Question

Balancing equations.

Answer 1

@Miracrown

Answer 2

Hi Parth. Need help?

Answer 3

KMnO4 + NaCl + H2SO4 ----> MnCl2 + Cl2 + K2SO4 + Na2SO4 + H2O

Let's track all the changes in oxidation states.

MnO4- -----> Mn 2+ (reduction)
Cl- ---> Cl2 (oxidation)

Now in the first equation.

Step 1: Balance all atoms other than H and O.
Step 2: Balance H and O. To do that,
- First balance O by adding H2O to the side insufficient in O.
- But now we have extra H's on one side. To balance that, add H+ ions to the side insufficient in hydrogen.
Step 3: Balance all the charge by adding electrons to the side that is more positively charged.

Let's do this.

REDUCTION
Balancing atoms other than O and H: Done.
Balancing O's: MnO4- ----> Mn2+ + 4H2O
Balancing H's: 8H+ + MnO4- -----> Mn2+ + 4H2O
Balancing charge: 8H+ + MnO4- + 5e -----> Mn2+ + 4H2O

OXIDATION
Cl- ----> Cl2
Atoms other than H and O: 2Cl- ----> Cl2
H and O atoms: no H and O atoms, so no need.
Charge: 2Cl- ----> Cl2 + 2e

REMARK: You can see how these oxidation and reduction halves work once they've been balanced. In reduction, there's a gain of electrons and you can see how electrons are gained by MnO4-. Now that we have our halves, we'll work with these halves by adding them. [Continued in the next post.]

Answer 4

@Preetha Nerp, I recently learned this and Mira is doing that currently, so I thought this deserves a post...

Answer 5

To review, here are the oxidation and reduction halves.

REDUCTION:
8H+ + MnO4- + 5e- -----> Mn2+ + 4H2O

OXIDATION:
2Cl- ----> Cl2 + 2e

What we now want to do when we add both the halves is to remove the electrons. We can achieve that by multiplying the reduction half by 2, oxidation half by 5, and adding both halves so that we have an equal number of electrons on both sides (here, 10) and they cancel each other out (just like in mathematics).

REDUCTION:
16H+ + 2MnO4- + 10e- ------> 2Mn2+ + 8H2O

OXIDATION:
10Cl- -----> 5Cl2 + 10e-

Now add both of them.

16H+ + 2MnO4- + 10Cl- ----> 2Mn2+ + 5Cl2 + 8H2O

We now have the balanced ionic equation! But that's not the same as balancing your original equation. But we've actually done most of the job, and it's not gonna be too difficult to balance.

Answer 6

Nice.

Answer 7

Send it to Abhisar as a tutorial!

Answer 8

Here's our original equation for reference:

KMnO4 + NaCl + H2SO4 ----> MnCl2 + Cl2 + K2SO4 + Na2SO4 + H2O

Look at our balanced ionic equation:

16H+ + 2MnO4- + 10Cl- ----> 2Mn2+ + 5Cl2 + 8H2O

In the above ionic equation, the ions are actually being given out by the compounds. MnO4- ion is given out by KMnO4, Cl- is given out by NaCl, H+ is given out by H2SO4 and so on. All we need to do is compare the coefficients of the ionic equation with that of the actual one. Here's what we now get:

2KMnO4 + 10NaCl + 8H2SO4 ----> 2MnCl2 + 5Cl2 + 8H2O

But now the problem is that we don't really have any K or Na atoms on the right hand side. Any atoms that don't undergo change in oxidation state must not be included in ionic equations (except H and O - they MUST be included).

2KMnO4 + 10NaCl + 8H2SO4 ---> 2MnCl2 + 5Cl2 + 8H2O + K2SO4 + 7Na2SO4

Above, we balanced K and the sulphate ion, but we still haven't balanced Na and Cl. On the right-hand-side, we have four extra Na and four extra Cl. What does that tell us? Add 4NaCl to the left!

2KMnO4 + 14NaCl + 8H2SO4 ---> 2MnCl2 + 5Cl2 + 8H2O + K2SO4 + 7Na2SO4

Answer 9

@Preetha This isn't a tutorial - I just solved a problem. :|

Answer 10

Good job here :3

Answer 11

I also forgot to mention that after I obtained the ionic equation, it was purely a work of guess-and-check.

Answer 12

Wow. A very deatiled explanation on how to balance equations, I can't thank you enough PK. I'll print it out and go over it whenever I'm stuck at balancing equations. A bundle of thanks :)

Answer 13

A NOTE ON BALANCING IN DIFFERENT MEDIA

As you saw in the balancing technique, I was adding \(\rm H^+ \) ions to a side insufficient in hydrogen atoms and \(\rm H_2 O \) molecules to those insufficient in oxygen. What allowed me to do this? How can I just introduce compounds/ions to a reaction?

It was done because these ions/compounds actually exist, and they facilitate the reactions. As you saw, the above reaction could not have been facilitated without the presence of \(\rm H^+\) given by \(\rm H_2 SO_4\). Thus, the acidity/basicity of a medium affects some reactions.

Whenever you see an \(\rm H^+\) coming into play, it means that the reaction is in an acidic medium and whenever you see an \(\rm OH^-\), it means that the reaction is in a basic medium. EARLIER, WE HAD BEEN BALANCING IN ACIDIC MEDIUM BECAUSE WE WERE ADDING H+ IONS.
- - - - - - - - - - -
STEPS TO BALANCE REACTIONS IN A BASIC MEDIUM

Step 0: Get an equation to balance.\[\rm Cr(OH)_3 + IO_3^- \longrightarrow I^- + CrO_4^{2- }\tag{alkaline medium}\]**THE EASY WAY BUILDING UP FROM PREVIOUS KNOWLEDGE:**
Step 1: Balance the reaction in acidic medium. (Do it yourself for a little practice.)

OXIDATION HALF REACTION\[\rm Cr(OH)_3 + H_2O\longrightarrow CrO_4^{2-} + 5H^+ + 3e^{-}\]REDUCTION HALF REACTION\[\rm IO_3^- + 6H^+ + 6e^- \longrightarrow I^- + 3H_2 O \]BALANCED REACTION IN ACIDIC MEDIUM\[\rm 2Cr(OH)_3 + IO_3^- \longrightarrow 2CrO4^{2-} + I^- + H_2 O + 4H^+\]Step 2: Add as many \(\rm OH^-\) to both sides as there are \(\rm H^+\)s and keep in mind the reaction \(\rm H^+ + OH^-\). Here, there are 4 \(\rm H^+\) and so add \(\rm 4 OH^-\) to both sides. Your final reaction should look like this:\[\rm 2Cr(OH)_3 + IO_3^- + 4OH^- \longrightarrow 2CrO4^{2-} + I^- + 5H_2O \]**THE DIFFICULT WAY - EASY WHEN YOU GET USED TO IT**:
Step 1: Balance atoms that are not O and H.
Step 2: Balancing H and O.
- First, balance O by adding sufficient amount of \(\rm H_2 O\) to the side insufficient in oxygen.
- Second, balance H by adding \(\rm OH^-\) to the side WITH EXTRA H (as many as there are extra H's) and adding that many \(\rm H_2 O\) to the other one. This works since \(\rm H_2 O\) has one extra hydrogen.
Step 3: Balancing charge.

Here's the technique in play.

OXIDATION HALF REACTION\[\rm Cr(OH)_3 + H_2 O + 5OH^- \longrightarrow CrO_4^{2-} + 5H_2 O + 3e^-\]REDUCTION HALF REACTION\[\rm IO_3^- + 6H_2 O + 6e^- \longrightarrow I^- + 3H_2 O + 6OH^- \]Multiply the first equation by 2 and add both equations to get:\[\rm 2Cr(OH)_3 + IO_3^- + 4OH^- \longrightarrow 2CrO4^{2-} + I^- + 5H_2O\]Same as the previous answer.

Answer 14

Cool! Redox is fun .. :D @ParthKohli - You will probably benefit a lot from all this in 11th. :D