Question

Write a balanced overall reaction from these unbalanced half-reactions. 

 Cu → cu²⁺  

 Ag⁺ → Ag  

balanced overall reaction:  

For a particular redox reaction NO is oxidized to NO₃^-  and Ag⁺ is reduced to Ag. Complete and balance the equation for this reaction in basic solution. Phases are optional. 

Balance the following equation in basic conditions. Phases are optional. 

Answer 1

General guidance

Concepts and reason

A redox reaction can be balanced via half-reaction method. The steps involved in balancing a chemical equation for a redox reaction (in basic solution) are given below:

1.Identify the species that are undergoing oxidation and reduction; separate the reaction into oxidation-half reaction and reduction-half reaction.

2.Balance all the atoms except oxygen and hydrogen.

3.Add protons to balance hydrogens:

4.Add electrons to balance the charge as follows:

5.Scale the half-reactions to make the number of electrons equal in both half-reactions.

6.Cancel the electrons after adding the two reactions.

7.Add ${\\rm{O}}{{\\rm{H}}^ - }$ on both sides of the equation to balance the protons, since the reaction medium is basic.

8.Combine ${{\\rm{H}}^ + }$and ${\\rm{O}}{{\\rm{H}}^ - }$ to form ${{\\rm{H}}_{\\rm{2}}}{\\rm{O}}$and cancel the common terms.

Fundamentals

According to law of conservation of mass \u201cmatter can neither be created nor destroyed\u201d. Therefore, the number of each type of atom on each side of a chemical equation must be the same.

Step-by-step

Step 1 of 3

Part 1

Write the balanced half-reactions as follows:

The balanced overall reaction is,

Part 1

The overall reaction is given below:

Explanation | Common mistakes | Hint for next step

Balance the given redox reaction as follows:

1.Write the half-reactions as follows

2.The oxidation state of copper increases from 0 to $+ 2$; therefore, copper is oxidized by losing two electrons. The oxidation state of Silver decreases from $+ 1$ to 0; therefore, Ag is reduced by accepting an electron. Represent the half-reactions along with the number of electrons associated with them as follows:

3.Balance the electrons in the equations, by multiplying the half-reaction involving silver by two, as follows:

4.Add the two reaction as follows:

5.Cancel the electrons from the equation to get the balanced redox reaction, as follows:

Step 2 of 3

Part 2

Write the balanced overall reaction as given below:

Part 2

The overall balanced equation is given below:

Explanation | Common mistakes | Hint for next step

Balance the given redox reaction via half-reaction method as follows:

1.Write the given redox reaction as follows:

2.Nitrogen of NO is oxidized from $+ 2$ oxidation state to $+ 5$ oxidation state in ${\\rm{NO}}_3^ -$; Ag is reduced by accepting an electron. Separate the half-reactions as follows:

3.All the atoms except oxygen are balanced; therefore, add ${{\\rm{H}}_{\\rm{2}}}{\\rm{O}}$ to balance oxygen.

4.Add protons to balance hydrogens:

5.Add electrons to balance the charge as follows:

6.Multiply the half-reaction involving silver by three, to make the number of electrons equal in both half-reactions.

7.Cancel the electrons after adding the two reactions.

8.Add ${\\rm{O}}{{\\rm{H}}^ - }$ on both sides of the equation to balance the protons, since the reaction medium is basic.

9.Combine ${{\\rm{H}}^ + }$and ${\\rm{O}}{{\\rm{H}}^ - }$ to form ${{\\rm{H}}_{\\rm{2}}}{\\rm{O}}$and cancel the common terms:

Step 3 of 3

Part 3

The balanced redox reaction is as follows:

Part 3

The overall balanced equation is given below:

${\\rm{ 2CoC}}{{\\rm{l}}_2} + {\\rm{N}}{{\\rm{a}}_2}{\\rm{O}}_2^{} + 2{{\\rm{H}}_{\\rm{2}}}{\\rm{O}} + 2{\\rm{O}}{{\\rm{H}}^ - } \\to 2{\\rm{Co}}{\\left( {{\\rm{OH}}} \\right)_{\\rm{3}}} + 2{\\rm{N}}{{\\rm{a}}^ + } + 4{\\rm{C}}{{\\rm{l}}^ - }$

Balance the given redox reaction via half-reaction method as follows:

1.Write the given redox reaction as follows:

2.Cobalt is oxidized, and oxygen is getting reduced in the given redox reaction. ${\\rm{C}}{{\\rm{l}}^ - }$ and ${\\rm{N}}{{\\rm{a}}^ + }$ are spectator ions in the reaction. Omit the spectator ions (for now) and separate ${\\rm{Co}}{\\left( {{\\rm{OH}}} \\right)_{\\rm{3}}}$ to ${\\rm{C}}{{\\rm{o}}^{3 + }}$ and ${\\rm{3O}}{{\\rm{H}}^ - }$ to clearly understand the redox reaction:

${\\rm{C}}{{\\rm{o}}^{2 + }} + {\\rm{O}}_2^{2 - } \\to {\\rm{C}}{{\\rm{o}}^{3 + }} + {\\rm{3O}}{{\\rm{H}}^ - }$

3.Separate the redox reaction into two half-reactions as follows:

$\\begin{array}{l}\\\\{\\rm{C}}{{\\rm{o}}^{2 + }} \\to {\\rm{C}}{{\\rm{o}}^{3 + }}\\\\\\\\{\\rm{O}}_2^{2 - } \\to {\\rm{3O}}{{\\rm{H}}^ - }\\\\\\end{array}$

4.All the atoms except oxygen and hydrogen are balanced; therefore, add ${{\\rm{H}}_{\\rm{2}}}{\\rm{O}}$ to balance oxygen.

$\\begin{array}{l}\\\\{\\rm{C}}{{\\rm{o}}^{2 + }} \\to {\\rm{C}}{{\\rm{o}}^{3 + }}\\\\\\\\{\\rm{O}}_2^{2 - } + {{\\rm{H}}_{\\rm{2}}}{\\rm{O}} \\to {\\rm{3O}}{{\\rm{H}}^ - }\\\\\\end{array}$

5.Add protons to balance hydrogens:

$\\begin{array}{l}\\\\{\\rm{C}}{{\\rm{o}}^{2 + }} \\to {\\rm{C}}{{\\rm{o}}^{3 + }}\\\\\\\\{\\rm{O}}_2^{2 - } + {{\\rm{H}}_{\\rm{2}}}{\\rm{O}} + {{\\rm{H}}^ + } \\to {\\rm{3O}}{{\\rm{H}}^ - }\\\\\\end{array}$

6.Add electrons to balance the charge as follows:

$\\begin{array}{l}\\\\{\\rm{C}}{{\\rm{o}}^{2 + }} \\to {\\rm{C}}{{\\rm{o}}^{3 + }} + {{\\rm{e}}^ - }\\\\\\\\{\\rm{O}}_2^{2 - } + {{\\rm{H}}_{\\rm{2}}}{\\rm{O}} + {{\\rm{H}}^ + } + 2{{\\rm{e}}^ - } \\to {\\rm{3O}}{{\\rm{H}}^ - }\\\\\\end{array}$

7.Multiply the half-reaction involving cobalt by two, to make the number of electron equal in both half-reactions.

$\\begin{array}{l}\\\\{\\rm{2C}}{{\\rm{o}}^{2 + }} \\to 2{\\rm{C}}{{\\rm{o}}^{3 + }} + 2{{\\rm{e}}^ - }\\\\\\\\{\\rm{O}}_2^{2 - } + {{\\rm{H}}_{\\rm{2}}}{\\rm{O}} + {{\\rm{H}}^ + } + 2{{\\rm{e}}^ - } \\to {\\rm{3O}}{{\\rm{H}}^ - }\\\\\\end{array}$

8.Cancel the electrons after adding the two reactions.

${\\rm{ 2C}}{{\\rm{o}}^{2 + }} + {\\rm{O}}_2^{2 - } + {{\\rm{H}}_{\\rm{2}}}{\\rm{O}} + {{\\rm{H}}^ + } \\to 2{\\rm{C}}{{\\rm{o}}^{3 + }} + {\\rm{3O}}{{\\rm{H}}^ - }$

9.Add ${\\rm{O}}{{\\rm{H}}^ - }$ on both sides of the equation to balance the protons, since the reaction

medium is basic. Combine ${{\\rm{H}}^ + }$ and ${\\rm{O}}{{\\rm{H}}^ - }$to form ${{\\rm{H}}_{\\rm{2}}}{\\rm{O}}$.

${\\rm{ 2C}}{{\\rm{o}}^{2 + }} + {\\rm{O}}_2^{2 - } + 2{{\\rm{H}}_{\\rm{2}}}{\\rm{O}} \\to 2{\\rm{C}}{{\\rm{o}}^{3 + }} + 4{\\rm{O}}{{\\rm{H}}^ - }$

10.To combine ${\\rm{C}}{{\\rm{o}}^{3 + }}$ and ${\\rm{O}}{{\\rm{H}}^ - }$ to ${\\rm{Co}}{\\left( {{\\rm{OH}}} \\right)_{\\rm{3}}}$, 6 moles of ${\\rm{O}}{{\\rm{H}}^ - }$are required, since there are 2 moles of ${\\rm{C}}{{\\rm{o}}^{3 + }}$. Add ${\\rm{2O}}{{\\rm{H}}^ - }$to both sides of the reaction and combine 2${\\rm{C}}{{\\rm{o}}^{3 + }}$and ${\\rm{6}}\\,{\\rm{O}}{{\\rm{H}}^ - }$to form ${\\rm{2Co}}{\\left( {{\\rm{OH}}} \\right)_{\\rm{3}}}$.

${\\rm{ 2C}}{{\\rm{o}}^{2 + }} + {\\rm{O}}_2^{2 - } + 2{{\\rm{H}}_{\\rm{2}}}{\\rm{O}} + 2{\\rm{O}}{{\\rm{H}}^ - } \\to 2{\\rm{Co}}{\\left( {{\\rm{OH}}} \\right)_{\\rm{3}}}$

11.Add the spectator ions to the above equation; make sure that they are balanced.

${\\rm{ 2CoC}}{{\\rm{l}}_2} + {\\rm{N}}{{\\rm{a}}_2}{\\rm{O}}_2^{} + 2{{\\rm{H}}_{\\rm{2}}}{\\rm{O}} + 2{\\rm{O}}{{\\rm{H}}^ - } \\to 2{\\rm{Co}}{\\left( {{\\rm{OH}}} \\right)_{\\rm{3}}} + 2{\\rm{N}}{{\\rm{a}}^ + } + 4{\\rm{C}}{{\\rm{l}}^ - }$

Answer

Part 1

The overall reaction is given below:

Part 2

The overall balanced equation is given below:

Part 3

The overall balanced equation is given below:

${\\rm{ 2CoC}}{{\\rm{l}}_2} + {\\rm{N}}{{\\rm{a}}_2}{\\rm{O}}_2^{} + 2{{\\rm{H}}_{\\rm{2}}}{\\rm{O}} + 2{\\rm{O}}{{\\rm{H}}^ - } \\to 2{\\rm{Co}}{\\left( {{\\rm{OH}}} \\right)_{\\rm{3}}} + 2{\\rm{N}}{{\\rm{a}}^ + } + 4{\\rm{C}}{{\\rm{l}}^ - }$

Answer only

Part 1

The overall reaction is given below:

Part 2

The overall balanced equation is given below:

Part 3

The overall balanced equation is given below:

${\\rm{ 2CoC}}{{\\rm{l}}_2} + {\\rm{N}}{{\\rm{a}}_2}{\\rm{O}}_2^{} + 2{{\\rm{H}}_{\\rm{2}}}{\\rm{O}} + 2{\\rm{O}}{{\\rm{H}}^ - } \\to 2{\\rm{Co}}{\\left( {{\\rm{OH}}} \\right)_{\\rm{3}}} + 2{\\rm{N}}{{\\rm{a}}^ + } + 4{\\rm{C}}{{\\rm{l}}^ - }$