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Balance the equation by oxidation number method : (Cr2O7-2)+(Fe+2)+(H+)=(Fe+3)+(Cr+3)+H2O |
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Answer» Cr 2 O 72−+H + ⟶Fe 3++Cr 3+ +H 2O(1) Oxidation: Fe 2+→Fe 3+ REDUCTION: Cr+62 O 72− ⟶ C+3 r 3+ (2) Balancing the atomsFe 2+⟶Fe 3+ Cr 2O 72−+14H + ⟶2Cr 3+ +7H 2 O(3) BALANCE the chargeFe 2+ ⟶Fe 3+ +e − ⟶(1) Cr 2 O 72− +14H ++6e − ⟶2Cr 3++7H 2O⟶(2)(4) EQUATION (1) ×6+ Equation (2) ×1 Balancing electrons. 6Fe 2+ +Cr 2 O 72−+14H + +6e − ⟶6Fe 3+ +2Cr 3+ +6e – +7H 2 O(5) Simplifying;6Fe 2+ +Cr 2 O 72− +14H ⟶6Fe 3+ +2Cr 3+ +7H 2 O_____________________________ |
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