Cu3PO4 + H+ + Cr2072 → Cut2 + H3PC
When the above redox reaction is balanced,
using simplest whole numbers, how many
stoichiometric coefficients will be odd digits
Answers
Answer:
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Explanation:
Given reaction:
xI
2
+yOH
−
→IO
3
−
+zI
−
+3H
2
O
in a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction.
Balancing a chemical reaction as:
Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:
I
2
0
+OH
−
→
I
+5
O
3
−
+
I
−
−1
+3H
2
O
Step 2: Identify the atoms that are oxidized and those that are reduced as:
Reduction:
I
2
0
→
I
−
−1
Oxidation:
I
2
0
→
I
+5
O
3
−
Step 3: oxidation-number change is:
Reduction:
I
2
0
→2
I
−
−1
: Gain of 2 electron
Oxidation: :
I
2
0
→2
I
+5
O
3
−
: Loss of total 10 electrons
Step 4: Balance the total change in oxidation number as:
Reduction: :
I
2
0
→2
I
−
−1
×5: Gain of 10 electron total
Oxidation: :
I
2
0
→2
I
+5
O
3
−
×1: Loss of 10 electron
Reduction: : 5
I
2
0
→10
I
−
−1
: Gain of 10 electron total
Oxidation: :
I
2
0
→2
I
+5
O
3
−
: Loss of 10 electron
Step 5: Balance O atoms in oxidation reaction by adding H
2
O and then balance H by H
+
as:
Oxidation: :
I
2
0
+6H
2
O→2
I
+5
O
3
−
+12H
+
Step 6: For base catalysed reaction add OH
−
to both side to neutralize H
+
as:
Oxidation: :
I
2
0
+6H
2
O+12OH
−
→2
I
+5
O
3
−
+12H
+
+12OH
−
or,
Reduction: : 5
I
2
0
→10
I
−
−1
Oxidation: :
I
2
0
+12OH
−
→2
I
+5
O
3
−
+6H
2
O
Thus overall reaction is:
5
I
2
0
+
I
2
0
+12OH
−
→10
I
−
−1
+2
I
+5
O
3
−
+6H
2
O
OR 6
I
2
0
+12OH
−
→10
I
−
−1
+2
I
+5
O
3
−
+6H
2
O
or 3
I
2
0
+6OH
−
→5
I
−
−1
+
I
+5
O
3
−
+3H
2
O
compairing the above balanced reaction with given reaction we get:
xI
2
+yOH
−
→IO
3
−
+zI
−
+3H
2
O
thus x=3,y=6,z=5