Question

Silver and a zinc voltameter are connected are connected in series and a current is passed through them for a time t, liberating W g of zinc. The weight of silver deposited is nearly:

• A. $Wg$
• B. $1.7Wg$
• C. $2.4Wg$
• D. $3.3Wg$

Answer 1

The correct option is D $3.3Wg$

Mass of the substance deposited (or liberated) due to a current I passed through them for a time $t$, is given by

$m=ZIt$

where, $Z$ is electro-chemical equivalent and it depends on atomic weight

$\left(A\right)$ and valency$\left(V\right)$, it is given by

$Z=\frac{A}{FV}$

where, $F$ is Faraday's constant.

Therefore, for silver and zinc we can write

$m_{Ag}=\frac{A_{Ag}}{F{V}_{Ag}}$ and $m_{Zn}=\frac{A_{Zn}}{F{V}_{Zn}}$

Dividing these equations, we get

$\frac{m_{Ag}}{m_{Zn}}=\frac{\frac{A_{Ag}}{F{V}_{Ag}}}{\frac{A_{Zn}}{F{V}_{Zn}}}$

Using, $A_{Ag}=108$, $V_{Ag}=1$, $A_{Zn}=65.39$, $V_{Zn}=2$ and $m_{Zn}=W$ we get

$\frac{m_{Ag}}{W}=\frac{\frac{108It}{F\left(1\right)}}{\frac{65.39It}{F\left(2\right)}}$

$\frac{m_{Ag}}{W}=\frac{108\left(2\right)}{65.39}$

$m_{Ag}=3.303Wg\approx 3.3Wg$