Complete and balance the equation for this reaction in acidic solution. zns+no3-

The equation given in question is as follow,

                                  ZnS + NO₃⁻    →    Zn²⁺ + S + NO

Oxidation state of elements in reactant side are as follow,
                                      Zn  =  +2
                                      S    =  -2
                                      N    =  +5
                                      O   =  -2
Oxidation state of elements in product side are as follow,
                                      Zn  =  +2
                                      S    =  0
                                      N    =  +2
                                      O   =  -2
So, In this reaction Nitrogen is Reduced while Sulfur is Oxidized.

Now Split the reaction into two half cell reactions,

Reduction Reaction,
                                ZnS     →     S  +  2e⁻
Oxidation Reaction,
                               NO₃⁻  +  3e⁻   →     NO

As the oxygen atoms are not balance, So, in acidic medium add H⁺ on the side having greater number of Oxygen atom and H₂O on the side having less number of Oxygen atoms,
Hence,
                                 H⁺  +   NO₃⁻  + 3e⁻    →     NO  +  H₂O
Now Balance the reaction,

                                 4H⁺  +   NO₃⁻  + 3e⁻    →     NO  +  2H₂O

So, write both half cell equations as,
 
                                             ZnS     →     S  +  2e⁻    (1)
                               4H⁺  +   NO₃⁻  + 3e⁻    →     NO  +  2H₂O  (2)

Multiply eq. 1 with 3 and eq. 2 with 2 to equalize the electron.
So,
                                         3ZnS     →     3S  +  6e⁻   

                       8H⁺  +   2NO₃⁻  + 6e⁻    →     2NO  +  4H₂O 
                     (e⁻ cancelled)

                            3ZnS + 2NO₃⁻ + 8H⁺   → 3Zn + 3S + 2NO + 4H₂O

The kinetic energy of the liquid increases. hope this !

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