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

Curren'S Math For Meds: Dosages & Sol
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Chapter6: Oral Medication Labels And Dosage Calculation
Section: Chapter Questions
Problem 9.4P
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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

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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