{Sr}(\mathrm{OH})_{2}\) is added to a solution of \(\mathrm{HNO}_{3}\).

Step by step solution

01

- Analyze the Redox Reaction for Mg and NiCl2

To find the oxidized and reduced atoms, compare the oxidation numbers before and after the reaction. Initially, Mg has an oxidation state of 0 and Ni has an oxidation state of +2 (Cl has -1 each). After the reaction, Mg has an oxidation state of +2 (forming MgCl2 where Cl is -1 each) and Ni has an oxidation state of 0. Thus, Mg is oxidized (increasing from 0 to +2), and Ni is reduced (decreasing from +2 to 0). The oxidizing agent is NiCl2 (it receives electrons) and the reducing agent is Mg (it donates electrons).

02

- Analyze the Redox Reaction for PCl3 and Cl2

Originally, P has an oxidation state of +3 (in PCl3, Cl is -1 each) and Cl2 is 0. In PCl5, P has an oxidation state of +5 (Cl is -1 each). Phosphorus is oxidized (from +3 to +5) and chlorine is reduced (from 0 to -1). The oxidizing agent is Cl2 and the reducing agent is PCl3.

03

- Analyze the Redox Reaction for C2H4 and O2

Carbon has an oxidation state of -2 in C2H4 (hydrogen is +1 each) and oxygen is 0 in O2. In CO2, carbon is +4 (oxygen is -2 each). Carbon is oxidized (from -2 to +4) and oxygen in O2 is reduced. The oxidizing agent is O2 and the reducing agent is C2H4.

04

- Analyze the Redox Reaction for Zn and H2SO4

Zinc begins with an oxidation state of 0 and ends with +2 in ZnSO4 (where S is +6 and O is -2 each, hydrogen is +1). Hydrogen starts at +1 in H2SO4 and ends at 0 in H2. Zinc is oxidized and hydrogen is reduced. The oxidizing agent is H2SO4 and the reducing agent is Zn.

05

- Analyze the Redox Reaction for K2S2O3 and I2

In K2S2O3, each S has an average oxidation state of +2 (S2O3 2- overall) and I2 is 0. In K2S4O6, S has an average oxidation state of +2.5 (+10/4) indicating no change in the oxidation state of sulfur, while in KI, iodine has an oxidation state of -1. Iodine is reduced but sulfur is neither oxidized nor reduced. The oxidizing agent is I2 and there is no reducing agent in the conventional sense as sulfur's average oxidation state does not change.

06

- Analyze the Redox Reaction for Cu and HNO3

Copper begins with an oxidation state of 0 and ends with +2 in Cu(NO3)2. Nitrogen starts at +5 in HNO3 and in NO, nitrogen has +2. Copper is oxidized and nitrogen is reduced. The oxidizing agent is HNO3 and the reducing agent is Cu.

07

- Balance the Acid-Base Reaction for HCl and Ca(OH)2

The reaction is a neutralization where HCl reacts with Ca(OH)2 to form CaCl2 and water. Balanced equation: HCl + Ca(OH)2 -> CaCl2 + 2H2O.

08

- Balance the Acid-Base Reaction for Sr(OH)2 and HNO3

Sr(OH)2 is a strong base and HNO3 is a strong acid. They react to form Sr(NO3)2 and water. Balanced equation: Sr(OH)2 + 2HNO3 -> Sr(NO3)2 + 2H2O.

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Key Concepts

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

Oxidation and Reduction

Oxidation and reduction are two types of chemical reactions that often occur simultaneously. In the realm of chemistry, these reactions are intimately connected, and understanding them is crucial for grasping various chemical processes.

In an oxidation reaction, an element loses electrons, which can be viewed as an increase in the oxidation state. On the flip side, reduction involves an element gaining electrons, resulting in a decrease in its oxidation state. These two reactions are collectively known as redox reactions.

To identify which atoms are oxidized or reduced, we look at the change in their oxidation states. In the given example of \( \mathrm{Mg}(s) + \mathrm{NiCl}_{2}(aq) \rightarrow \mathrm{MgCl}_{2}(aq) + \mathrm{Ni}(s) \), magnesium (Mg) starts with an oxidation state of 0 and ends with +2, indicating that Mg has been oxidized. Conversely, nickel (Ni) drops from a +2 oxidation state to 0, showcasing that Ni has been reduced.

Oxidizing and Reducing Agents

Understanding oxidizing and reducing agents is pivotal in deciphering redox reactions. An oxidizing agent, also known as an oxidant, accepts electrons from another substance, thus being reduced. Conversely, a reducing agent, or reductant, donates electrons to another species, becoming oxidized in the process.

For example, in the reaction where \( \mathrm{PCl}_{3}(l) + \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(s) \), \( \mathrm{Cl}_{2} \) acts as the oxidizing agent because it gains electrons and \( \mathrm{PCl}_{3} \) serves as the reducing agent since it loses electrons. These roles are essential for predicting the behavior of substances during chemical reactions and for the purposes of controlling reactions in a lab setting.

Balancing Redox Equations

Balancing redox equations ensures both mass and charge conservation, crucial for accurately predicting reaction outcomes. There are different methods to balance redox equations, but a common one includes dividing the equation into separate oxidation and reduction half-reactions, balancing each for mass and charge, and then combining back into a full equation.

Consider the reaction where \(3 \mathrm{Cu}(s) + 8 \mathrm{HNO}_{3}(aq) \rightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(aq) + 2 \mathrm{NO}(g) + 4 \mathrm{H}_{2} \mathrm{O}(l)\). We must ensure that the number of atoms for each element and the total charge is the same on both sides of the equation. Only then we can consider the equation to be properly balanced. This process may involve adding coefficients, like the '3' in front of \(\mathrm{Cu}\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\), reflecting stoichiometric ratios essential for quantitative analysis in chemistry.

Acid-Base Reactions

Acid-base reactions are another fundamental reaction type in chemistry, typified by the neutralization process where an acid and a base react to form a salt and usually water. In such reactions, acids are proton donors, while bases are proton acceptors.

For instance, when a solution of \(\mathrm{Sr}(\mathrm{OH})_{2}\) is added to \(\mathrm{HNO}_{3}\), you get a reaction producing \(\mathrm{Sr}(\text{NO}_{3})_{2}\) and water, showing the neutralization between \(\mathrm{Sr}(\mathrm{OH})_{2}\), acting as the base, and \(\mathrm{HNO}_{3}\), the acid.

Balancing these reactions involves making sure that the number of hydrogen ions (\(H^{+}\)) equals the number of hydroxide ions (\(OH^{-}\)) produced or consumed, resulting in water molecules, which also must be balanced with the other products of the reaction.