Question

Given the equation

2NO₃^- + 3S²⁻ → 2NO + 3S + 4H₂O

how many H+'s should be added to the left side to balance the hydrogen atoms?

Answer 1

answer follow as below

3S2- -------> 3S

balance oxidation no.

3S2- --------> 3S + 6e- ....(1)

this is the balanced oxidation half equation

reduction half :-

2NO3- ---------> 2NO

balance oxidation no.

2NO3- + 6e- --------> 2NO

balance charge by adding H+

2NO3- + 6e- + 8H+ ----------> 2NO

now balance H and O by adding H2O

2NO3- + 6e- + 8H+ -----------> 2NO + 4H2O ....(2)

this is the balanced reduction half equation

add (1) and (2)

3S2- --------> 3S + 6e-

2NO3- + 6e- + 8H+ -----------> 2NO + 4H2O

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6e- will cancel out from both sides

3S2- + 2NO3- + 8H+ ----------> 3S + 2NO + 4H2O

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