4 - Redox Reactions and n factor Questions Answers

In first case reactions involved are$$\begin{gathered} \mathrm{Reduction} \mathrm{of} {\mathrm{Fe}}^{+++}, {\mathrm{Fe}}^{+++}\to {\mathrm{Fe}}^{++} \\ \mathrm{Titration} \mathrm{step}, 6{\mathrm{Fe}}^{++}+{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\to 6{\mathrm{Fe}}^{+++}+2{\mathrm{Cr}}^{+++} \\ \mathrm{By} \mathrm{using} \mathrm{titration} \mathrm{reaction}, \mathrm{solve} \mathrm{the} \mathrm{question} \end{gathered}$$

 In second case reaction involved is${\mathrm{As}}_4{\mathrm{O}}_6+4{{\mathrm{I}}_3}^{-}\to {\mathrm{As}}_4{\mathrm{O}}_{10}+12{\mathrm{I}}^{-}$Now solve

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Third reaction is the sum of 1st and 2nd so energy of 3rd will be the sum of energies of 1st and 2nd.

delta G = n₁FE₁ + n₂FE₂ = 1F(0.771) + 2F(0.447) = F(0.771+0.894) = 1.665F = 1.665*96368 = 160452.7 J = 38203.03 cal = 38.20303 cal

so total energy of the reaction 

Fe³⁺(aq) + 3e^- -------> Fe(s) is 38.203 cal

and value of n = 3 so per mole energy = 38.203/3 

2 EO₂ + 4 KOH + O₂ → 2 K₂EO₄ + 2 H₂O

and

3 K₂EO₄ + 2 CO₂ → 2 KEO₄ + 2 K₂CO₃ + EO₂

 

so oxide will be EO₂