Answer:
Time take to deposit Ni is 259.02 sec.
Explanation:
Given:
Current [tex]I = 14.9[/tex] A
Faraday constant [tex]= 96485[/tex] [tex]\frac{C}{mole}[/tex]
Molar mass of Ni [tex]= 58.69[/tex] [tex]\frac{g}{mole}[/tex]
Mass of Ni [tex]= 0.94[/tex] g
First find the no. moles in Ni solution,
Moles of Ni [tex]= \frac{0.94}{58.69}[/tex]
[tex]= 0.02[/tex] mol
From the below reaction,
[tex]Ni^{2+} + 2e[/tex] ⇆ [tex]Ni_{(s)}[/tex]
Above reaction shows "1 mol of [tex]Ni^{2+}[/tex] requires 2 mol of electron to form 1 mol of [tex]Ni_{(s)}[/tex] "
So for finding charge flow in this reaction we write,
[tex]=[/tex] [tex]0.02 \times \frac{2 }{1 } \times 96485 \frac{C}{mol}[/tex]
Charge flow [tex]= 3859.4[/tex] C
For finding time of reaction,
[tex]I = \frac{q}{t}[/tex]
Where [tex]q =[/tex] charge flow
[tex]t = \frac{q}{I}[/tex]
[tex]t = \frac{3859.4}{14.9}[/tex]
[tex]t = 259.02[/tex] sec
Therefore, time take to deposit Ni is 259.02 sec.
