400.0 mL dari 3.75 mM Fe2+ dititrasi dengan 20.0 mM MnO4-  pada pH 0 dalam 1 M H2SO4. Hitung potensialsel vs elektroda standard kalomel pada volume titran 13 ml, 15 ml, dan 17 ml.

Data:          Fe3+ + e <--> Fe2+ Eo= 0,68 V dalam 1 M H2SO4

       MnO4-  + 8H+ + 5e <--> mn2+ +4H2O Eo=1,507V 

E kalomel = 0,241 V




We are dealing with a redox titration problem involving the titration of Fe²⁺ by MnO₄⁻ in acidic conditions, specifically in 1 M H₂SO₄ at pH 0. The goal is to calculate the cell potential at specific volumes of titrant added: 13 mL, 15 mL, and 17 mL. 

To solve this, we'll break it down into the following steps:

1. **Identify the Redox Reaction**:
   \[
   5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O
   \]
   This is a 5:1 stoichiometric reaction between Fe²⁺ and MnO₄⁻.

2. **Determine the Initial Number of Moles of Fe²⁺ and MnO₄⁻**:

   - The initial concentration of Fe²⁺ is 3.75 mM and the volume is 400.0 mL.
     \[
     \text{moles of Fe}^{2+} = 3.75 \times 10^{-3} \, \text{mol/L} \times 0.4000 \, \text{L} = 1.5 \times 10^{-3} \, \text{mol}
     \]

   - The concentration of MnO₄⁻ is 20.0 mM, and we will calculate moles for each titrant volume.

     For \( V_{\text{titran}} = 13 \, \text{mL} \):
     \[
     \text{moles of MnO}_4^- = 20.0 \times 10^{-3} \, \text{mol/L} \times 0.013 \, \text{L} = 2.6 \times 10^{-4} \, \text{mol}
     \]

     For \( V_{\text{titran}} = 15 \, \text{mL} \):
     \[
     \text{moles of MnO}_4^- = 20.0 \times 10^{-3} \, \text{mol/L} \times 0.015 \, \text{L} = 3.0 \times 10^{-4} \, \text{mol}
     \]

     For \( V_{\text{titran}} = 17 \, \text{mL} \):
     \[
     \text{moles of MnO}_4^- = 20.0 \times 10^{-3} \, \text{mol/L} \times 0.017 \, \text{L} = 3.4 \times 10^{-4} \, \text{mol}
     \]

3. **Determine the Number of Moles of Fe²⁺ and Fe³⁺ Remaining**:
   Based on the stoichiometric ratio, for every mole of MnO₄⁻, 5 moles of Fe²⁺ are oxidized.

   - For \( V_{\text{titran}} = 13 \, \text{mL} \):
     \[
     \text{moles of Fe}^{2+} \text{ consumed} = 5 \times 2.6 \times 10^{-4} = 1.3 \times 10^{-3} \, \text{mol}
     \]
     \[
     \text{moles of Fe}^{2+} \text{ remaining} = 1.5 \times 10^{-3} - 1.3 \times 10^{-3} = 2.0 \times 10^{-4} \, \text{mol}
     \]
     \[
     \text{moles of Fe}^{3+} \text{ formed} = 1.3 \times 10^{-3} \, \text{mol}
     \]

   - For \( V_{\text{titran}} = 15 \, \text{mL} \):
     \[
     \text{moles of Fe}^{2+} \text{ consumed} = 5 \times 3.0 \

400.0 mL dari 3.75 mM Fe2+ dititrasi dengan 20.0 mM MnO4- pada pH 0 dalam 1 M H2SO4. Hitung potensial sel vs elektroda standard kalomel pada volume titran 13 ml, 15 ml, dan 17 ml. Data: Fe3+ + e <--> Fe2+ Eo= 0,68 V dalam 1 M H2SO4, MnO4- + 8H+ + 5e <--> Mn2+ + 4H2O Eo=1,507V, E kalomel = 0,241 V

This tutorial provides a step-by-step solution to calculate the cell potential during the redox titration of 400.0 mL 3.75 mM Fe²⁺ with 20.0 mM MnO₄⁻ in 1 M H₂SO₄ at pH 0. The solution includes calculations for titrant volumes of 13 mL, 15 mL, and 17 mL, using the Nernst equation and redox reaction principles.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation