write a balanced overall reaction from these unbalanced half reactions cu cu2 ag ag
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Question “write a balanced overall reaction from these unbalanced half reactions cu cu2 ag ag”
Write a balanced overall reaction from these unbalanced half-reactions. Cu rightarrow cu2+ Ag+ rightarrow Ag balanced overall reaction: For a particular redox reaction NO is oxidized to and Ag+ is reduced to Ag. Complete and balance the equation for this reaction in basic solution. Phases are optional. NO+ Ag+ rightarrow + Ag Balance the following equation in basic conditions. Phases are optional. CoCI2 + Na2O2 rightarrow Co(OH)3+C1-+ Na+
Answer
General guidance
Half-reaction can balance a redox reaction. Below are the steps required to balance a chemical reaction for a redox (in a basic solution).
1. Identify the species under oxidation or reduction. Separate the reactions into oxidation/half reaction and reduction/half reaction.
2Balance all atoms, except hydrogen and oxygen.
3 Add protons to balance hydrogens
4 Add electrons to balance charge as follows:
5 Scale half-reactions so that the number of electrons in each half-reaction is equal.
6 Cancel the electrons after the addition of the two reactions.
7.Add {\\rm{O}}{{\\rm{H}}^ - }
8.Combine {{\\rm{H}}^ + }
The law of conservation is that matter cannot be created or destroyed. The number of atoms on each side must be equal.
Step-by-step
Step 1 from 3
Common mistakes
The following redox reaction can be balanced:
1.Recite the half-reactions like this

2. Copper’s oxidation state increases from 0 up to $math_tag_2

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

Add the two reactions as follows:

5. To achieve the balanced redox reaction:

Step 2 from 3
Common mistakes
You can balance the given redox reaction using half-reaction as follows:
1. Write the following redox reaction:

2.Nitrogen from NO is oxidized starting at + 2

3.All atoms, except oxygen, are equal; add {{\\rm{H}}_{\\rm{2}}}{\\rm{O}}

4 Add protons to balance hydrogens

5 Add electrons to balance charge as follows:

6. Multiply the silver-reaction half-reaction by three to equal the number of electrons in the two half-reactions.

7. Cancel the electrons after the addition of the two reactions.

8.Add {\\rm{O}}{{\\rm{H}}^ - }

9.Combine {{\\rm{H}}^ + }

Step 3 from 3
Below is the overall balanced equation:
You can balance the given redox reaction using half-reaction as follows:
1. Write the following redox reaction:

2.Cobalt has been oxidized and oxygen has been reduced by the given redox reactions. {\\rm{C}}{{\\rm{l}}^ - } {\\rm{C}}{{\\rm{o}}^{2 + }} + {\\rm{O}}_2^{2 - } \\to {\\rm{C}}{{\\rm{o}}^{3 + }} + {\\rm{3O}}{{\\rm{H}}^ - }
3.Separate redox reaction into 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 atoms, except hydrogen and oxygen, are balanced. Add {{\\rm{H}}_{\\rm{2}}}{\\rm{O}} \\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 in order to balance the charge.
\\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 with two to get the number of electrons in each half-reaction.
\\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 both 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}}^ - }
Basic is medium Combine $math_tag_1
{\\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 + }} {\\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 spectatorions to the equation and ensure that they are equal.
{\\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
Below is the overall balanced equation:
Only answer
Below is the overall balanced equation:
Conclusion
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