Question

If the same quantity of electricity is passed through solutions of ${AgNO}_3$, ${CuSO}_4$ and ${CrCl}_3$ the ratio of the respective weights of products at cathode will be : $(Cu=63.5,Ag=108,Cr=52)$

• A. $108:63.5:52$
• B. $54:63.5:104$
• C. $108:31.75:17.3$
• D. $108:127:104$

Answer 1

The correct option is C $108:31.75:17.3$

${Ag}^{+}+e^{-}\to Ag{Cu}^{2+}+{2e}^{-}\to Cu{Cr}^{3+}+3e^{-}\to Cr$

Let the quantity of electricity passed be $F$ then,

for eg $Ag$ $F$ moles deposited,

$Cu$ $\frac{F}{2}$ moles deposited

$Cr$ $\frac{F}{3}$ moles deposited

We know, no. of moles$=\frac{weight}{molarmass}$

for $Ag,F=\frac{W_1}{108}\Rightarrow W_1=108FCu,\frac{F}{2}=\frac{W_2}{63.5}\Rightarrow W_2=\frac{63.5F}{2}Cr,\frac{F}{3}=\frac{W_3}{52}\Rightarrow W_3=\frac{52F}{3}{W}_1:W_2:W_3=108F:\frac{63.5F}{2}:\frac{52F}{3}=108:31.75:17.33$.