Chemistry - Electrochemistry Question with Solution | TestHub

(A) During electrolysis using inert electrodes:

Anode: ${\mathrm{H}}_2\mathrm{O}(\ell )⟶\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})+2{\mathrm{H}}^{+}(\mathrm{aq})+2{\mathrm{e}}^{-}$

Cathode: $\mathrm{C}{\mathrm{u}}^{2+}(\mathrm{aq})+2{\mathrm{e}}^{-}⟶\mathrm{Cu}(\mathrm{s})$

${\mathrm{n}}_{\mathrm{F}}={\mathrm{n}}_{{\mathrm{H}}^{+}}=\left(\frac{100}{1000}\right)(0.1)=10^{-2}$

$\mathrm{Q}=965\mathrm{C}=(0.8\mathrm{I})(\mathrm{t})$

$(\mathrm{t})=\frac{965}{(.8)(0.965)}=\frac{10000}{8}=1250\sec$

(B) $2\mathrm{CuS}{\mathrm{O}}_4(\mathrm{aq})+4\mathrm{KI}(\mathrm{aq})⟶\mathrm{C}{\mathrm{u}}_2{\mathrm{I}}_2(\mathrm{s})+{\mathrm{I}}_2+2{\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})$

Equivalent wt of $\mathrm{CuS}{\mathrm{O}}_4=\frac{\text{ Molar mass }}{1}$

(C) ${\mathrm{I}}_2(\mathrm{aq})+2\mathrm{N}{\mathrm{a}}_2{\mathrm{S}}_2{\mathrm{O}}_3(\mathrm{aq})⟶2\mathrm{N}{\mathrm{a}}_2{\mathrm{S}}_4{\mathrm{O}}_6(\mathrm{aq})+2\mathrm{NaI}(\mathrm{aq})$

Eq.wt of $\mathrm{N}{\mathrm{a}}_2{\mathrm{S}}_2{\mathrm{O}}_3=\frac{\text{ Molar mass }}{1}$

(D)

$$\begin{array}{rl}{\left({\mathrm{n}}_{\mathrm{CuS}{\mathrm{O}}_4}\right)}_{\text{Total }}& =\left(\frac{{\mathrm{n}}_{\mathrm{F}}}{2}\right)+{\mathrm{n}}_{\text{hypo }}\\ & =\left(\frac{10^{-2}}{2}\right)+\left(0.04\times \frac{35}{1000}\right)\\ & =5\times 10^{-3}+1.4\times 10^{-3}\\ & =6.4\times 10^{-3}\\ \left[\mathrm{CuS}{\mathrm{O}}_4\right]& =\frac{6.4\times 10^{-3}}{0.1}=0.064\mathrm{M}\end{array}$$