Question

A student performs redox titration between 10.0 mL of hypochlorous acid, HCIOlaq) and chromlumilll) nitrate; Cr(NO3)3aq) The half-reactions are given below: 7 Hzo() 2 Cr3+(aq) Cr2072-(aq) 14 At(aq) HcIO(aq) 2 Hi(aq) + 2 e C12(g) HzO() Give the overall balanced oxidation-reduction reaction; What is the oxidation number of chromium in Cr2o72 Give one Indication that this Is oxidation-reduction reaction: Explain your reasoning: As the reaction proceeds the solution becomes more acidic and the pH ofthe solution decreases. Explain why: The titration uses chromium(IlI) nitrate yet nitrate does not appear in the half-reaction: Explain the role of nitrate in the reaction The 10.0 mL of hypochlorous acid requires 26.5 mL of 0.120 M chromium(Ill) nitrate to reach equivalence_ Determine the initial molarity of hypochlorous acid. Determine the final molarity of Cr2072 once the reaction has gone completion: (Assume the volumes are additive )

Answer 1

From the stoichiometry of the reaction, the final molarity of Cr2O7^2- in the solution is 0.044 M.

1) The balanced reaction equation is;

4Cr^3+(aq) + 14HClO(aq) + 14H^+(aq) -----> 28H^+(aq) + 7Cl2(g) + 2Cr2O7^2-(aq)

2) The oxidation number of chromium in Cr2O7^2- is obtained as follows;

2x + 7(-2) = -2

2x - 14 = -2

2x = -2 + 14

2x = 12

x = 6

3) This is an oxidation - reduction reaction because the oxidation number of Cr increased from + 3 to +6.

4) This is because, the reaction produces hydrogen ions causing the pH to decrease.

5) Nitrate does not appear in the balanced reaction equation hence it is a spectator ion.

6) Number of moles of chromium nitrate = 26.5/1000 L × 0.120 M = 0.00318 moles

Since 4 moles of Cr^3+ reacts with 14 moles of hypochlorous acid

0.00318 moles of  Cr^3+ reacts with  0.00318 moles × 14 moles/4 moles =

0.0011 moles

Molarity of hypochlorous acid = 0.0011 moles/0.01 L = 0.11 M

7) If  4 moles of Cr^3+  yields 2 moles of  Cr207^2-

0.00318 moles of  Cr^3+ yields  0.00318 moles ×  2 moles / 4 moles

= 0.00159 moles

Final molarity of   Cr207^2- =  0.00159 moles/(10 mL + 26.5 mL) × 10^-3

= 0.044 M.

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