Question

Identify the atoms that are oxidized and reduced, the change in the oxidation state for each, and oxidising and reducing agents in each of the following equations:
(a)Mg + NiCl2 ->MgCl2 + Ni
(b)PCl3 + Cl2 -> PCl5
(c)C2H4 + 3O2 -> 2CO2 + 2H2O
(d)Zn + H2SO4 -> H2 + ZnSO4
(e)2K2S2O3 + I2- > K2S4O6+2KI
(f)3Cu + 8HNO3 = 3Cu(NO3)2 + 4H2O + 2NO

Answer 1

Answer:

Explanation:

a) $Mg + NiCl_2 \rightarrow MgCl_2 + Ni$

Oxidation state of Mg changes to 0 to +2.

Oxidation state of Ni changes to +2 to 0.

Oxidation state of Cl does not change.

So, Mg is oxidized, Ni is reduced.

Mg acts as reducing agent and $NiCl_2$ acts as oxidizing agent.

b) $PCl_3 + Cl_2 \rightarrow PCl_5$

Oxidation state of P changes to +3 to +5.

Oxidation state of Cl changes to 0 to -1.

So,

P is oxidized and Cl is reduced.

$PCl_3$ acts as reducing agent and $Cl_2$ acts as oxidizing agent.

c) $C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$

Oxidation state of C changes to -2 to +4.

Oxidation state of O changes to 0 to -2.

Oxidation state of H does not change.

So,

C is oxidized and O is reduced.

$C_2H_4$ acts as reducing agent and $O_2$ acts as oxidizing agent.

d) $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$

Oxidation state of Zn changes to 0 to +2.

Oxidation state of H changes to +1 to 0.

Oxidation state of S and O do not change.

So, Zn is oxidized, H is reduced.

Zn acts as reducing agent and $H_2SO_4$ acts as oxidizing agent

e) $K_2S_2O_3 + I_2 \rightarrow K_2S_4_O6 + 2KI$

Oxidation state of S changes to +2 to +2.5.

Oxidation state of I changes to 0 to -1.

Oxidation state of K and O do not change.

So, S is oxidized, I is reduced.

$K_2S_2O_3$ acts as reducing agent and $I_2$ acts as oxidizing agent.

f) $3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O+2NO$

Oxidation state of Cu changes to 0 to +2.

Oxidation state of N changes to N +5 to +2.

Oxidation state of H and O do not change.

So, Cu is oxidized, N is reduced.

Cu acts as reducing agent and $HNO_3$ acts as oxidizing agent