Calculate voltage, cost and time of nickel coating

To nickel with a bath of $$\ce{NiSO4}$$ a cathodic current of $$\pu{0.3 A/cm2}$$ is applied with a square plate of side $$\pu{5 cm}$$ is used to nickel a certain substrate. Due to the formation of $$\ce{H2}$$ the current efficiency is $$75\%.$$ If the resistance of the solution is $$\pu{0.4 \Omega}$$ and the price of the current $$\pu{1 c€/kWh},$$ determine the voltage to be applied, the energy cost and the required time to nickel a total surface of $$\pu{2 m2}$$ with a coating thickness of $$\pu{0.03 cm}.$$
Data: $$M(\ce{Ni}) = \pu{58.71 g/mol}$$; $$\delta(\ce{Ni}) = \pu{8.9 g/cm3}.$$

First I’m going to show you what I got

I’m sure this is wrong because I didn’t use given data. But I don’t know where should I use mass of nickel.

I’ve never seen this type of exercise before and I don’t where should I start from.

• This is just a friendly suggestion but in such questions I always recommend writing down all known quantities before starting so that one doesn't end up jumbled in all the data – user78585 Nov 19 '19 at 20:45

The volume of nickel plated is surface area times the thickness. That's $$2m^2 \times \pu{0.03 cm}$$.

We'll use cubic cm as the units so it is: $$2m \times 1m \times \pu{0.03 cm} = \pu{200 cm} \times \pu{100 cm} \times \pu{0.03 cm} = \pu{600 cm3}$$

This helps to find the mass as the density is given as $$\pu{8.9 g/cm3}$$.

The mass = density $$\times$$ volume = $$\pu{8.9 g/cm3} \times \pu{600 cm3} = \pu{5.34 kg}$$

Convert this to moles of Nickel: $$\displaystyle \frac{5.34 \times \pu{10^3 g}}{\pu{58.71 g/mol}} = \pu{90.96 mol}$$. We'll come back to this.

Now for the current, $$\pu{0.3 A/cm2}$$ applied to a $$5 \times \pu{5 cm}$$ square plate, meaning that the total current on the plate at a given time is $$\pu{0.3 A} \times \pu{25 cm2} = \pu{7.5 A}$$ or $$\pu{7.5 C/s}$$. However, only a fraction of that actually reduces the nickel ($$75\%$$) so $$\pu{5.625 A}$$.

The voltage is then $$V = IR = 5.625 \times 0.4$$

The time calculation involves the actual nickel. $$I = \displaystyle \frac{q}{t}$$
Where $$q =$$ total charge

We have $$\pu{90.96 mol}$$ of Nickel reduced.

$$\ce{Ni^{2+} + 2e^- -> Ni}$$

That's $$\pu{2 mol}$$ of electrons per mole of Nickel. So we need $$\pu{181.92 mol}$$ of electrons.

The charge per mol of electrons is Faraday's constant of $$\pu{96485.3 C/mol}$$ The total charge is just $$\pu{181.92 mol} \times \pu{96485.3 C/mol}$$.

Then we rearrange the formula of current for time $$\displaystyle t = \frac{q}{I} = \frac{\pu{181.92 mol} \times \pu{96485.3 C/mol}}{\pu{5.625 C/sec}}$$ $$\text{time} = \pu{3120463 seconds}$$ or $$\pu{866 hours}$$.

That's sort of the basic idea I'd imagine. I may have made a mistake somewhere in between. The time though while seems long is in line with this: https://sciencing.com/calculate-electroplating-7597391.html

There, they plated just one mole of Cu and with a higher current. That took 2 hours. We have a lot more of $$\ce{Ni}$$ and smaller current.

Then $$\text{Energy} = \text{power} \times \text{time} = VIt$$ and you can calculate the cost from there.

• Thank you so much for your nice explanation!! – Hansoo Nov 20 '19 at 14:59