Redox Reaction in Acidic Media

Given that: $\ce{Cr_2O_7^{2-} + Cl- -> Cr^3+ + Cl2}$

1) Reduction half: $\ce{Cr_2O_7^{2-} -> Cr^3+}$

• Balance $\ce{Cr}$ atoms: $\ce{Cr_2O_7^{2-} -> 2 Cr^3+}$
• Balance $\ce{O}$ atoms: $\ce{Cr_2O_7^{2-} -> 2 Cr^3+ + 7H2O}$
• Balance $\ce{H}$ atoms: $\ce{Cr_2O_7^{2-} +14H+ -> 2 Cr^3+ + 7H2O}$
• Balance charges: $\ce{Cr_2O_7^{2-} +14H+ + 6e- -> 2 Cr^3+ + 7H2O}$

2) Oxidation half: $\ce{Cl- -> Cl2}$

• Balance $\ce{Cl}$ atoms: $\ce{2Cl- -> Cl2}$
• Balance charges: $\ce{2Cl- -> Cl2 + 2e-}$

3) Two reactions together: $\ce{Cr_2O_7^{2-} + 6Cl- +14H+ -> 2 Cr^3+ + 3Cl2} + \boxed{7H_2O}$

The number of water molecules needed is $\boxed{7}$ .

We are allowed to add any $\ce{H^+}$ and any $\ce{H2O}$ we wish to the equation. Consider the oxygen atoms: there are seven on the lefthand side, which must be balanced by seven on the righthand side. That requires $\boxed{7}$ water molecules.

Perhaps in the balancing process we wish to avoid fractions and place a coefficient $n$ in front of the $\ce{Cr2O7^{2-}}$ . Then we would need $7n$ water molecules. However, this will certainly be a non-zero multiple of 7; the only non-zero multiple of seven in the list of choices is 7.