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Elden [556K]
2 years ago
7

Balance the following redox equation and identify the element oxidized, the element reduced, the oxidizing agent, and the reduci

ng agent. Show all of the work used to solve the problem. Br- + MnO2 yields Br2 + Mn2+
Chemistry
1 answer:
earnstyle [38]2 years ago
6 0

The charge of Br changed from –1 to 0, therefore it is the element which is oxidized. Since it is oxidized then Br is also the reducing agent.

 

The charge of Mn changed from +4 to +2 therefore it is the element which is reduced. Since Mn is reduced, then MnO2 is the oxidizing agent.

 

The half –reactions are:

Br: 2Br --> Br2 + 2e-

Mn: MnO2 --> Mn2+

First balance oxygen by adding H2O:

MnO2 --> Mn2+ + 2H2O

Then balance hydrogen by adding H+ ions:

4H+ + MnO2 --> Mn2 + 2H2O

Then the appropriate electrons:

4e- + 4H+ + MnO2 --> Mn2 + 2H2O

 

Multiply the half-reaction of Br by 2 because the half-reaction of Mn has 4 electrons.

4Br --> 2Br2 + 4e-

 

Combine the two half reactions and cancel common factors:

4Br-  +  4H+  +  MnO2 --> 2Br2  +  Mn2  +  2H2O

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Answer:

                     2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Explanation:

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Step 2: Balance Carbon  Atoms;

As there are 4 carbon  atoms on left hand side and 1 carbon atoms on right hand site therefore, to balance them multiply CO₂ on right hand side by 4 i.e.

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Step 3: Balance Hydrogen Atoms;

There are 10 hydrogen atoms on left hand side and 2 hydrogen atom on right hand site therefore, to balance them multiply H₂O on left hand side by 5 i.e.

                            C₄H₁₀ + O₂ → 4 CO₂ +  5 H₂O

Step 4: Balance Oxygen Atoms:

Now there are 2 oxygen atoms in reactant side and 13 in product side. So, multiply O₂ by 6.5 i.e.

                            C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O

Step 5: Remove fraction coefficients as,

Multiply whole equation by 2 to get rid of fractions i.e.

                            2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

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2 years ago
Other questions:
vlabodo [156]

Answer:

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Chemical formula : C_{7}H_{6}O

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