a) Identify the substance oxidised, substance reduced, oxidising agent and reducing agent in the following reaction: 

M n O subscript 2 space plus 4 space H C l space minus negative greater than space M n C l subscript 2 space plus 2 H subscript 2 O plus C l subscript 2

b) In the reaction Z n plus C u S O subscript 4 minus negative greater than Z n S O subscript 4 plus space C u, what is oxidised and what is reduced? 

Asked by adalroshan2464 | 12th May, 2019, 03:19: PM

Expert Answer:

(a) In this reaction, We need to find oxidation numbers in each compound
 
          MnO subscript 2 plus 4 HCl rightwards arrow MnCl subscript 2 plus 2 straight H subscript 2 straight O plus Cl subscript 2
In MnO2,Oxidation number of Mn will be  x-4=0
                                                            x=+4
In MnCl2,Oxidation number of Mn will be  x-2=0
                                                             x=+2
So from conversion to MnO2 to MnCl2, Oxidation number is reduced which means  MnO2 is reduced to MnCl2 and  this process is reduction.
MnOis reduced so it will oxidise other compounds.
So MnO2 =Oxidising agent 
When HCl is converted to Cl2, it looses hydrogen so when hydrogen is released it means this is Oxidation process and HCl is oxidised which means it will reduce other compounds
So HCl=Reducing agent
 
 
(b)
 
Z n plus C u S O subscript 4 rightwards arrow Z n S O subscript 4 plus C u

0 space space space space space space space space space plus 2 space space space space space space space plus 2 space space space space space space space space 0
 
Zn is oxidised in ZnSO4 and CuSO is reduced in Cu.

Answered by Ravi | 14th May, 2019, 12:28: PM