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Chapter 21: Problem 11

Consider the following balanced redox reaction: \(2 \mathrm{CrO}_{2}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+6 \mathrm{ClO}^{-}(a q) \longrightarrow\) \(2 \mathrm{CrO}_{4}^{2-}(a q)+3 \mathrm{Cl}_{2}(g)+4 \mathrm{OH}^{-}(a q)\) (a) Which species is being oxidized? (b) Which species is being reduced? (c) Which species is the oxidizing agent? (d) Which species is the reducing agent? (e) From which species to which does electron transfer occur? (f) Write the balanced molecular equation, with \(\mathrm{Na}^{+}\) as the spectator ion.

Short Answer

Expert verified
(a) CrO₂⁻ (b) ClO⁻ (c) ClO⁻ (d) CrO₂⁻ (e) From CrO₂⁻ to ClO⁻ (f) 2 NaCrO₂(aq) + 2 H₂O(l) + 6 NaClO(aq) → 2 Na₂CrO₄(aq) + 3 Cl₂(g) + 4 NaOH(aq)

Step by step solution

01

- Identify Oxidation States in Reactants and Products

Assign oxidation states to all elements in the reactants and products. For Cr in CrO₂⁻: Cr has an oxidation state of +4. In CrO₄²⁻: Cr has an oxidation state of +6. For Cl in ClO⁻: Cl has an oxidation state of +1. In Cl₂: Cl has an oxidation state of 0.
02

- Determine Oxidation and Reduction

Oxidation is the loss of electrons and increase in oxidation state. Reduction is the gain of electrons and decrease in oxidation state. Cr changes from +4 to +6 (oxidized), Cl changes from +1 to 0 (reduced).
03

- Identify the Oxidizing and Reducing Agents

The species being reduced (gains electrons) is the oxidizing agent. ClO⁻ gains electrons and is reduced, so it is the oxidizing agent. The species being oxidized (loses electrons) is the reducing agent. CrO₂⁻ loses electrons and is oxidized, so it is the reducing agent.
04

- Determine Electron Transfer

Electrons are transferred from the reducing agent to the oxidizing agent. Electrons move from CrO₂⁻ to ClO⁻.
05

- Write the Balanced Molecular Equation with Na⁺

To account for the spectator ion Na⁺, rewrite the balanced equation so that every anionic species is joined with Na⁺: 2 NaCrO₂(aq) + 2 H₂O(l) + 6 NaClO(aq) → 2 Na₂CrO₄(aq) + 3 Cl₂(g) + 4 NaOH(aq)

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Key Concepts

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

Oxidation States
Oxidation states are numerical representations of the oxidation level of an element in a compound.
They help in identifying which species undergoes oxidation and which undergoes reduction in a redox reaction.
By assigning oxidation states, we can track the movement of electrons during a chemical reaction.

Here are some key rules for assigning oxidation states:
  • The oxidation state of an element in its pure form is zero. For example, in \(Cl₂\), chlorine has an oxidation state of 0.
  • For monoatomic ions, the oxidation state is equal to the charge of the ion. For example, the oxidation state of \(Na^+\) is +1.
  • In compounds, fluorine always has an oxidation state of -1.
  • Oxygen usually has an oxidation state of -2, except in peroxides or when bonded to fluorine.
  • Hydrogen generally has an oxidation state of +1 when bonded to nonmetals and -1 when bonded to metals.

In the given reaction:
* Chromium in \(CrO₂^-\) switches from +4 to +6 in \(CrO₄^{2-}\). This indicates oxidation.
* Chlorine in \(ClO^-\) changes from +1 to 0 in \(Cl₂\), indicating reduction.
Identifying these shifts lets us determine which elements gain or lose electrons.
Oxidizing Agent
The oxidizing agent is the species that gains electrons in a redox reaction.
It is always reduced during the reaction.
Knowing the oxidizing agent helps us understand what drives the reaction to occur.

In the given reaction:
  • The species that gains electrons is \(ClO^-\)
  • As \(ClO^-\) gains electrons, its oxidation state decreases from +1 to 0, forming \(Cl₂\).
  • Therefore, \(ClO^-\) is the oxidizing agent in this reaction.


Always remember, the oxidizing agent is reduced to drive the oxidation of another species.
This agent facilitates the overall redox reaction by accepting electrons from another element or compound.
Reducing Agent
The reducing agent is the species that loses electrons in a redox reaction.
It is oxidized in the process.
Understanding the reducing agent helps us trace how the reaction proceeds and how electron transfer occurs.

In the given reaction:
  • The species that loses electrons is \(CrO₂^-\)
  • As \(CrO₂^-\) loses electrons, its oxidation state increases from +4 to +6, forming \(CrO₄^{2-}\).
  • Therefore, \(CrO₂^-\) is the reducing agent in this reaction.


Reduction and oxidation reactions are always coupled.
The reducing agent supplies the electrons that the oxidizing agent accepts.
Together, they enable the reaction to take place by enabling the movement of electrons from one species to another.

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Most popular questions from this chapter

In Appendix \(D,\) standard electrode potentials range from about \(+3 \mathrm{~V}\) to \(-3 \mathrm{~V}\). Thus, it might seem possible to use a half- cell from each end of this range to construct a cell with a voltage of approximately 6 V. However, most commercial aqueous voltaic cells have \(E^{\circ}\) values of \(1.5-2 \mathrm{~V}\). Why are there no aqueous cells with significantly higher potentials?

In an electric power plant, personnel monitor the \(\mathrm{O}_{2}\) content of boiler feed water to prevent corrosion of the boiler tubes. Why does Fe corrode faster in steam and hot water than in cold water?

Car manufacturers are developing engines that use \(\mathrm{H}_{2}\) as fuel. In Iceland, Sweden, and other parts of Scandinavia, where hydroelectric plants produce inexpensive electric power, the \(\mathrm{H}_{2}\) can be made industrially by the electrolysis of water. (a) How many coulombs are needed to produce \(3.5 \times 10^{6} \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas at \(12.0 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C} ?\) (Assume that the ideal gas law applies.) (b) If the coulombs are supplied at \(1.44 \mathrm{~V}\), how many joules are produced? (c) If the combustion of oil yields \(4.0 \times 10^{4} \mathrm{~kJ} / \mathrm{kg},\) what mass of oil must be burned to yield the number of joules in part (b)?

(a) How do the relative magnitudes of \(Q\) and \(K\) relate to the signs of \(\Delta G\) and \(E_{\text {cell }} ?\) Explain. (b) Can a cell do work when \(Q / K>1\) or \(Q / K<1 ?\) Explain. 21.53 A voltaic cell consists of \(\mathrm{A} / \mathrm{A}^{+}\) and \(\mathrm{B} / \mathrm{B}^{+}\) half-cells, where A and \(B\) are metals and the A electrode is negative. The initial \(\left[\mathrm{A}^{+}\right] /\left[\mathrm{B}^{+}\right]\) is such that \(E_{\text {cell }}>E_{\text {cell }}^{\circ}\) (a) How do \(\left[\mathrm{A}^{+}\right]\) and \(\left[\mathrm{B}^{+}\right]\) change as the cell operates? (b) How does \(E_{\text {cell }}\) change as the cell operates? (c) What is \(\left[\mathrm{A}^{+}\right] /\left[\mathrm{B}^{+}\right]\) when \(E_{\text {cell }}=E_{\text {cell }}^{\circ} ?\) Explain. (d) Is it possible for \(E_{\text {cell }}\) to be less than \(E_{\text {cell }}^{\circ} ?\) Explain.

Can one half-reaction in a redox process take place independently of the other? Explain.
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