balance these redox equations ### Balancing Redox Reactions#### Acidic ConditionsThe reaction under acidic conditions is written in blue:\[ \text{Cr}_2\text{O}_7^{2-} (\text{aq}) + \text{C}_2\text{O}_4^{2-} (\text{aq}) \rightarrow \text{Cr}^{3+} (\text{aq}) + \text{CO}_2 (\text{g}) \]- **Reactants**: Dichromate ion (\(\text{Cr}_2\text{O}_7^{2-}\)) and oxalate ion (\(\text{C}_2\text{O}_4^{2-}\))- **Products**: Chromium ion (\(\text{Cr}^{3+}\)) and carbon dioxide (\(\text{CO}_2\))This process is denoted as occurring in **acidic** conditions.#### Basic ConditionsThe reaction under basic conditions is written in green:\[ \text{MnO}_4^{-} (\text{aq}) + \text{C}_2\text{O}_4^{2-} (\text{aq}) \rightarrow \text{MnO}_2 (\text{s}) + \text{O}_3^{2-} (\text{aq}) \]- **Reactants**: Permanganate ion (\(\text{MnO}_4^{-}\)) and oxalate ion (\(\text{C}_2\text{O}_4^{2-}\))- **Products**: Manganese dioxide (\(\text{MnO}_2\)) and carbonate ion (\(\text{O}_3^{2-}\))This process is denoted as occurring in **basic** conditions. ### ExplanationBalancing redox reactions requires an understanding of the changes in the oxidation states of the elements involved. The dichromate and permanganate ions are common oxidizing agents, and oxalate is typically a reducing agent. These reactions demonstrate how the conditions (acidic or basic) can affect the products of redox reactions.

a) The reaction given is, => Cr2O72- (aq) + C2O42- (aq) --------> Cr3+ (aq) + CO2 (g) In the…

Answered: 2- 3+ acıdıc 2- basic

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Concept explainers

Question

balance these redox equations

### Balancing Redox Reactions

#### Acidic Conditions

The reaction under acidic conditions is written in blue:

\[ \text{Cr}_2\text{O}_7^{2-} (\text{aq}) + \text{C}_2\text{O}_4^{2-} (\text{aq}) \rightarrow \text{Cr}^{3+} (\text{aq}) + \text{CO}_2 (\text{g}) \]

- **Reactants**: Dichromate ion (\(\text{Cr}_2\text{O}_7^{2-}\)) and oxalate ion (\(\text{C}_2\text{O}_4^{2-}\))
- **Products**: Chromium ion (\(\text{Cr}^{3+}\)) and carbon dioxide (\(\text{CO}_2\))

This process is denoted as occurring in **acidic** conditions.

#### Basic Conditions

The reaction under basic conditions is written in green:

\[ \text{MnO}_4^{-} (\text{aq}) + \text{C}_2\text{O}_4^{2-} (\text{aq}) \rightarrow \text{MnO}_2 (\text{s}) + \text{O}_3^{2-} (\text{aq}) \]

- **Reactants**: Permanganate ion (\(\text{MnO}_4^{-}\)) and oxalate ion (\(\text{C}_2\text{O}_4^{2-}\))
- **Products**: Manganese dioxide (\(\text{MnO}_2\)) and carbonate ion (\(\text{O}_3^{2-}\))

This process is denoted as occurring in **basic** conditions.

### Explanation

Balancing redox reactions requires an understanding of the changes in the oxidation states of the elements involved. The dichromate and permanganate ions are common oxidizing agents, and oxalate is typically a reducing agent. These reactions demonstrate how the conditions (acidic or basic) can affect the products of redox reactions.

Expert Solution

Step 1

a) The reaction given is,

=> Cr₂O₇^2- (aq) + C₂O₄^2- (aq) --------> Cr³⁺ (aq) + CO₂ (g)

In the above reaction, the oxidation state of Cr initially in Cr₂O₇^2- is 6+ (because each O is in 2+ oxidation state and there is 2- charge) and finally its in 3+ oxidation state.

And C is initially in 3+ oxidation state in C₂O₄^2- and finally its in 4+ oxidation state in CO₂

Hence the oxidation of C in C₂O₄^2- and reduction of Cr in Cr₂O₇^2- is taking place in the reaction.

Step 2

Hence the half reactions taking place can be written as,

Oxidation : C₂O₄^2- (aq) -------> CO₂ (g) + e^-

Balancing : Since we have 2 C in LHS. Hence making it 2 in RHS also.

=> C₂O₄^2- (aq) -------> 2 CO₂ (g) + e^-

Now we have 2- charge in LHS. Hence making it 2 - in RHS also.

=> C₂O₄^2- (aq) -------> 2 CO₂ (g) + 2 e^-

Since both number of each elements and the charge is equal in both side of the reaction now.

Hence the reaction is now balanced.

Step 3

Reduction : Cr₂O₇^2- (aq) + e^- ------> Cr³⁺ (aq)

Balancing : Since we have 2 Cr in LHS. Hence making it 2 in RHS also.

=> Cr₂O₇^2- (aq) + e^- ------> 2 Cr³⁺ (aq)

Now we have 7 O in LHS. Hence balancing it in RHS using water.

=> Cr₂O₇^2- (aq) + e^- ------> 2 Cr³⁺ (aq) + 7 H₂O (l)

Now we have 14 H in the RHS. Hence balancing it in LHS using proton because the medium is acidic.

=> Cr₂O₇^2- (aq) + 14 H⁺ (aq) + e^- ------> 2 Cr³⁺ (aq) + 7 H₂O (l)

Since total charge in RHS is 6+. Hence making it 6+ in LHS also.

=> Cr₂O₇^2- (aq) + 14 H⁺ (aq) + 6 e^- ------> 2 Cr³⁺ (aq) + 7 H₂O (l)

Since both number of each elements and the charge is equal in both side of the reaction now.

Hence the reaction is now balanced.

Step 4

Multiplying the oxidation half reaction with 3 and then adding it with reduction half reaction to get overall balanced redox reaction by cancelling 6 electrons as,

=> Cr₂O₇^2- (aq) + 14 H⁺ (aq) + 3 C₂O₄^2- (aq) -------> 6 CO₂ (g) + 2 Cr³⁺ (aq) + 7 H₂O (l)

Step by step

Solved in 8 steps

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions