(b) ClO−(aq) + Cr(OH)4−(aq) --------à CrO42-(aq) + Cl−(aq)(in basic solution)
2. Two half reactions are:
Oxidation: Cr(OH)4-(aq) --------à CrO42-(aq) (Cr oxidized from +3 to +6)
Reduction: ClO-(aq) ------------à Cl-(aq) (Cl reduced from +1 to -1 )
3. Mass balance atoms except for O and H
Already mass balanced
4. Now mass balance for O and H
Oxidation: Cr(OH)4-(aq) ----------à CrO42-(aq) + 4 H+ (aq)
Reduction: ClO-(aq) + 2 H+(aq) -----à Cl-(aq) + H2O (aq)
5. Balance each half reaction for charge using e- Oxidation:
Cr(OH)4-(aq) --------à CrO42-(aq) + 4 H+ (aq) + 3 e-
Reduction: ClO-(aq) + 2 H+(aq) + 2 e- ----------à Cl-(aq) + H2O (aq)
6.Reaction in basic solution: Therefore add same number of OH ions to both sides to neutralize H+ ions.
Oxidation:
Cr(OH)4-(aq) + 4 OH-(aq) ------à CrO42-(aq) + 4 H+(aq) + 4 OH-(aq) + 3 e-
Cr(OH)4-(aq) + 4 OH-(aq) -------à CrO42-(aq) + 4 H2O (l) + 3 e-
Reduction:
ClO-(aq) + 2 H+(aq) + 2 OH-(aq) + 2 e- -------à Cl-(aq) + H2O (aq) + 2 OH-(aq)
ClO- (aq) + 2 H2O (l) + 2 e- --------à Cl-(aq) + H2O (l) + 2 OH-(aq)
ClO-(aq) + H2O (l) + 2 e- ----------à Cl-(aq) + 2 OH-(aq)
7. Balanced the electrons.
(Cr(OH)4-(aq) + 4 OH-(aq) -----à CrO42-(aq) + 4 H2O (l) + 3 e-) x 2
(ClO-(aq) + H2O (l) + 2 e- --------à Cl-(aq) + 2 OH-(aq)) x 3
2 Cr(OH)4 + 3 ClO- + 2OH- = 2 CrO42-+ 3 Cl- + 5 H2O
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