Chapter 16: Redox Titrations

What is a Jones reductor and what is it used for?

Why don't${\mathbf{Cr}}^{\mathbf{3}\mathbf{+}}$and ${\mathbf{TiO}}^{\mathbf{2}\mathbf{+}}$interfere in the analysis of ${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$when a Walden reductor, instead of a Jones reductor, is used for pre-reduction?

From information in Table 16-3, explain how you would use ${\mathbf{KMnO}}_{\mathbf{4}}$to find the content of ${\left({\mathrm{NH}}_4\right)}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{8}}$in a solid mixture with${\mathbf{\left(}{\mathbf{NH}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}$What is the purpose of phosphoric acid in the procedure?

Write balanced half-reactions in which $Mn{O}_4^{-}$acts as an oxidant at

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{pH}=0; \\ \left(\mathrm{b}\right)\mathrm{pH}=10; \\ \left(\mathrm{c}\right)\mathrm{pH}=15. \end{gathered}$$

Whenof unknown were passed through a Jones reductor, molybdate ion $\mathbf{\left(}{\mathbf{MoO}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{\right)}$was converted into ${\mathbf{Mo}}^{\mathbf{3}\mathbf{+}}$. The filtrate required 16.43mLof $\mathbf{0}\mathbf{.}\mathbf{01033}{\mathbf{MKMnO}}_{\mathbf{4}}$to reach the purple end point.

${\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{+}{\mathbf{Mo}}^{\mathbf{3}\mathbf{+}}\mathbf{\to }{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}}\mathbf{+}{\mathbf{MoO}}_{\mathbf{2}}^{\mathbf{2}\mathbf{+}}$

A blank required 0.04mL. Balance the reaction and find the molarity of Molybdate in the unknown.

When 25.00mLof unknown were passed through a Jones reductor, molybdate ion$\left(Mo{O}_4^{2-}\right)$was converted into. The filtrate required 16.43mLof$\mathbf{0}\mathbf{.}\mathbf{01033}{\mathbf{MKMnO}}_{\mathbf{4}}$to reach the purple end point.

role="math" localid="1663608295687" ${\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{+}{\mathbf{Mo}}^{\mathbf{3}\mathbf{+}}\mathbf{\to }{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}}\mathbf{+}{\mathbf{MoO}}_{\mathbf{2}}^{\mathbf{2}\mathbf{+}}$

A blank required. Balance the reaction and find the molarity of Molybdate in the unknown.

A 25.00mL volume of commercial hydrogen peroxide solution was diluted to 250.0mL in a volumetric flask. Then 250.0 of the diluted solution were mixed with 200mL of water and 20mLof $\mathbf{3}{\mathbf{MH}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$ and titrated with $\mathbf{0}\mathbf{.}\mathbf{02123}{\mathbf{MKMnO}}_{\mathbf{4}}$. The first pink color was observed with 27.66mL of titrant. A blank prepared from water in place of${\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}$ required 0.04Ml to give visible pink color. Using the${\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}$ reaction in Table16 - 3, find the molarity of the commercial ${\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}$.

A titration of 50.0 mLof unknown ${\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}$with 0.100MCeat $\mathbf{25}\mathbf{°}\mathbf{C}$ , monitored with Pt and calomel electrodes, gave data in the table. Prepare a Gran plot and decide which data lie on a straight line. Find the x-intercept of this line, which is the equivalence volume. Calculate the molarity of ${\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}$in the unknown.

Two possible reactions of ${\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}$with${\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}$to produce${\mathbf{O}}_{\mathbf{2}}$andare${\mathbf{Mn}}^{\mathbf{+}}$

Scheme:$$\begin{gathered} {\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\to }{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}} \\ {\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\to }{\mathbf{O}}_{\mathbf{2}} \end{gathered}$$

Scheme:$$\begin{gathered} {\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\to }{\mathbf{O}}_{\mathbf{2}}\mathbf{+}{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}} \\ {\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\to }{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \end{gathered}$$

(a) Complete the half reactions for both schemes by adding ${\mathbf{e}}^{\mathbf{+}}$and ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$and ${\mathbf{H}}^{\mathbf{+}}$write a balanced net equation for each scheme.

(b) Sodium peroxyborate tetrahydrate, ${\mathbf{NaBO}}_{\mathbf{3}}\mathbf{\cdot }\mathbf{4}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{(}\mathbf{FM}\mathbf{153}\mathbf{.}\mathbf{86}\mathbf{)}$produces ${\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}$when dissolved in acid ${\mathbf{BO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\to }{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}{\mathbf{BO}}_{\mathbf{3}}^{\mathbf{-}}$. To decide whether Scheme 1 or 2 Schemeoccurs student at the U.S. Naval academy weighed $\mathbf{0}\mathbf{.}\mathbf{123}{\mathbf{gNaBO}}_{\mathbf{3}}\mathbf{.}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$into a 100mLvolumetric flask added 20mLof $\mathbf{1}\mathbf{M}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$and diluted to the mark with ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$. Then they titratedof this solution with$\mathbf{0}\mathbf{.}\mathbf{01046}{\mathbf{MKMnO}}_{\mathbf{4}}$until the first pale pink color persisted. How may mL of${\mathbf{KMnO}}_{\mathbf{4}}$are required in Scheme 1and 2 Scheme?

(The Scheme 1stoichiometry was observed).

A sample containing ${\mathbf{La}}^{\mathbf{3}\mathbf{+}}$ was treated with sodium oxalate to precipitate${\mathbf{La}}_{\mathbf{2}}\mathbf{(}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}{\mathbf{)}}_{\mathbf{3}}$ which was washed, dissolved in acid and titrated with 18.04mL of$\mathbf{0}\mathbf{.}\mathbf{006363}{\mathbf{MKMnO}}_{\mathbf{4}}$ . Write the titration reaction and find$\mathbf{\left[}{\mathbf{La}}^{\mathbf{3}\mathbf{+}}\mathbf{\right]}$ in the unknown.