How many electrons are transferred in the redox reaction taking place in basic solution:
Cr(s) + ClO4-(aq) --> ClO3-(aq) + Cr3+(aq)?
let us calculate the oxidation state of Cl in ClO4-
let oxidation state be x
charge over each oxide is -2
so we can right
x+(4*-2)=-1 as the overall charge is -1
x-8=-1
x=-1+8=7
so oxidation state of Cl is +7
let us calculate the oxidation state of lr in CrO3-1
let oxidation state be y
charge over each oxide is -2
so we can right
y+(3*-2)=-1 as the molecule is negatively charge or overall charge is -1
y-6=-1
y=-1+6=5
so oxidation state of Cl is +5
we see oxidation state of Cl goes from +7 to +5 there fore it gain of electron and is getting reduced
the change oxidation state for chromium # from0 so it is losing electron and is getting oxidised
half cell oxidation
Cr
Cr3+
charge on left side is 0 and right side is +3 so if we add +3e on right side we can balance the charge
we get
Cr
Cr3+ +3e equation 1
half cell reduction
ClO4-ClO3-
we see the number of oxygen atom on left side is 4 and right side is 3
so in basic condition
for every excess of oxygen 1 H2O added to side which have more oxygen atom and 2OH- molecule on the side which is deficient in it
therefore in this case 1H2O added to left side and 2OH- molecule on the right side to balance O atoms
so we get
ClO4- +1H2O ClO3-
+2OH-
we see all atoms are balance (mass balanced equation)
charge on left side is -1 and right side is -2 +(-1)=-3 so if we add +2e on left side we can balance the charge
we get
ClO4- +1H2O + 2e ClO3-
+2OH- equation 2
Cr
Cr3+ +3e equation 1
now multiply equation 1 with 2 and equation 2 with 3 to make the electron count same
3ClO4- +3H2O + 6e 3ClO3-
+6OH- equation 4
2Cr
2Cr3+ +6e equation 3
net equation adding equation 3 and 4
3ClO4- +3H2O +2Cr 3ClO3-
+6OH +2Cr3+
thus 6 electrons are transferred in the redox reaction taking place in basic solution
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