Question

A metal surface of 100 cm$^2$ area has to be coated with nickel layer of thickness 0.001 mm. A current of 2 A was passed through a solution of Ni(NO$_3$)$_2$ for 'x' seconds to coat the desired layer. The value of x is ________ (Nearest integer) ($\rho_{Ni}$ (density of Nickel) is 10 g mL, Molar mass of Nickel is 60 g mol$^{-1}$ F = 96500 C mol$^{-1}$)

Answer 1

161

Answer 2

1. Calculate the volume of the nickel layer: Area = 100 cm$^2$ Thickness = 0.001 mm = $0.001 \times 10^{-1}$ cm = $10^{-4}$ cm Volume ($V$) = Area $\times$ Thickness = 100 cm$^2$ $\times$ $10^{-4}$ cm = $10^{-2}$ cm$^3$

2. Calculate the mass of nickel to be deposited: Density of Nickel ($\rho_{Ni}$) = 10 g mL$^{-1}$ = 10 g cm$^{-3}$ Mass ($m$) = Volume $\times$ Density = $10^{-2}$ cm$^3$ $\times$ 10 g cm$^{-3}$ = 0.1 g

3. Determine the charge required for deposition using Faraday's laws: The electrolysis of Ni(NO$_3$)$_2$ involves the reduction of Ni$^{2+}$ ions: Ni$^{2+}$(aq) + 2e$^-$ $\rightarrow$ Ni(s) This means $n=2$ moles of electrons are required to deposit 1 mole of nickel. Molar mass of Nickel ($M$) = 60 g mol$^{-1}$ Faraday's constant ($F$) = 96500 C mol$^{-1}$ Using Faraday's first law: $m = \frac{M \times Q}{n \times F}$ Rearranging to find charge ($Q$): $Q = \frac{m \times n \times F}{M}$ $Q = \frac{0.1 \text{ g} \times 2 \times 96500 \text{ C mol}^{-1}}{60 \text{ g mol}^{-1}} = \frac{19300}{60} \text{ C} = \frac{965}{3} \text{ C}$

4. Calculate the time taken (x): The relationship between charge ($Q$), current ($I$), and time ($t$) is $Q = I \times t$. Given current $I = 2$ A. $x = \frac{Q}{I} = \frac{965/3 \text{ C}}{2 \text{ A}} = \frac{965}{6} \text{ s}$ $x \approx 160.8333...$ s

5. Round to the nearest integer: The value of $x$ rounded to the nearest integer is 161.