Balancing Redox Equations?

Question

I'm having difficulty balancing this redox equation. please help me.

HgS(s) + NO^3-(aq) + Cl-(aq) -----> HgCl4^2- (aq) + NO(g) + S(s)

NOTE: this is in an ACIDIC medium.


thanks. 10 pts for the best answer.

Answer 1

First identify the species that is getting oxidized and reduced. It is Hg that is getting oxidized and N that is getting reduced.


Writing the equations involved in oxidation and reduction


S(-2) ----> S(0) + 2e- ...(I)

N(+5) + 3e(-)----> N(+2) ..(II)

To balance the number of electrons on the RHS and the LHS, multiply #(I) by 3 and #(IV) by 2, and add. So, we have

2N(+5) + 3S(-2) ----> 3S(0) + 2N(+2)
Adding the spectacular ions, we have
3HgS + 2NO3(-) + 12Cl(-) ----> 3HgCl4(-2) + 2NO + 3S

The only atoms remaining to balance is the O atoms. To balance them, introduce H+ in the product side and add H2O to the reactant side

3HgS + 2NO3(-) + 12Cl(-) + 8H(+)----> 3HgCl4(-2) + 2NO + 3S + 4H2O

{The last step is carried out because we are dealing with an acidic solution}

Answer 2

3HgS + 2(NO3)- + 12Cl- + 8H+ -------> 3[HgCl4]2- + 2NO + 3S + 4H2O

It's by looking at the oxidation number changes of S and N that you get the ratio of 3:2, and then it's plain sailing!