Question

How would I do a calculation for an electroplating of solid metal onto the cathode in an electrolytic cell if, say, I am given current and the time that it took? What helps increase the mass yielded?

Answer 1

Here's my explanation

Explanation:

According to Ohm's law the current passed through a material is directly proportional to the voltage.

You can see the mathematical equation that describes this proportionality

`I = V/R`

`V = IR`

Where `I` is the current in the units of "amperes"
Where `V` is voltage
and `R` is the resistance of the material in units of "ohms"

The other equation

`E = Jsigma`

Where E is electric field at the given location
J is current density
σ (sigma) is a material-dependent parameter called the conductivity

`R =\rho\frac l A`

R is the electrical resistance of a uniform specimen of the material
`l` is the length of the piece of material
A is the cross-sectional area of the specimen

Simplifying

`\rho = \R\frac A l`

The resistance of a given material increases with length, but decreases with increasing cross-sectional area

If the `A = 1m^2` and `l=1` the resistance of the conductor is equal to the the ρ

ρ depends on the conductor and is a constant

Here's a table of ρ of the conductor at `20^@C`

http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/rstiv.html

Note that ρ is different for conductors at different temperature

ρ increases as the temperature increases
ρ decreases as the temperature decreases

`sigma = 1/ρ`

Where `sigma` is conductivity
ρ is resistivity

There are many version of the Ohm's law

`P = (ΔV^2) / R`

`DeltaV` is the potential difference between the two points in the conductor

`P = I^2 • R`

Now lets come to the main problem.

`ΔV =( ΔPE )/ Q`

Where Q is charge
PE is energy

By this you can understand that Voltage is proportional to charge which is proportional to coulombs

As

`C = V * Q`
There is a relationship between V and C

As the V increases and the Q is thought to be constant the C increases. Means more voltage more coulombs

Now let me show you an example

Suppose that `2C`/s is applied to a solution of `CuSO4` solution for 2 seconds.

The reaction

= `2H_2O + 2CuSO_4 rarr O_2 + 2H_2SO4 + 2Cu`

Net ionic reaction

`2H_2O + Cu^(2+) rarr 2H_2 + 2Cu`

Each mole of `Cu^(2+)` produces one mole of `Cu`

`2H_2O` is reduced to `H_2`

`H_2O + rarr H_2 + 2e^-`
`Cu^(2+) + 2e^-) rarr Cu`

Thus
`2H_2O rarr 2H_2 + 4e^-`
`2Cu^(2+) + 4e^-) rarr Cu`

The whole reaction

`2H_2O + 2Cu^(2+) + 4e^-) rarr 2H_2 + Cu + 4e^-)`
`2H_2O + 2Cu^(2+) rarr 2H_2 + Cu`

As 2C/s is applied for 2s

`(2C)/cancel(s) xx 2cancel(s)`

= `4C`

Use the Faraday's constant to calculate the moles of electrons lost and gained in total.

`( 4 cancel"Coulombs")xx ("1 mole of electrons") / (96500 cancel"Coulombs") = "0.00004145077 mole of electrons"`

1 mole of `Cu^(2+)` is reduced per 2 mole electrons

Thus moles of `Cu` formed

`"1 mol of Cu "/(2 molcancel( e^-)) xx "0.00004145077 mol of" cancel(e^-)`

`1/2 xx 0.00004145077 = "0.00002072538 mol of Cu"`

Now see what happens if we apply more charge

Consider the charge as 3C/s for 2s

Do the same calculations

As 3C/s is applied for 2s

`(3C)/cancel(s) xx 2cancel(s)`

= `6C`

`( 6 cancel"Coulombs")xx ("1 mole of electrons") / (96500 cancel"Coulombs") = "0.00006217616 mole of electrons"`

`"1 mol of Cu "/(2 molcancel( e^-)) xx "0.00004145077 mol of" cancel(e^-)=" 0.00003108808 mol of Cu"`

`"0.00003108808 mol of Cu" > "0.00002072538 mol of Cu"`

Thus more the current,voltage and coulomb the more the mol of copper formed thus more the mass of copper formmed

Plotting a graph

If my handwriting is too bad

y = mol of Cu formed from reduction of `Cu^(2+)`
x = mol of electrons used in the process

Answer 2

One can impart current to force a nonspontaneous chemical reaction, such as electroplating solid onto the cathode of an electrolytic cell. Increasing the voltage increases the mass plated by stoichiometry.

Here is a calculation example.

Suppose we have to plate `"Cu"`, and we supply `"6.30 A"` of current through a `"Cu"("NO"_3)_2` solution for `"14.0 min"`. What mass is plated?

The preliminary calculation for `"6.30 A"` of current supplied would be:

`14.0 cancel"min" xx (60 cancel"s")/cancel"1 min" xx (6.30 cancel"C")/cancel"s" xx cancel("1 mol e"^(-))/(96485 cancel"C") xx (2 cancel"mol Cu")/cancel("1 mol e"^(-)) xx "63.55 g Cu"/cancel"1 mol Cu"`

`=` `color(green)("6.97 g Cu")`

From Ohm's law, since

`V = IR`,

where `V` is the voltage available in `"V"`, `I` is the current supplied in `"C/s"`, and `R` is the resistance in `"V"cdot"s/C"` (or `Omega`), we have a direct proportionality relationship between voltage and current.

Since voltage is directly proportional to the current supplied, a higher-voltage supply with the same resistance would increase the amount of mass plated.