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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

enter image description here

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.

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  • $\begingroup$ 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 $\endgroup$ – user78585 Nov 19 '19 at 20:45
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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.

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

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