how to balance electrons in ion-electron method?balancing the electrons of a chemical equation
H2S(g)+ NO3-(aq) ===============> NO(g) + S(s)
Balancing a redox reaction by electron method first we need to find the medium of reaction, whether it is acidic or basic medium.
Let us consider the above reaction and consider it to be in acidic medium.
First we need to divide the reaction into two half reaction:
H2S =======> S
NO3(-) =====> NO
Now we need to balanced the it, if there is less oxygen we need to add equal number of water (H2O) and then H+ to balanced the Hydrogen atom
H2S =======> S + 2 H(+)
NO3(-) + 4 H(+)======> NO + 2H2O
Now the elements are balanced. Now we need to balanced the charge by adding electron.
H2S =======> S + 2 H(+) + 2e [as there are two proton so we add 2 electron]
NO3(-) + 4 H(+)+3e======> NO + 2H2O [as there is 1 -ve charge in NO3(-) and 4 H+, so the total charge is +3 so we added 3 electron]
Now we need to take LCM, to make equal number of electron, Let's multiply the 1st reaction by 3 and then 2nd by 2
3 H2S =======> 3 S + 6 H(+) + 6 e
2 NO3(-) + 8 H(+)+6e======> 2 NO + 4 H2O
======================================= [adding these two equation]
3 H2S + 2 NO3(-) + 2 H(+) =======> 3 S + 2 NO + 4 H2O
this is the balanced reaction in acidic condition. If it is basic condition, we need to add equal number of (OH-) for H(+) as in basic condition there will be OH(-) not H(+).
3 H2S + 2 NO3(-) + 2 H(+) + 2 OH(-) =======> 3 S + 2 NO + 4 H2O + OH(-)
3 H2S + 2 NO3(-) + 2 H2O =======> 3 S + 2 NO + 4 H2O + OH(-) [ 2 H(+) + 2 OH(-) ==> 2 H2O}
3 H2S + 2 NO3(-) =======> 3 S + 2 NO + 2 H2O + OH(-)
this is balanced reaction in basic conditions.
Hope its has help you.