# Question bbed8

May 1, 2017

It has nothing to do with ions or anything like.

$J = \frac{I}{A \left({m}^{2}\right)}$

Where J is current density

$A$ is the cross sectional area in ${m}^{2}$

I is the charge in the total area

$1 A = \frac{\text{1 coulomb}}{\sec}$

This means more the current density more the current

As a simple example, assume the current density is uniform (equal density) across the cross section of a wire with area $3 {m}^{2}$. Suppose that the total current flow is $I = 3.2 A .$ What is the current density?

$J = \frac{3.2 A}{3 {m}^{2}}$

$1.067 \text{A/} {m}^{2}$

Increasing the current

zappppp!!!!!!!

Now assume the current density is uniform (equal density) across the cross section of a wire with area $3 {m}^{2}$. Suppose that the total current flow is $I = 18 A .$ What is the current density

$J = \frac{18 A}{3 {m}^{2}}$

$J = \frac{6 A}{m} ^ 2$

So increasing the current density means that you are actually increasing the current if you assume that the area of the conductor is remaining the same

In electrolysis if you apply more energy obviously the electrolysis would be faster. This is for every device that uses energy including us. More energy more work.Electrolysis is actually a reaction which cannot start on its own.But it can be considered as a reaction.

Like in reactions in electrolysis too bonds are broken are formed into new chemical bonds. But to break a bond you need to apply some energy and this amount of energy or the energy to break a bond is known as bond dissociation energies. This energy comes from the electrical source. So more energy the faster the process

Now let me show you an example

Suppose that $2 C$/s is applied to a solution of $C u S O 4$ solution for 2 seconds. Recall that 1A = 1C/s

The reaction

= $2 {H}_{2} O + 2 C u S {O}_{4} \rightarrow {O}_{2} + 2 {H}_{2} S O 4 + 2 C u$

Net ionic reaction

$2 {H}_{2} O + C {u}^{2 +} \rightarrow 2 {H}_{2} + 2 C u$

Each mole of $C {u}^{2 +}$ produces one mole of $C u$

$2 {H}_{2} O$ is reduced to ${H}_{2}$

${H}_{2} O + \rightarrow {H}_{2} + 2 {e}^{-}$
Cu^(2+) + 2e^-) rarr Cu 

Thus
$2 {H}_{2} O \rightarrow 2 {H}_{2} + 4 {e}^{-}$
2Cu^(2+) + 4e^-) rarr Cu 

The whole reaction

2H_2O + 2Cu^(2+) + 4e^-) rarr 2H_2 + Cu + 4e^-)
$2 {H}_{2} O + 2 C {u}^{2 +} \rightarrow 2 {H}_{2} + C u$

As 2C/s is applied for 2s

$\frac{2 C}{\cancel{s}} \times 2 \cancel{s}$ = $4 C$

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 $C {u}^{2 +}$ is reduced per 2 mole electrons

Thus moles of $C u$ formed

$\text{1 mol of Cu "/(2 molcancel( e^-)) xx "0.00004145077 mol of} \cancel{{e}^{-}}$

$\frac{1}{2} \times 0.00004145077 = \text{0.00002072538 mol of Cu}$

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

$\frac{3 C}{\cancel{s}} \times 2 \cancel{s}$

= $6 C$

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

$\text{1 mol of Cu "/(2 molcancel( e^-)) xx "0.00004145077 mol of" cancel(e^-)=" 0.00003108808 mol of Cu}$

$\text{0.00003108808 mol of Cu" > "0.00002072538 mol of Cu}$

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

mole to grams

Plotting a graph

If my handwriting is too bad

y = mol of Cu formed from reduction of $C {u}^{2 +}$
x = mol of electrons used in the process