When the following equation of a redox reaction in acidic solution is properly balanced, what are the coefficients for Cr2O72–, Fe2+ H+, Cr3+, Fe3+, and H2O, respectively? __Cr2O72– + __Fe2+ + __H+ --> __Cr3+ + __Fe3+ + __H2O (A) 1, 3, 14, 2, 3, 7; (B) 1, 6, 14, 2, 6, 7; (C) 2, 10, 14, 2, 10, 7; (D) 2, 12, 28, 4, 12, 14
When the following equation of a redox reaction in acidic solution is properly balanced, what are the coefficients for Cr2O72–, Fe2+ H+, Cr3+, Fe3+, and H2O, respectively? __Cr2O72– + __Fe2+ + __H+ --> __Cr3+ + __Fe3+ + __H2O (A) 1, 3, 14, 2, 3, 7; (B) 1, 6, 14, 2, 6, 7; (C) 2, 10, 14, 2, 10, 7; (D) 2, 12, 28, 4, 12, 14
Chapter10: Reconstitution Of Powdered Drugs
Section: Chapter Questions
Problem 30SST
Related questions
Question
When the following equation of a
__Cr2O72– + __Fe2+ + __H+ --> __Cr3+ + __Fe3+ + __H2O
(A) 1, 3, 14, 2, 3, 7;
(B) 1, 6, 14, 2, 6, 7;
(C) 2, 10, 14, 2, 10, 7;
(D) 2, 12, 28, 4, 12, 14
Expert Solution
Step 1
First Step: split the equation into two half-reactions(ignore hydrogen or oxygen)
- Cr2O72-→ Cr3+
- Fe2+ → Fe3+
STEP 2: bALANCE THE EQUATION
- Cr2O72-→ 2Cr3+
- HERE coefficient 2 IS NEEDED on the product side because Cr2O72- has an extra chromium
- Fe2+ → Fe3+
- This EQUATION is fine since there is 1 iron on each side.
- STEP 3: Balance all the oxygens by adding an H2O
- Cr2O72-→ 2Cr3+ + 7H2O
- , we add 7 water molecules to the products to balance BECAUSE Cr2o72 has 7 oxygen
- Fe2+ → Fe3+
STEP 4: Balance all the hydrogens by adding +14 H+ ions
- Cr2O72-+ 14H+ → 2Cr3+ + 7H2O
- Because of the 7 water molecules we added, we need 14 hydrogen ions to balance out the reactants
- Fe2+ → Fe3+
STEP 5: Balance the charges by adding electrons to side where is more positive
- Cr2O72-+ 14H+ + 6e- → 2Cr3+ + 7H2O
- The overall charge of the reactants is +12, because of the 14H+ and the one Cr2O72-
- The overall charge = +6 because of the 2Cr3+
- To balance the equation we add 6 electrons
- Fe2+ → Fe3+ + e-
- the product needs one electron because the reactants are +2 and the products are +3
- Cr2O72-+ 14H+ + 6e- → 2Cr3+ + 7H2O
- 6 (Fe2+ → Fe3+ + e-) → 6Fe2+ → 6Fe3+ + 6e-
- we multiply everything by 6
- 6Fe2+ Cr2O72-+ 14H+ + 6e- → 2Cr3+ + 7H2O + 6Fe3+ + 6e-
- 6Fe2+ Cr2O72-+ 14H+ → 2Cr3+ + 7H2O + 6Fe3+
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