Balancing using Oxidation States

Question

Cl2 KOH --> KClO3 KCl H2O

HNO3(aq) H2S(g) --> S(s) NO(g) H2O(l)

K2Cr2O7(aq) H2O(l) S(s) --> KOH(aq) Cr2O3(aq) SO2(g)

Can someone show me these using oxidation numbers..

Answer 1

Figure out the charges of each element

Find the two charge changes; one will be a loss of electrons, one will be a gain of electrons

Write the two half reactions remembering to balance atoms first.  Make the number of electrons in the two half reactions equal and then add them together (the electrons will cancel).

This gives you the coefficients for 4 of the substances in the equation; balance any remaining elements (usually H and O) the old way.

The first equation is very tricky:  the Cl on the reactant side is undergoing two different charge changes; usually it is two different elements

In the first equation on the reactant side:

charge of Cl in Cl2 = 0

On the product side

Cl = +5 (loss of electrons) in the KClO3

Cl = -1(gain of electrons) in the KCl

The two half reactions:

Cl(0)   =   Cl+5 + 5e

(Cl(0)  + 1e  =   Cl-1) 5

I multiplied the second equation by 5 to make the electrons equal

Adding them together you get:

6 Cl(0) = Cl+5 + 5Cl-1

This means you need 6 chlorine on the reactant side (3Cl2)

You need 1 KCl03 (that's where the Cl is +5)

You need 5 KCl (that's where the Cl is -1)

At this point you know you have 3Cl2, 1 KClO3 and 5 KCl; you shouldn't need to change these coefficients

Balance the K by puting a 6 for KOH

This gives you 6H, so put a 3 for H20

You have 6 O in the 6KOH and you have 6 O in the KClO3 + 3H20

Final equation:

3Cl2 + 6KOH = 1KClO3 + 5KCl =3H20

The second one is easier.  The two charge changes are N and S

On the reactant side the N is +5 on the product side it is +2

On the reactant side the S is -2 on the product side it is 0

Write the two half reactions and balance electrons

2(N+5 +3e = N+2)

3(S = 0 +2e)

Add them together:

2N+5 + 3S = 2N+2 + 3S

Put these coefficients back into the original equation

Then balance the H's and the O's

Final equation:

2HNO3 + 3H2S = 3S + 2NO +4H20

In the last equation the two charge changes are CR+6 to Cr+3 and S0 to S+4

Half reactions

2(Cr(2)+6 + 6e = Cr(2) +3)

3(S 0 = S+4 +4e)

2 goes in front of K2Cr2O7 and Cr2O3

3 goes in front of S and SO2

Balance the K, H and O

Final equation:

2K2Cr2O7 + 2H2O + 3S = 4KOH + 2Cr2O3 + 3SO2

Hope that helps.  #1 was a mean one to start with!

Answer 2

3Cl2+6KOH--------->KClO3+5KCl+3H2O

Answer 3

try to work on your assignment....