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(C) The electrochemical equivalent $Z$ is defined as $Z = \frac{E}{F}$,where $E$ is the equivalent weight and $F$ is Faraday's constant.Since $F$ is c

Vedclass · Accepted Answer

$1 : 31.75 : 108$

Vedclass · Answer

(C) The electrochemical equivalent $Z$ is defined as $Z = \frac{E}{F}$,where $E$ is the equivalent weight and $F$ is Faraday's constant.Since $F$ is constant,the ratio $Z_{1} : Z_{2} : Z_{3}$ is equal to the ratio of their equivalent weights $E_{1} : E_{2} : E_{3}$.For $H_{2}$ $(2H^+ + 2e^- ightarrow H_2)$: $E_{1} = \frac{	ext{Molar mass}}{	ext{n-factor}} = \frac{2}{2} = 1$.For $Cu$ $(Cu^{2+} + 2e^- ightarrow Cu)$: $E_{2} = \frac{	ext{Molar mass}}{	ext{n-factor}} = \frac{63.5}{2} = 31.75$.For $Ag$ $(Ag^+ + 1e^- ightarrow Ag)$: $E_{3} = \frac{	ext{Molar mass}}{	ext{n-factor}} = \frac{108}{1} = 108$.Therefore,$Z_{1} : Z_{2} : Z_{3} = E_{1} : E_{2} : E_{3} = 1 : 31.75 : 108$.

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