/*! 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 2 Write balanced equations for the... [FREE SOLUTION] | 91影视

91影视

Write balanced equations for the following half-reactions. Specify whether each is an oxidation or reduction. (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{O}_{2}(\mathrm{g})\) (in acid) (b) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})\) (in acid) (c) \(\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{g})\) (in acid) (d) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{2}(\mathrm{s})\) (in base)

Short Answer

Expert verified
(a) Oxidation, (b) Oxidation, (c) Reduction, (d) Reduction.

Step by step solution

01

Identify Oxidation States

For each reaction, identify the oxidation states of the elements involved in the transformation. This will help determine whether the reaction is an oxidation or a reduction.
02

Balance Oxygen Atoms with Water

Balance the oxygen atoms by adding water molecules (H鈧侽) to the side that has fewer oxygen atoms, according to the needs of each half-reaction.
03

Balance Hydrogen Atoms with Protons

In acidic solutions, balance the hydrogen atoms by adding protons (H鈦) to the side that has fewer hydrogen atoms. This helps in balancing the charge as well as mass.
04

Balance Charges with Electrons

Balance the charges of the half-reaction by adding electrons to either side. Add electrons to the side that has a more positive charge to make the charges equal.
05

Determine Oxidation or Reduction

Determine whether the balanced half-reaction represents oxidation (loss of electrons) or reduction (gain of electrons).
06

Example Solutions: Part (a)

- Reaction: H鈧侽鈧(aq) 鈫 O鈧(g) - Balancing Oxygen: It is already balanced in terms of oxygens. - Balancing Hydrogen: Add 2H鈦 to the right side: H鈧侽鈧 鈫 O鈧 + 2H鈦 - Balancing Charge: Add 2 electrons to the right side: H鈧侽鈧 鈫 O鈧 + 2H鈦 + 2e鈦 - This is an oxidation reaction.
07

Example Solutions: Part (b)

- Reaction: H鈧侰鈧侽鈧(aq) 鈫 CO鈧(g) - Balancing Oxygen: Add two H鈧侽 to the left side: H鈧侰鈧侽鈧 + 2H鈧侽 鈫 2CO鈧 - Balancing Hydrogen: Balance with H鈦: H鈧侰鈧侽鈧 鈫 2CO鈧 + 2H鈦 - Balancing Charge: Add 2 electrons to the right side: H鈧侰鈧侽鈧 鈫 2CO鈧 + 2H鈦 + 2e鈦 - This is an oxidation reaction.
08

Example Solutions: Part (c)

- Reaction: NO鈧冣伝(aq) 鈫 NO(g) - Balancing Oxygen: Add H鈧侽 to the right side: NO鈧冣伝 鈫 NO + 2H鈧侽 - Balancing Hydrogen: Add 4H鈦 to the left side to balance hydrogen: NO鈧冣伝 + 4H鈦 鈫 NO + 2H鈧侽 - Balancing Charge: Add 3 electrons to the left side: NO鈧冣伝 + 4H鈦 + 3e鈦 鈫 NO + 2H鈧侽 - This is a reduction reaction.
09

Example Solutions: Part (d)

- Reaction: MnO鈧勨伝(aq) 鈫 MnO鈧(s) - Balancing Oxygen: Add H鈧侽 to the right side: MnO鈧勨伝 鈫 MnO鈧 + 2H鈧侽 - Balancing Charge and Add OH鈦: Add 4OH鈦 to the left (since it's in basic conditions): MnO鈧勨伝 + 4OH鈦 鈫 MnO鈧 + 2H鈧侽 - Balancing Charge: Add 3e鈦 to the right side: MnO鈧勨伝 + 4OH鈦 鈫 MnO鈧 + 2H鈧侽 + 3e鈦 - This is a reduction reaction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Oxidation States
Understanding oxidation states is crucial in identifying how elements transform in a chemical reaction. They represent the hypothetical charge an atom would have if all bonds to atoms of different elements were completely ionic. In the context of your reactions, this means determining which elements are losing or gaining electrons. To do this, assign oxidation states to each element in a compound, based on standard rules such as:
  • Oxygen typically has an oxidation state of -2.
  • Hydrogen typically has an oxidation state of +1.
  • Elements in their elemental form have an oxidation state of 0.
Identifying these oxidation states allows you to figure out which species are being oxidized and which are being reduced. In oxidation, an element increases its oxidation state (loses electrons), while in reduction, an element decreases its oxidation state (gains electrons). This distinction is essential in accurately balancing half-reactions. By determining the oxidation states, you can decide where electrons are added or removed in subsequent steps.
Electron Transfer
Electron transfer is the heart of redox reactions and involves the movement of electrons from one reactant to another. In the example half-reactions, the movement of electrons defines whether a substance is oxidized or reduced. For oxidation reactions, electrons are products, indicating a loss of electrons (increase in oxidation state). For reduction reactions, electrons are reactants, indicating a gain of electrons (decrease in oxidation state).
Let's illustrate with a simple point:
  • Oxidation (OIL): Oxidation Is Loss of electrons.
  • Reduction (RIG): Reduction Is Gain of electrons.
By "seeing" where electrons go, you balance each half-reaction's overall charge. This balancing is vital, as it ensures that both mass and charge are conserved. Remember: the number of electrons lost in the oxidation process must equal the number gained in the reduction process. This connection keeps the entire redox equation balanced.
Chemical Balancing
Balancing chemical equations involves adjusting the number of atoms and charges on both sides of the reaction. In half-reactions, this means:
  • First, balance atoms other than O and H.
  • Next, balance oxygen atoms by adding water molecules.
  • Then, balance hydrogen atoms using protons (H鈦), appropriate for acidic solutions.
  • Finally, balance the overall charge with electrons.
Let's understand each step fully. When balancing oxygen in an acidic medium, add water (H鈧侽) to the side lacking oxygen. If you add water, it may introduce excess hydrogen, so reciprocally, balance with H鈦 ions. Once masses are balanced, adjust the charge imbalance by introducing electrons to either side as necessary. Recognizing and applying these steps ensures the equation respects the law of conservation of mass and charge.
Acid-Base Reactions
In the context of redox chemistry, half-reactions often occur in acidic or basic environments. This setting greatly influences the balancing process. In acidic conditions, you add H鈦 ions to balance any additional hydrogen atoms introduced when balancing oxygen with water. In basic settings, instead of directly adding H鈦 ions, use OH鈦 ions and additional water to account for these imbalances.
The environment鈥攁cidic or basic鈥攁lters the intermediaries used to achieve chemical balance without affecting electron accounting.
Here's a quick breakdown:
  • Acidic medium: Add H鈦 ions and water as needed to balance hydrogen and oxygen.
  • Basic medium: Use OH鈦 ions and water to balance instead of H鈦.
Understanding the medium is essential for balancing half-reactions correctly and is a core facet of mastering redox equations. Recognizing whether reactions occur in acid or base allows you to predict the correct products and maintain proper chemical stoichiometry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A Balance the following equations involving organic compounds. $$\begin{aligned} &\text { (a) } \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}(\mathrm{aq}) \rightarrow\\\ &&\mathrm{Ag}(\mathrm{s})+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}) \quad \text { (acid solution) } \end{aligned}$$ $$\begin{aligned} &\text { (b) } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) \rightarrow\\\ &&\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}) \quad+\mathrm{Cr}^{3+}(\mathrm{aq}) \quad \text { (acid solution) } \end{aligned}$$

A voltaic cell can be built using the reaction between Al metal and \(\mathrm{O}_{2}\) from the air. If the Al anode of this cell consists of \(84 \mathrm{g}\) of aluminum, how many hours can the cell produce 1.0 A of electricity, assuming an unlimited supply of \(\mathrm{O}_{2} ?\)

Balance the following redox equations. All occur in basic solution. (a) \(\mathrm{Al}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{Al}(\mathrm{OH})_{4}^{-}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\) (b) \(\mathrm{CrO}_{4}^{2-}(\mathrm{aq})+\mathrm{SO}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (c) \(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}(\mathrm{OH})_{2}(\mathrm{s}) \rightarrow\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\) (d) \(\mathrm{HS}^{-}(\mathrm{aq})+\mathrm{ClO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{S}(\mathrm{s})+\mathrm{Cl}^{-}(\mathrm{aq})\)

A "silver coulometer" (Study Question 94) was used in the past to measure the current flowing in an electrochemical cell. Suppose you found that the current flowing through an electrolysis cell deposited \(0.089 \mathrm{g}\) of Ag metal at the cathode after exactly 10 min. If this same current then passed through a cell containing gold(III) ion in the form of \(\left[\mathrm{AuCl}_{4}\right]^{-}\), how much gold was deposited at the cathode in that electrolysis cell?

The standard potential, \(E^{\circ},\) for the reaction of \(\mathrm{Zn}(\mathrm{s})\) and \(\mathrm{Cl}_{2}(\mathrm{g})\) is \(+2.12 \mathrm{V}\). What is the standard free energy change, \(\Delta_{\mathrm{r}} G^{\circ},\) for the reaction?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.