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?

2 Answers
Apr 30, 2017

Here's my explanation

Explanation:

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

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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

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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

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If my handwriting is too bad

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

Apr 30, 2017

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.