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• Redox reactions are oxidation-reduction chemical reactions in which the oxidation states of the reactants change. The term redox refers to the reduction-oxidation process.
• All redox reactions can be divided down into two types of reactions: reduction and oxidation.
• In a redox reaction, or Oxidation-Reduction process, the oxidation and reduction reactions always happen at the same time.

To balance the equation we have to write two half-reactions so we can determine the ratio of moles

$$\begin{gathered} \mathrm{Reduction}:{\mathrm{MnO}}_4^{-}+5{\mathrm{e}}^{-}+8{\mathrm{H}}^{+}鈬�{\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}/鈰�3 \\ \mathrm{Oxidation}:{\mathrm{Mo}}^{3+}+2{\mathrm{H}}_2\mathrm{O}鈬�{\mathrm{MoO}}_2^{2+}+3{\mathrm{e}}^{-}+4{\mathrm{H}}^{+}/鈰�5 \end{gathered}$$

To balance the number of electrons on the left and right side, we multiply the first reaction with 3 and the second reaction with 5 and we get:

$$\begin{gathered} \mathrm{Reduction}:3{\mathrm{MnO}}_4^{-}+15{\mathrm{e}}^{-}+24{\mathrm{H}}^{+}鈬�3{\mathrm{Mn}}^{2+}+12{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Oxidation}:5{\mathrm{Mo}}^{3+}+10{\mathrm{H}}_2\mathrm{O}鈬�5{\mathrm{MoO}}_2^{2+}+15{\mathrm{e}}^{-}+20{\mathrm{H}}^{+} \end{gathered}$$

By adding up these two reactions we get:

$3{\mathrm{MnO}}_4^{-}+15\mathrm{e}+24{\mathrm{H}}^{+}+5{\mathrm{Mo}}^{3+}+10{\mathrm{H}}_2\mathrm{O}鈬�3{\mathrm{Mn}}^{2+}+12{\mathrm{H}}_2\mathrm{O}+5{{\mathrm{MoO}}_2}^{2+}+15\mathrm{e}+20{\mathrm{H}}^{-}$

Net reaction is:

$3{{\mathrm{MnO}}_4}^{-}+5{\mathrm{Mo}}^{3+}+4{\mathrm{H}}^{+}鈬�3{\mathrm{Mn}}^{2+}+5{{\mathrm{MoO}}_2}^{2+}+2{\mathrm{H}}_2\mathrm{O}$

To find the molarity of molybdate, we have to calculate the molarity of ${\mathrm{KMnO}}_4$and put it in ratio with ${\mathrm{Mo}}^{3+}$. The data that is given:

$$\begin{gathered} \mathrm{V}\left(\mathrm{filtrate}\right)=16.43鈥�\mathrm{mL} \\ \mathrm{V}\left(\mathrm{blank}\right)=0.04鈥�\mathrm{mL} \\ \mathrm{c}\left({\mathrm{KMnO}}_4\right)=0.01033鈥�\mathrm{M} \\ \mathrm{V}\left(\mathrm{analyte}\right)=\mathrm{V}\left({\mathrm{Mo}}^{3+}\right)=25.00鈥�\mathrm{mL} \end{gathered}$$

We can calculate the volume of ${\mathrm{KMnO}}_4$by subtracting the volume of filtrate with the volume of blank:

$$\begin{gathered} \mathrm{V}\left({\mathrm{KMnO}}_4\right)=\mathrm{V}\left(\mathrm{filtrate}\right)-\mathrm{V}\left(\mathrm{blank}\right) \\ \mathrm{V}\left({\mathrm{KMnO}}_4\right)=(16.43-0.04)鈥�\mathrm{mL} \\ \mathrm{V}\left({\mathrm{KMnO}}_4\right)=16.39鈥�\mathrm{mL} \end{gathered}$$

The ratio of moles of ${\mathrm{MO}}^{3+}$and ${\mathrm{KMnO}}_4$and 5 : 3 is so we can write:

$$\begin{gathered} \mathrm{n}\left({\mathrm{Mo}}^{3+}\right)=5/3\mathrm{n}\left({\mathrm{KMnO}}_4\right) \\ \mathrm{c}\left({\mathrm{Mo}}^{3+}\right)鈰�\mathrm{V}\left({\mathrm{Mo}}^{3+}\right)=5/3鈰�\mathrm{c}\left({\mathrm{KMnO}}_4\right)鈰�\mathrm{V}\left({\mathrm{KMnO}}_4\right)/:\mathrm{V}\left({\mathrm{Mo}}^{3+}\right) \\ \mathrm{c}\left({\mathrm{Mo}}^{3+}\right)=\frac{5鈰�\mathrm{c}\left({\mathrm{KMnO}}_4\right)鈰�\mathrm{V}\left({\mathrm{KMnO}}_4\right)}{(3鈰�\mathrm{V}({\mathrm{Mo}}^{3+})} \\ \mathrm{c}\left({\mathrm{Mo}}^{3+}\right)=\frac{(5鈰�0.01033\mathrm{M}鈰�16.39鈥�\mathrm{mL})}{(3鈰�25鈥�\mathrm{mL})} \\ \mathrm{c}\left({\mathrm{Mo}}^{3+}\right)=0.0113鈥�\mathrm{M} \end{gathered}$$

Hence, the molarity of the molybdate is$\mathrm{c}\left({\mathrm{Mo}}^{3+}\right)=0.0113鈥�\mathrm{M}$