# Quantitative Aspects of Electrolysis

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 15:32

Michael FaradayThe electric charge carried by one mole of electrons, 9.648 670 x 104 C mol-1; abbreviated F. discovered in 1833 that there is always a simple relationship between the amount of substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula. produced or consumed at an electrodeIn an electrochemical cell, a surface on which oxidation or reduction occurs; an electrode conducts electric current into or out of a cell. during electrolysis and the quantity of electrical charge Q which passes through the cell. For example, the half-equation

Ag+ + e → Ag      (1)

tells us that when 1 mol Ag+ is plated out as 1 mol Ag, 1 mol e must be supplied from the cathodeThe electrode in an electrochemical cell where reduction occurs; the negatively charged electrode in a vacuum tube.. Since the negative charge on a single electronA negatively charged, sub-atomic particle with charge of 1.602 x 10-19 coulombs and mass of9.109 x 1023 kilograms; electrons have both wave and particle properties; electrons occupy most of the volume of an atom but represent only a tiny fraction of an atom's mass. is known to be 1.6022 × 10–19 C, we can multiply by the Avogadro constant to obtain the charge per moleThat chemical amount of a substance containing the same number of units as 12 g of carbon-12. of electrons. This quantity is called the Faraday constantThe electric charge carried by one mole of electrons, 9.648 670 x 104 C mol-1; abbreviated F., symbol F:

F = 1.6022 × 10–19 C × 6.0221 × 1023 mol–1 = 9.649 × 104 C mol–1

Thus in the case of Eq. (1), 96 490 C would have to pass through the cathode in order to deposit 1 mol Ag. For any electrolysis the electrical charge Q passing through an electrode is related to the amount of electrons ne– by

$F=\frac{Q}{n_{e^{-}}}$

Thus F serves as a conversion factorA relationship between two units of measure that is derived from the proportionality of one quantity to another; for example, the mass of a substances is proportional to its volume and the conversion factor from volume to mass is density. between ne– and Q.

EXAMPLE 1 Calculate the quantity of electrical charge needed to plate 1.386 mol Cr from an acidic solutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. of K2Cr2O7 according to half-equation

H2Cr2O7(aq) + 12H+(aq) + 12e → 2Cr(s) + 7 H2O(l) (2).

Solution According to Eq. (2), 12 mol e is required to plate 2 mol Cr, giving us a stoichiometric ratio S(e/Cr). Then the Faraday constant can be used to find the quantity of charge. In road-map form

nCr $\xrightarrow{S\text{(}e^{-}\text{/Cr)}}$ ne$\xrightarrow{F}$ Q

Q = 1.386 mol Cr × $\frac{\text{12 mol }e^{-}}{\text{2 mol Cr}}$ × $\frac{\text{9}\text{.649 }\times \text{ 10}^{\text{4}}\text{ C}}{\text{ 1 mol }e^{-}}$ = 8.024 × 105 C

Often the electrical current rather than the quantity of electrical charge is measured in an electrolysis experiment. Since a coulomb is defined as the quantity of charge which passes a fixed point in an electrical circuit when a current of one ampere flows for one second, the charge in coulombs can be calculated by multiplying the measured current (in amperes) by the time (in seconds) during which it flows:

Q = It      (3)

In this equation I represents current and t represents time. If you remember that

coulomb = 1 ampere × 1 second      1 C = 1 A s

you can adjust the time units to obtain the correct result.

EXAMPLE 2 Hydrogen peroxide, H2O2, can be manufactured by electrolysis of cold concentratedIncreased the concentration of a mixture or solution (verb). Having a large concentration (adjective). sulfuric acid. The reaction at the anodeThe electrode in an electrochemical cell where oxidation occurs. The positively charged electrode in a vacuum tube. is

2H2SO4 → H2S2O8 + 2H+ + 2e      (4a)

When the resultant peroxydisulfuric acid, H2S2O8, is boiled at reduced pressureForce per unit area; in gases arising from the force exerted by collisions of gas molecules with the wall of the container., it decomposes:

2H2O + H2S2O8 → 2H2SO4 + H2O2      (4b)

Calculate the massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object. of hydrogen peroxide produced if a current of 0.893 flows for 1 h.

Solution The productA substance produced by a chemical reaction. of current and time gives us the quantity of electricity, Q. Knowing this we easily calculate the amount of electrons, ne–. From half-equation (4a) we can then find the amount of peroxydisulfuric acid. Equation (4b) then leads to nH2O2 and finally to mH2O2. The road map to describe this logic is as follows:

$I\xrightarrow{t}Q\xrightarrow{F}n_{e^{-}}\xrightarrow{S_{e}}n_{\text{H}_{\text{2}}\text{S}_{\text{2}}\text{O}_{\text{8}}}\xrightarrow{S}n_{\text{H}_{\text{2}}\text{O}_{\text{2}}}\xrightarrow{M}m_{\text{H}_{\text{2}}\text{O}_{\text{2}}}$

so that

$m_{\text{H}_{\text{2}}\text{O}_{\text{2}}}=\text{0}\text{.893 A }\times \text{ 3600 s }\times \text{ }\frac{\text{1 mol }e^{-}}{\text{96 490 C}}\text{ }\times \text{ }\frac{\text{1 mol H}_{\text{2}}\text{S}_{\text{2}}\text{O}_{\text{8}}}{\text{2 mol }e^{-}}$

$=\frac{\text{1 mol H}_{\text{2}}\text{O}_{\text{2}}}{\text{1 mol H}_{\text{2}}\text{S}_{\text{2}}\text{O}_{\text{8}}}\text{ }\times \text{ }\frac{\text{34}\text{.01 g H}_{\text{2}}\text{O}_{\text{2}}}{\text{1 mol H}_{\text{2}}\text{O}_{\text{2}}}$

= 05666 $\frac{\text{A s}}{\text{C}}$ × g H2O2 = 0.5666 g H2O2