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.

  • $\begingroup$ 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, 2019 at 20:45

1 Answer 1


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.

  • $\begingroup$ Thank you so much for your nice explanation!! $\endgroup$
    – Hansoo
    Nov 20, 2019 at 14:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.