Question

A. State two advantages of \[{{\text{H}}_{\text{2}}}{\text{ - }}{{\text{O}}_{\text{2}}}\] fuel cell over ordinary cell.
B.Silver is electrodeposited on a metallic vessel of total surface area 900 ${\text{c}}{{\text{m}}^{\text{2}}}$ by passing a current of $0.5$ ampere for two hours. Calculate the thickness of silver deposited. (Density of silver${\text{10}}{\text{.5 g/cc}}$, atomic mass of silver = 108, one Faraday = 96500 coulombs)

Answer 1

Hint:A fuel cell is an electrochemical cell that converts the chemical energy of a fuel (which is often hydrogen) and an oxidising agent (oxygen) into electrical energy through a pair of redox reactions. The amount of silver deposited can be calculated using Faraday’s Laws of Electrolysis (1 mole of silver is deposited by 1F charge.)

Complete step by step answer:
Fuel cells are more advantageous over the ordinary cells as:
Fuel cells can produce electricity continuously for as long as the fuel, which is hydrogen, and the oxidant, which is oxygen are supplied.
These cells do not cause pollution and the source of the fuel and the oxidant are both environment-friendly.
Silver dissociates on the passage of electricity as follows:
${\text{Ag }} \rightleftarrows {\text{A}}{{\text{g}}^{\text{ + }}}{\text{ + }}{{\text{e}}^{\text{ - }}}$
One faraday is defined as the electrical charge on 1 mole of electrons, therefore the charge for n moles of electrons is nF. Where 1 faraday = 96500 coulombs
Thus the quantity of electricity required to deposit one mole of silver,
Q = 96500 coulombs.
The quantity of electricity passed is equal to the product of the current supplied and the duration for which it was supplied.
Therefore, charge = $0.5 \times 2 \times 60 \times 60 = 3600$ Coulombs
As given, the molar mass of silver is 108 ${\text{g/ mol}}$.
So one Faraday of charge = 96500 coulombs, should be able to deposit 108 ${\text{g/ mol}}$ of silver.
Therefore, 3600 coulombs of electricity would produce, $\dfrac{{108}}{{96500}} \times 3600 = 4.03$ grams.
The volume of silver deposited = $\dfrac{{{\text{mass}}}}{{{\text{density}}}}$=$\dfrac{{4.03}}{{10.5}} = 0.384$c.c., Density of silver${\text{10}}{\text{.5 g/cc}}$.
Therefore thickness of the silver deposited = $\dfrac{{{\text{volume}}}}{{{\text{surface area}}}} = \dfrac{{0.384}}{{900}} = 4.27 \times {10^{ - 4}}$ cm.
So the thickness of silver deposited is $4.27 \times {10^{ - 4}}$ cm.

Note:
Electrodeposition or electroplating is a broad term for a number of electrical process that includes electrocoating, cathodic electrodeposition, anodic electrodeposition, and electrophoretic coating, or electrophoretic painting.