Question

1o examples of Redox reaction​

Answer 1

Answer:

Explanation:

Problem :  MnO−4+I−→I2+Mn2+  

Steps to balance :

1) Separate the half-reactions that undergo oxidation and reduction.

Oxidation:  I−→I2  

This is the oxidation half because the oxidation state changes from -1 on the left side to 0 on the right side. This indicates a gain in electrons.

Reduction:  MnO−4→Mn2+  

This is the reduction half because the oxidation state changes from +7 on the left side to +2 on the right side. This indicates a reduction in electrons.

2) In order to balance this half reaction we must start by balancing all atoms other than any Hydrogen or Oxygen atoms.

Oxidation:  2I−→I2  

In order to balance the oxidation half of the reaction you must first add a 2 in front of the I on the left hand side so there is an equal number of atoms on both sides.

Reduction:  MnO−4→Mn2+  

For the reduction half of the reaction, you can notice that all atoms other than Hydrogen and Oxygen are already balanced because there is one manganese atom on both sides of the half reaction.

3) Balance Oxygen atoms by adding H2O to the side of the equation that needs Oxygen. Once you have completed this step add H+ to the side of the equation that lacks H atoms necessary to be balanced.

Oxidation:  2I−→I2  

Because there are no Oxygen or Hydrogen atoms in this half of the reaction, it is not required to perform any kind of balancing.

Reduction:  MnO−4→Mn2++4H2O  

The first step in balancing this reaction using step 3 is to add4  H2O atoms in order to balance the Oxygen atoms with the 4 on the other side of MnO4-

Reduction:  MnO−4+8H+→Mn2++4H2O  

Now that the Oxygen atoms have been balanced you can see that there are 8 H atoms on the right hand side of the equation and none on the left. Therefore, you must add 8 H+ atoms to the left hand side of the equation to make it balanced.

4) Now that the two half reactions have been balanced correctly one must balance the charges in each half reaction so that both the reduction and oxidation halves of the reaction consume the same number of electrons.

Answer 2

These are your 10 examples:
>2Mg + O2 - 2MgO
>2Zn + 2HCl - ZnCl2 + H2
>3Fe + 4H2O - Fe3O4 + 4H2
>CuO + H2 - Cu + H2O
>2Cu + O2 - 2CuO
>MnO2 + 4HCl - MnCl2 + Cl2 + 2H2O
>Pb + CuSO4 - PbSO4 + Cu
>Ch4 + O2 - Co2 + H2O
>Mg + 2HCl - MgCl2 + H2
>2K + 2H2O - 2KOH + H2

Hope this helps you..
Thank you