balance it in detail..........
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This is the oxidation of ethanal to ethanoic acid. I will assume acidic conditions.
Ist set up the 1/2 equations:
CH3C−1HO→CHCOO+1H
You can see from the oxidation numbers in red that the H atom in question goes from -1 to +1. This means it must give out 2 electrons.
We add water on the left and H+ ions and electrons on the right ⇒
CH3CHO+H2O→CHCOOH+2H++2e...........(1)
Now for the dichromate(VI):
+6Cr2O2−7→2+3Cr3+
So we have:
(2×+6)=+12→(2×+3)=+6
i.e +12→+6 so this requires 6 electrons to effect this change.
We add H+ ions on the left to mop up the oxygens⇒
Cr2O2−7+14H++6e→2Cr3++7H2O..............(2)
We now need to x (1) by 3 to get the electrons to balance⇒
3CH3CHO+3H2O→3CHCOOH+6H++6e..........(3)
Now we add both sides of (2)and(3) together:
Cr2O2−7+14H++6e+3CH3CHO+3H2O→2Cr3++7H2O+3CH3COOH+6H++6e
This cancels down to:
Cr2O2−7+8H++3CH3CHO→2Cr3++4H2O+3CH3COOH
therefore.Answer:
Cr2O2−7+8H++3CH3CHO→2Cr3++4H2O+3CH3COOH..plz Mark it in Brainliest..
Ist set up the 1/2 equations:
CH3C−1HO→CHCOO+1H
You can see from the oxidation numbers in red that the H atom in question goes from -1 to +1. This means it must give out 2 electrons.
We add water on the left and H+ ions and electrons on the right ⇒
CH3CHO+H2O→CHCOOH+2H++2e...........(1)
Now for the dichromate(VI):
+6Cr2O2−7→2+3Cr3+
So we have:
(2×+6)=+12→(2×+3)=+6
i.e +12→+6 so this requires 6 electrons to effect this change.
We add H+ ions on the left to mop up the oxygens⇒
Cr2O2−7+14H++6e→2Cr3++7H2O..............(2)
We now need to x (1) by 3 to get the electrons to balance⇒
3CH3CHO+3H2O→3CHCOOH+6H++6e..........(3)
Now we add both sides of (2)and(3) together:
Cr2O2−7+14H++6e+3CH3CHO+3H2O→2Cr3++7H2O+3CH3COOH+6H++6e
This cancels down to:
Cr2O2−7+8H++3CH3CHO→2Cr3++4H2O+3CH3COOH
therefore.Answer:
Cr2O2−7+8H++3CH3CHO→2Cr3++4H2O+3CH3COOH..plz Mark it in Brainliest..
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