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 H
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1. User Answers pstnonsonjoku
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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