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Grade: 11
        
balance redox reaction by oxidation number method for following reaction cr2o7 +so3 – cr +so4 video send now
 
 
 
 
 
 
 
 
 
 
 
 
   
 
 
 
 
 
11 months ago

Answers : (1)

Arun
15841 Points
							

You need to first break this into half-equations 

Cr2O7{2-}(aq) + SO3{2-}(aq) → Cr{3+} (aq) + SO4{2-}(aq) 

The first half reaction is: 
Cr2O7{2-}(aq) → Cr{3+} (aq) 

The seond half reaction is: 
SO3{2-}(aq) → SO4{2-}(aq) 

You need to make each side balanced now. 
You need to make the coefficient on the right side 2 because on the left side there are 2 Cr 
Then you do this by adding either H+ ions of H2O to sides when needed 
For the first one: 
Cr2O7{2-}(aq) → 2Cr{3+} (aq) 

You know that you have 7 oxygens on the left side, so you also need 7 on the right side. 
So to fix this, you add 7 H2Os to the right side. But then you have 14 Hs unaccounted for on the left side! so to fix this, you add 14H+ to the left side. 
So you get 
14H+ + Cr2O7{2-}(aq) → 2Cr{3+} (aq) + 7H2O 
Then you want to calculate the total charge on each side, because that needs to stay the same. 
On the left side, the total charge is +12. On the right side, the total charge is +6. You can add electrons, because that doesn't change any properties of the atoms. So to make each side equal, you need to add 6 electrons to the left side. 
14H+ + Cr2O7{2-}(aq) + 6e- → 2Cr{3+} (aq) + 7H2O 

Now for the second half reaction: 
SO3{2-}(aq) → SO4{2-}(aq) 
You need the same number of oxygens, so you add one H2O to the left side. But then you have an uneven number of hydrogens, so you add 2 H+s to the right side. 
H2O + SO3{2-}(aq) → SO4{2-}(aq) + 2H+ 
Then you add the total charge of each side, and you get -2 on the left side and 0 on the right. 
So you want to add 2 electrons to the right side to make the charges equal. 
H2O + SO3{2-}(aq) → SO4{2-}(aq) + 2H+ + 2e- 

Now you want to combine both half reactions together...but wait! In order to do that, you need the amount of electrons to cancel out, but for the first reaction there are 6 on the left side, but only 2 on the right side of the second reaction. You can fix this by multiplying the entire second reaction by 3 (which will make the amount of electrons on the right side 6). 

3(H2O + SO3{2-}(aq) → SO4{2-}(aq) + 2H+ + 2e-) 
= 3H2O + 3SO3{2-}(aq) → 3SO4{2-}(aq) + 6H+ + 6e- 

Now, to combine them, all you have to do is add them together and cancel things out. 
14H+ + Cr2O7{2-}(aq) + 6e- → 2Cr{3+} (aq) + 7H2O 
3H2O + 3SO3{2-}(aq) → 3SO4{2-}(aq) + 6H+ + 6e- 

You can cancel out some of the H+ because they are on both the left and right sides, as well as the electrons and some waters. 
You get: 
8H+ Cr2O7{2-}(aq) → 2Cr{3+} (aq) + 4H2O 
3SO3{2-}(aq) → 3SO4{2-}(aq) 

Now you add them and get: 

8H+ Cr2O7{2-}(aq) + 3SO3{2-}(aq) → 2Cr{3+} (aq) + 3SO4{2-}(aq) + 4H2O 

And that is your final answer!

11 months ago
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