Find the mass of silver deposited when a current of 1.5 ampere is passed through silver nitrate solution for 5 min.
(Atomic mass of $A g$ is $108$ g)

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Solution

Molar mass of $A g$ $= 108$ g mol $^{- 1}$
1 Faraday $= 96500$ C
$Q = i \times t = 1.5$ $\times 5 \times 60 = 450$ Coulombs
Weight of the substance deposited $= Z Q$
$Z = \frac{M}{n F}$ n-factor of $A g N O_{3}$ =1
$Z = \frac{108}{1 \times 96500}$
Weight of the substance= $\frac{108}{96500} \times 450 = \frac{54 \times 90}{9650} = 0.5$ g

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Find the mass of silver deposited when a current of 1.5 ampere is passed through silver nitrate solution for 5 min.(Atomic mass of Ag is 108 g)

Molar mass of Ag =108 g mol−11 Faraday =96500 CQ=i×t=1.5 ×5×60=450 CoulombsWeight of the substance deposited =ZQZ=MnF n-factor of AgNO3=1Z=1081×96500Weight of the substance=10896500×450=54×909650=0.5 g

Find the mass of silver deposited when a current of 1.5 ampere is passed through silver nitrate solution for 5 min.
(Atomic mass of $A g$ is $108$ g)