Solveeit Logo
Login

Question

Question: A current of 40 microamperes is passed through silver nitrate solution for 16 minutes using platinum...

A current of 40 microamperes is passed through silver nitrate solution for 16 minutes using platinum electrodes. Also, 50% of cathode is occupied by a single atom thick silver layer. Calculate the total surface area of the cathode is one silver atom occupies 5.5×\times 10−16cm2{10^{ - 16}}c{m^2}surface area.

Explanation

Solution

Hint: To calculate the deposited silver, firstly we have to follow Faraday’s first law of electrolysis. We need to understand what it states and then by using its formula, it would be easy to reach our solution.

Complete step-by-step answer:
Faraday’s Laws of electrolysis are based upon the electrochemical research published by Michael Faraday in 1833. The Laws state that the amount of material liberated/produced from an electrode during the reaction is directly proportional to the total conducted charge.
Faraday’s first law of electrolysis is defined as the quantity of elements separated by passing an electric current through a molten or dissolved salt is proportional to the quantity of electric charge passed through the circuit.
The mass of substance(m)\left( m \right) deposited or liberated at an electrode is directly proportional to the quantity of electricity or charge(Q)\left( Q \right)passed.
In this equation k is equal to the electrochemical constant,
i.e. m=k.qm = k.q
Or m=e.Qm = e.Q
Now, as we know that the mass of substance(m)\left( m \right) deposited or liberated at an electrode is directly proportional to the quantity of electricity or charge(Q)\left( Q \right) passed.
m=k.qm = k.q
m=e.Qm = e.Q

Mass of silver deposits may be calculated according to Faraday’s first law of electrolysis.
W = ItE96500W\, = \,\dfrac{{ItE}}{{96500}}
=40×10−6×60×16×10896500= \dfrac{{40 \times {{10}^{ - 6}} \times 60 \times 16 \times 108}}{{96500}}
= 42.976× 10−6g= \,42.976 \times \,{10^{ - 6}}g
Total no. of deposited ‘Ag’ atoms
= 42.976× 10−6108×6.023×1023= \,\dfrac{{42.976 \times \,{{10}^{ - 6}}}}{{108}} \times 6.023 \times {10^{23}}
=2.3967× 1017= 2.3967 \times \,{10^{17}} atom
Surface occupied by deposited silver =Number of silver atoms ×\times Area occupied by single atom
⇒  2.3967×5.5×10−16 ⇒  131.818 cm2  \Rightarrow \,\,2.3967 \times 5.5 \times {10^{ - 16}} \\\ \Rightarrow \,\,131.818\,c{m^2} \\\
Since, deposited silver occupies 50% of total area of cathode,
Hence, total surface area of cathode
⇒  2×131.818  ⇒  263.636cm2  \Rightarrow \,\,2 \times 131.818\, \\\ \Rightarrow \,\,263.636c{m^2} \\\

NOTE – We have used Faraday’s first law in this question, which states that when an electric current is passed through dissolved or molten salt, the quantity of elements separated is proportional to the quantity of electric charge passed through the circuit. Hence, we used this law to reach our result.