Question

1. NO2- + SnO32- ---> SnO22- + NO3-

In the above redox reaction, use oxidation numbers to identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent.

name of the element oxidized:

name of the element reduced:

formula of the oxidizing agent:

formula of the reducing agent:

2. Si + 2N2H4+ H2O+ 2OH- ---> 4NH3 + SiO32-

In the above redox reaction, use oxidation numbers to identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent.

name of the element oxidized:

name of the element reduced:

formula of the oxidizing agent:

formula of the reducing agent:

Answer 1

1)

1)

Oxidation number of each element in NO2-1 is

O=-2

N=+3

Oxidation number of each element in SnO3-2 is

O=-2

Sn=+4

Oxidation number of each element in SnO2-2 is

O=-2

Sn=+2

Oxidation number of each element in NO3-1 is

O=-2

N=+5

Oxidation number of each element in reactant is

O=-2

N=+3

Sn=+4

Oxidation number of each element in product is

O=-2

Sn=+2

N=+5

N in NO2- has oxidation state of +3

N in NO3- has oxidation state of +5

So, N in NO2- is oxidised to NO3-

Sn in SnO3-2 has oxidation state of +4

Sn in SnO2-2 has oxidation state of +2

So, Sn in SnO3-2 is reduced to SnO2-2

name of the element oxidized: Nitrogen

name of the element reduced: Tin

formula of the oxidizing agent: SnO32-

formula of the reducing agent: NO2-

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