Balance the following reaction. If a reactant or product is not present, put a zero (0) in the blank. H+(aq) + H2O(l) + ClO-13(aq) + e-1 <--> Cl-1(g) + H+(aq) + H2O(l)
Balance the following reaction. If a reactant or product is not present, put a zero (0) in the blank.
H+(aq) + H2O(l) + ClO-13(aq) + e-1 <--> Cl-1(g) + H+(aq) + H2O(l)
ClO3- (aq ) ---------> Cl-1 (aq)
Oxidation state of Chlorine decreases from +5 (ClO3- ) to -1 ( Cl-1 ). So it is a reduction half reation.
Since there are three O atom on left hand side, add 3 H2O molecules to the right hand side.
ClO3- (aq ) ---------> Cl-1 (aq) + 3H2O (l)
Now since there are 6 H atom on right hand side, add 6 H+ on left hand side
ClO3- (aq ) +6 H+ (aq) ---------> Cl-1 (aq) + 3H2O (l)
Now balance the charge.
Left hand side =+ 5
Right hand side = -1
So add 5 electron to left-hand side
ClO3- (aq ) +6 H+ (aq) +5e- ---------> Cl-1 (aq) + 3H2O (l)
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