Question

Balance by redox
CrBr3 + NaOH + Cl2 ----> NaCrO4 + NaBrO + NaCl + H2O
please help :)

Answer 1

please help

Answer 2

Cr(+3)Br3(-1) + Na(+1)O(-2)H(+1) + Cl2(0) ----> Na(+1)Cr(+7)O4(-2) + Na(+1)Br(+1)O(-2) + Na(+1)Cl(-1) + H2(+1)O(-2)

The oxidation number is enclosed by parenthesis.

Answer 3

i'm not good at chem, but i think you should find their oxidation states, then make sure that sum of oxidation states on left = that on right :)

Answer 4

Is it correct?

Answer 5

I already did that but I can't balance it correctly.
Did I made a mistake in identifying their oxidation number?

Answer 6

@lgbasallote

Answer 7

Can you help me?

Answer 8

this is just balancing right? no extra thingies?

Answer 9

balancing by state change method. we will identify the oxidizing numbers then balance

Answer 10

...oh

Answer 11

not balancing by inspection

Answer 12

well im not that good in chemistry to know all those. not my specialty sorry

Answer 13

ohh ok thanks for trying :)

Answer 14

Ok this is something I can help with after class, but first...

1. @sasogeek is correct, net electron gain & loss must equal out on both sides

2. @lgbasallote is correct, you need to balance the equation, and I would do this first (easier). Not sure how you were introduced to this in class, but you should balance moles of atoms first (most of the time)...

3. Chromium is your wildcard (meaning you should do it LAST for charges), but I can tell you from experience with this element that +7 would be HIGHLY unstable. Unless you're dealing with intermediary reactions, you won't have this charge on the valence electrons. Stable charges are +1, +2, +3, +4, +5, and +6. The +3 cation charge is the most common because of the natural D-orbital configuration for this element (most common = most stable)

4. Memorize:
The substance which loses electrons is oxidized. (L.E.O.)
The substance which gains electrons is reduced. (G.E.R.)
Can you identify what is being reduced and what is being oxidized?

5. When writing charges, you want to pay attention to treating polyatomic anions as whole groups. So OH\(^-\) , SO\(_4 \ ^{2-}\) , etc. The reason why you should get into the practice of this is because of larger organize polyatomic anions and cations which have resonance structures (i.e.: benzene rings, nucleotides, etc.)

I'll be back to check to see if you can figure out the balancing portion first.

Answer 15

thanks for the info

Answer 16

I already balanced it :)