Question

explain the extraction of metal law in activity series.​

Answer 1

in the activity series the metals at the end of the activity series are known as low reactive

The unbalanced reaction is :

MnO4- + I- → MnO2 + I2 (in basic medium)

Step 1: Assign oxidised state for each and every element. Identify oxidation and reduction.

[ Mn+7O4-2 ]- + I- → Mn +4O2-2 + I20

Here, Mn+7 undergoes oxidation to form Mn +4 and I- undegoes reduction to form I2 0 .

Step 2: Balance the atom which undergoes oxidation number change.

Mn+7O4- +2 I- → Mn+4O2 + I20

Step 3: Equalise the increase and decrease in the oxidation number by multiplying with suitable number.

2MnO4- +6 I- →2MnO2 +3 I2

Here, Mn+7 decreases by 3 to form Mn +4 and 2I- increase by 2 to form I2 0 .

So, we equalise by making the decrease and increase value 6.

Step 4: Balancing oxygen. Oxygen can be balanced by adding sufficient number of H2O molecules to the side deficient of oxygen.

2MnO4- +6 I- →MnO2 +3 I2 + 4H2O

Step 5: Balancing of hydrogen. In basic medium hydrogen can be balanced by adding sufficient H2O molecules to the side deficient of hydrogen and the same number of OH- ions to the opposite side.

2MnO4- +6 I- + 8H2O → MnO2 +3 I2 + 4H2O + 8OH-

⇒ 2MnO4 - +6 I- + 4H2O → MnO2 +3 I2 + + 8OH-

Answer 2

metals are roasted first and then oxidised