You are titrating a 50.0 mL solution of 0.050 M hydrogen peroxide (H2O2) with 0.100 M potassium permanganate (KMnO4) using a platinum electrode. The standard electrode potential for the platinum electrode is E∘=0 V. Calculate the potential of the solution After adding excess potassium permanganate 75.0 mL).
Question
You are titrating a 50.0 mL solution of 0.050 M hydrogen peroxide (H2O2) with 0.100 M potassium permanganate (KMnO4) using a platinum electrode. The standard electrode potential for the platinum electrode is E∘=0 V. Calculate the potential of the solution After adding excess potassium permanganate 75.0 mL).
Solution
The reaction between hydrogen peroxide and potassium permanganate in acidic solution is as follows:
2MnO4- + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2
From the stoichiometry of the reaction, we can see that 2 moles of MnO4- react with 5 moles of H2O2.
First, we need to calculate the number of moles of H2O2 and KMnO4.
Number of moles of H2O2 = Molarity * Volume(L) = 0.050 M * 0.050 L = 0.0025 moles Number of moles of KMnO4 = Molarity * Volume(L) = 0.100 M * 0.075 L = 0.0075 moles
Since 2 moles of MnO4- react with 5 moles of H2O2, the number of moles of H2O2 required to react with 0.0075 moles of MnO4- is (0.0075 moles * 5) / 2 = 0.01875 moles.
Since the number of moles of H2O2 present (0.0025 moles) is less than the number of moles of H2O2 required (0.01875 moles), all the H2O2 will react and there will be excess MnO4-.
The half-reactions are:
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O E° = +1.51 V H2O2 → O2 + 2H+ + 2e- E° = +0.68 V
The overall cell potential E°cell is given by the equation:
E°cell = E°cathode - E°anode
In this case, MnO4- is the oxidizing agent (it gains electrons, so it's the cathode) and H2O2 is the reducing agent (it loses electrons, so it's the anode). Therefore,
E°cell = E°MnO4-/Mn2+ - E°H2O2/O2 = +1.51 V - (+0.68 V) = +0.83 V
Therefore, the potential of the solution after adding excess potassium permanganate is +0.83 V.
Similar Questions
Solve: 1. You are titrating a 50.0 mL solution of 0.100 M iodide ions (I−) with 0.100 M lead(II) nitrate (Pb(NO₃)₂) using a lead electrode. The standard electrode potential for the lead electrode is E∘=−0.126 V. Calculate the potential of the solution at the following points: 4.5 marks a) Before adding any lead(II) nitrate. b) After adding 25.0 mL of lead(II) nitrate. c) At the equivalence point. d) After adding excess lead(II) nitrate (e.g., 75.0 mL). 2. You are titrating a 50.0 mL solution of 0.050 M hydrogen peroxide (H2O2) with 0.100 M potassium permanganate (KMnO4) using a platinum electrode. The standard electrode potential for the platinum electrode is E∘=0 V. Calculate the potential of the solution at the following points: 4.5 marks a) Before adding any potassium permanganate. b) After adding 25.0 mL of potassium permanganate. c) At the equivalence point. d) After adding excess potassium permanganate (e.g., 75.0 mL).
Problem: You are titrating a 50.0 mL solution of an unknown concentration of chloride ions (Cl⁻) with 0.100 M silver nitrate (AgNO₃) using a silver/silver chloride electrode. The standard electrode potential for the silver/silver chloride electrode is E∘=0.222 V. Calculate the potential of the solution at the following points: 1. Before adding any AgNO₃. 2. After adding 10.0 mL of AgNO₃. 3. At the equivalence point. 4. After adding excess AgNO₃ (e.g., 60.0 mL). Solve: 1. You are titrating a 50.0 mL solution of 0.100 M iodide ions (I−) with 0.100 M lead(II) nitrate (Pb(NO₃)₂) using a lead electrode. The standard electrode potential for the lead electrode is E∘=−0.126 V. Calculate the potential of the solution at the following points: 4.5 marks a) Before adding any lead(II) nitrate. b) After adding 25.0 mL of lead(II) nitrate. c) At the equivalence point. d) After adding excess lead(II) nitrate (e.g., 75.0 mL). 2. You are titrating a 50.0 mL solution of 0.050 M hydrogen peroxide (H2O2) with 0.100 M potassium permanganate (KMnO4) using a platinum electrode. The standard electrode potential for the platinum electrode is E∘=0 V. Calculate the potential of the solution at the following points: 4.5 marks a) Before adding any potassium permanganate. b) After adding 25.0 mL of potassium permanganate. c) At the equivalence point. d) After adding excess potassium permanganate (e.g., 75.0 mL).
calculate molarity of the standardized potassium permanganate solution in redox titration of iron 2 into iron 3. molecular weight of hydrated ferrous ammonium sulfate is 392.14 g/mol. also use the ration of iron 2 to permanganate ion from equation: 5Fe + Mn04 + 8H gives 5Fe + Mn + 4H2O. VOLUME OF KMnO4 used in titration is 7.65 ml and mass of hydrated ferrous ammonium sulfate is 0.2911 g.
A chemist prepares a solution of potassium permanganate KMnO4 by measuring out 7.6mg of potassium permanganate into a 50.mL volumetric flask and filling the flask to the mark with water.Calculate the concentration in /molL of the chemist's potassium permanganate solution. Be sure your answer has the correct number of significant digits.
Calculate the mass (in g) of potassium permanganate (KMnO4) required to make a 250 mL of a 0.951 M solution. Answer to 3 decimal places.MW(KMnO4) = 158.04 g.mol-1
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.