Question

silver is electro deposit on a metallic vessel of surface area 800cm^2 by passing a current of 0.2A for 3 hours . calculate the thickness of silver deposit . density of silver is 10.47gm/cc.​

Answer 1

$\fbox{\texttt{\green{Answer:}}}$

Thickness of silver deposited = $\bold{2.88\times{}10^{-4}\:cm}$

$\fbox{\texttt{\pink{Given:}}}$

Total surface area of the vessel = 800 cm²

Density of silver = 10.47 g/cc

Atomic mass of Ag = 107.92 amu

$\fbox{\texttt{\blue{To\:find:}}}$

Thickness of silver deposited.

$\fbox{\texttt{\red{How\:to\:Find:}}}$

Quantity of electricity passed = 0.2 × 3 × 60 × 60 C = 2160 C

For deposition of Ag, reaction is :

$\bold{Ag^++e^----)Ag}$

96500 C deposit Ag = 107.92 g

2160 C will deposit Ag =

$\bold{\frac{107.92}{96500}\times{}2160\:g=2.4156\:g}$

Volume deposited =

$\bold{\frac{Mass}{Density}=\frac{2.4156}{10.47}cc=0.2307\:cc}$

$\rule{200}2$

$\therefore$Thickness of silver deposited =

$\bold{\frac{Volume}{Area}=\frac{0.2307}{800}=2.88\times{}10^{-4}\:cm}$