Slayt 1 -

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Iron ores are minerals with a high iron content. One of them hematite, Fe 2 O 3 is chemically very similar to iron rust. The reaction which iron metal is produced from hematite in a blast furnace is described as: Fe 2 O 3(s) + 3CO (g) 2 Fe (l) + 3 CO 2(g) We can think of the CO(g) as taking O atoms away from Fe 2 O 3 to produce CO 2 (g) and the free element iron. A commonly used term to describe a reaction in which a substance loses O atoms, is « reduction» and gains O atoms is «oxidation». An oxidation and reduction must always occur together, and the reaction in which they do is called an

oxidation-reduction reaction.

Oxidation State (O.S) Changes




-2 Fe 2 O 3(s) + 3CO (g)

0 +4

-2 2 Fe (l) + 3 CO 2(g) The O.S of oxygen is -2 everywhere it appears. That of iron(shwon in red) changes. It decreases from + 3 in Fe 2 O 3 to 0 in the free element, Fe. The O.S of carbon also changes. It increases from +2 in CO to +4 in CO 2 . In terms of oxidation state changes, in an

oxidation process

the O.S of some element


and in


the O.S of an element



Identifying Oxidation- Reduction Reactions

Indicate whether each of the following is an oxidation-reduction reaction a) b) MnO 2(k) + 4 H + (aq) H 2 PO 4 (aq) + 2 Cl (aq) + OH (aq)  HPO 4  2 Mn (aq) 2+ + H (aq) 2 O + 2 H (s) 2 O (s) + Cl 2(g) Solution: a)The O.S of Mn in MnO 2 MnO 2 decreases from +4 to +2 in Mn is reduced to Mn 2+ . The O.S of O remains 2+ --2 .

throughout the reaction, and that of H at +1. The O.S of Cl increases from -1 in Cl to 0 in Cl 2 . Cl is oxidized to Cl 2 reaction is an oxidation-reduction reaction.


b) The O.S of H is +1 on both sides of the equation. The O.S

of O remains -2 throughout the reaction. The O.S of phosphorus is +5 in both H 2 PO 4 and HPO 4 2 . There are no changes in O.S . This is not an oxidation-reduction reaction. It is infact an acid-base reaction.

Oxidation and Reduction Half reactions

The reaction illustrated below is an oxidation-reduction reaction Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) We can show this by evaluating changes in the O.S states. But we may think of the reaction as involving two half-reactions occuring at the same time. The overall or net reaction is the sum of these two reactions: Oxidation: Reduction: Net: Zn (k)  Zn 2+ (aq) + 2e Cu 2+ (aq) + 2e  Cu (k) Zn (k) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (k) In the half-equation the O.S of Zn increases from 0 to +2, corresponding to a loss of two electrons by each Zinc atom. In the second half-equation the O.S of Copper decreases from +2 to 0 corresponding to the gain of 2e by each Cu 2+ (aq) . To summarize;

Oxidation and Reduction Half reactions


, is a process in which the O.S of some element increases, and in which electrons appear on the right in a half-equation


, is a process in which the O.S of some element decreases, and in which the electrons appear on the left in a half equation.

Oxidation and reduction half-reactions must always occur together, and the total number of electrons associated with the oxidation must equal to the total number associated with the reduction.


The same principles of equation balancing apply to oxidation-reduction (redox equations are balance for numbers of atoms and balance for electric charge.Several

methods are possible but we emphasize the one described below: The-Half reaction(Ion-Electron Method) In this method of balancing a redox equation we:  write and balance seperate half-equations for oxidation and reduction.

 adjust coefficients in the two-half equations so that the same number of electrons appears in each half equation  add together the two half-equations to obtain the balanced net equation.

Balancing the Equation for a Redox Reaction in Acidic Solution

• Example: Write the balanced equation for the reaction: • • SO 3 2 (aq) + MnO 4 (aq)  SO 4 2 (aq) + Mn 2+ (aq)

Step 1.

Write the «skeleton» half-equations based on the species undergoing oxidation and reduction. The O.S of sulfur increases from +4 in SO 3 2 to +6 in SO 4 2 . The O.S. Of Mn decreases from +7 in MnO 4 Mn 2+ . The skeleton half-equations are: to +2 in SO 3 2 (aq)  SO 4 2 (aq) MnO 4 (aq)  Mn 2+ (aq)

Step 2.

Balance each half-equation atomically in this order • Atoms other than H and O • O atoms by adding H 2 O with the appropriate coefficient • H atoms by adding H + with the appropriate coefficient • The other atoms( S and Mn) are already balanced in the skeleton half equations. To balance O atoms we add one H 2 O molecule to the left of the first half-equation and four to the right of the second.

Balancing the Equation for a Redox Reaction in Acidic Solution

SO 3 2 (aq) + H 2 O (l)  SO 4 2 (aq) MnO 4 (aq)  Mn 2+ (aq) + 4 H 2 O (l) To balance H atoms we add two H + to the right of the first half-equation and eight to the left of the second SO 3 2 (aq) + H 2 O (l)  SO 4 2 (aq) + 2H + (aq) MnO 4 (aq) + 8 H + (aq)  Mn 2+ (aq) + 4 H 2 O (l)

Step 3.

Balance each half-equation electrically. Add the number of electrons necessary to get the same electric charge on both sides of each half equation.The half-equation in which the electrons appear on the right side is the oxidation half reaction. The other half equation with electrons on the left is the reduction half-equation.

Oxidation: SO 3 2 (aq) + H 2 O (l)  SO 4 2 (aq) + 2H + (aq) + 2e (net charge on each side, -2) Reduction: MnO 4 (aq) + 8 H + (aq) + 5e  Mn 2+ (aq) + 4 H 2 O (l) (net charge on each side, +2)

Balancing the Equation for a Redox Reaction in Acidic Solution

Step 4.

Obtain the net redox reaction by combining the half-equations. Multiply through the oxidation half-equation by 5 and through the reduction half-equation by 2. This results in 10 e on each side of the net equation: 5 SO 3 2 (aq) + 5 H 2 O (l)  5 SO 4 2 (aq) + 10 H + (aq) + 10e 2MnO 4 (aq) + 16 H + (aq) + 10e  2 Mn 2+ (aq) + 8 H 2 O (l) 5 SO 3 2 (aq) + 2 MnO 4 (aq) + 5 H 2 O (l) + 16 H + (aq)  5 SO 4 2 (aq) + 2 Mn 2+ (aq) + 8 H 2 O (l) + 10 H + (aq)

Step 5.

Simplify. The net equation should not contain the same species on both sides. Subtract 5 H 2 O from each side of the equation in step 4. This leaves 3 H 2 O on the right. Also subtract 10 H + from each side, leaving 6 on the left.

5 SO 3 2 (aq) + 2 MnO 4 (aq) + 6 H + (aq)  5 SO 4 2 (aq) + 2 Mn 2+ (aq) + 3 H 2 O (s)

Step 6.

Verify. Check the final net equation to ensure that is balanced both atomically and electrically.e.g the net charge on each side of the equation is: (5x2-) +(2 x 1-) + (6 x 1+) = (5 x 2-) + (2 x 2+) = -6.

Balancing the Equation for a Redox Reaction in Basic Solution

Balance the equation for the reaction in which chromate ion oxidizes sulfide ion in basic solution to produce free sulfur and chromium(III) hydroxide: CrO 4 2 (aq) + S 2 (aq) + OH Cr(OH) 3 (s) + S(s) +H 2 O • Solution : Initially, we treat the half-reactions as if they occur in acidic solution, and then we adjust them for a basis solution: • STEP 1. write the two skeleton half-equations and balance them for Cr and S atoms • CrO 4 2 (aq) Cr(OH) 3 (s) • S 2 (aq) S(s) • STEP 2. Balance each half-equation for H and O atoms. Note that the second half-equation has no H or O atoms • CrO 4 2 (aq) + 5 H + (aq) Cr(OH) 3 (s) + H 2 O • S 2 (aq) S(s)

Balancing the Equation for a Redox Reaction in Basic Solution

STEP 3. Balance the half-equations for electric charge by adding the appropriate numbers of electrons Reduction: CrO 4 2 (aq) + 5 H + (aq) + 3 e Oxidation: S 2 (aq) S(s) + 2 e Cr(OH) 3 (s) + H 2 O STEP 4. Change from an acidic to a basic medium by adding OH ions and eliminating H + .

The oxidation half equation is unaffected because it has no H + ions. Add 5 OH ions to each side of the reductionhalf equation; combine H + and OH into H 2 O; eliminate H 2 O from the right side of the half-equation.

CrO 4 2 + 5 OH (aq) + 5 H + (aq) + 5 OH (aq) (aq) + 3 e Cr(OH) 3 (s) + H 2 O CrO 4 2 CrO 4 2 (aq) + 5 H 2 O + 3 e (aq) + 4 H 2 O + 3 e Cr(OH) Cr(OH) 3 3 (s) + H (s) + 2 O + 5 OH 5 OH (aq) (aq)

Balancing the Equation for a Redox Reaction in Basic Solution

STEP 5. Combine the half-equations to obtain the net redox equation. (multiply the reduction half-equation by 2 and the oxidation half reaction by 3) 2 CrO 4 2 OH (aq) (aq) + 8 H 2 O + 6 e 3 S 2 (aq) 3 S(s) + 6 e 2 Cr(OH) 3 (s) + 10

2 CrO 4 2 OH (aq) + 8 H (aq) + 3 S(s) 2 O + 3 S 2 (aq) 2 Cr(OH) 3 (s) + 10


In a redox reaction, the substance that makes it possible for some other substances to be oxidized is called oxidizing agent or oxidant. In doing so, the

oxidizing agent

is itself reduced. Similarly, the substance that causes some other substances to be reduced is called the reducing agent or reductant. In the reaction the reducing agent itself is oxidized.

• An

oxidizing agent(oxidant)

: • contains an element whose

O.S. decreases

in a redox reaction • «


» electrons(electrons are found on the left side of the half-equation • A

reducing agent(reductant)

: • contains an element whose

O.S. increases

in a redox reaction • «


» electrons(electrons are found on the right side of the half-equation


Example: Hydrogen peroxide, H 2 O 2 is a versatile chemical. Its uses include bleaching wood pulp and fabrics and substituting for chlorine in water purification. One reason for its versatility is that it can be either an oxidizing or reducing agent. For the following reactions, identify whether hydrogenperoxide is an oxidizing or reducing agent.

a) H 2 O 2 b) 5 H 2 O + 2 Fe 2+ (aq) + 2 H + (aq) 2 H 2 O + 2 Fe 3+ 2 (aq) + 2 MnO Mn 2+ (aq) + 5 O 2 (g) 4 (aq) + 6 H + 8 H 2 O +2 Solution: a) Fe 2+ is oxidized to Fe 3+ and

H 2 O 2

makes it possible; H 2 O 2 b) MnO 4 is an

oxidizing agent

is reduced to Mn 2+ and H 2 O 2 . makes this possible;

H 2 O 2

is a

reducing agent


Two kinds of interactions are possible between metal atoms on the electrode and metal ions in solution 1. A metal ion M n+ may collide with the electrode, gain n electrons and be converted to a metal atom M, the ion is reduced 2. A metal atom M on the electrode may lose n electrons and enter the solution as the ion M n+ . The metal atom is oxidized.


Electrode potential

is a property proportional to the density of negative electric charge. To measure a difference in potential, we need to connect two half-cells in a special way. The flow of electric current is in the form of migration of ions from the greater negative charge to the other electrode.(here from Cu to Ag)


The net reaction that occurs as electric current flows through the electrochemical cell is:

Anode:Oxidation Cathode:Reduction Net

• The reading on the voltmeter is (0,463 V) significant. It is the potential difference between two half cells. Because this potential difference is the driving force » for electrons, it is often called the «

electromotive force(emf)

of the cell or the

cell potential .

Why does copper not displace Zn 2+ from solution?

• If we set up an electrochemical cell Zn(s) / Zn 2+ half cell and Cu 2+ /Cu(s) half cell, we find that electrons flow from the Zn to the Cu. The spontaneous net reaction in the electrochemical cell is • Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) • This is a spontaneous reaction and the reverse of this reaction does not occur spontaneously.



• A

cell diagram

shows the components of an electrochemical cell in a symbolic way. We will use the following conventions in writing cell diagrams: •

The anode

, the electrode at which the oxidation occurs, is placed at the left side of the diagram •

The cathode

, the electrode at which reduction occurs, is placed at the right side of the diagram.

• A

boundary between different phases

(e.g, an electrode and a solution) is represented by a single vertical line


• The

boundary between half-cell components

, usually a

salt bridge

, is represented by a double vertical line


• The electrochemical cells which produce electricity as a result of chemical reactions are called

galvanic cells voltaic



Representing a Redox reaction Through a Cell Diagram Aluminium metal displaces zinc(II) ion from aqueous solution a) Write oxidation and reduction half equations and a net equation for this redox reaction b) Write a cell diagram for a voltaic cell in which this reaction occurs.

Solution: a) The term solution as Al 3+ « displaces» means the Aluminium goes into and Zn 2+ comes out of the solution as zinc metal.

b) Al(s) is oxidized to Al 3+ in the anode half-cell(left) and Zn 2+ is reduced to Zn(s) in the cathode half-cell(right) (aq)


To do a calculation of cell voltages, we choose a particular half-cell to which we assign a potential of zero. We then compare other half cells to this reference. The commonly accepted reference is the standard hydrogen electrode


Standard electrode potential, E˚

, measures the tendency for a reduction process to occur at an electrode.

• To determine the value of E˚ for an electrode, we compare it with a standard hydrogen electrode(SHE). In the voltaic cell indicated below the measured potential difference is 0,337 V, with electrons flowing from H the Cu electrode.

2 to


A standard cell potential, E cell ˚

, is the potential difference or voltage of a cell formed from two standard electrodes. The net reaction that occurs in the voltaic cell is The cell reaction indicates that Cu 2+ (1M) is more easily reduced than is H + (1M). The standard electrode potential representing the reduction of Cu 2+ (aq) to Cu(s) is assigned a positive value.


When a standard hydrogen electrode is combined with a standard zinc electrode, electrons flow in the opposite direction, that is from the zinc to the hydrogen electrode .


In summary, the potential of the standard hydrogen electrode is set at 0.

Any electrode at which a reduction half-reaction shows a greater tendency to occur than does the reduction of H + (1M) to H 2 (g) has a


value for its reduction potential,E ˚.

Any electrode at which a reduction half-reaction shows a lesser tendency to occur than does the reaction of H + (1M) to H 2 (g) has a


potantial , E˚.

The value value for its standard reduction of E˚ does not depend on the amount of substances involved . E˚ values are unaffected by multiplying half-equations by constant coefficients.




A new battery system currently under study for possible use in electric vehicles is the zinc-chlorine battery. The net reaction producing electricity in this cell is What is the E cell ˚ of this voltaic cell?

Cadmium is found in small quantities wherever zinc is found. Unlike zinc, which in trace amounts is an essential element, cadmium is an environmental poison. To determine cadmium ion concentrations by electrical measurements the standard reduction potential for the Cd 2+ /Cd electrode is needed. Th voltage of the following voltaic cell is measured: Cd(s) / Cd 2+ (1 M) // Cu 2+ (1 M) / Cu(s) What is the standard reduction potential for the Cd 2+ /Cd electrode?

Solution: =E Cd 2+ / Cd

Determining a Free Energy Change from A Cell Potential

. Determine ΔG˚ for the reaction

n mol e


= electric current(ampere) x time(s) / 96485 C For the reaction Cu


(aq) + 2 e



The mass of Cu accumulated on the electrode= n/2 x M


(atomic mass of Cu)

Our main criterion for spontaneous change is that ΔG<0 . However, according to the equation if ΔG< 0, then ΔG< 0 E cell >0.

If E cell >0 , a reaction occurs

spontaneously direction

in the


If E cell <0, the reaction occurs

spontaneously direction


in the


If E cell =0, a reaction is at equilibrium.

If a cell reaction is reversed, E cell changes sign

Making Qualitative Predictions with Electrode Potential Data

Solution Because the E˚value for the reduction of S 2 O 8 2 (aq) is larger than that for the reduction of Cr 2 O 7 2 (aq) , S 2 O 8 2 (aq) should be the better oxidizing agent.




˚ and K


R=8,314 Jmol -1 K -1 at 25˚C


What is the value of the equilibrium constant K eq for the reaction between copper metal and iron(III) ions in aqueous solution at 25˚C?










Example : What is the value of E cell for the voltaic cell pictured below?

E cell = 0,011 V


Will the cell reaction proceed spontaneously as written for the following cell?





Any voltaic cell in which the net cell reaction involves only a change in the concentration of some species( here H + ) is called a

concentration cell


A concentration cell consists of

two half-cells with identical electrodes


different ion concentrations

. Because the electrodes are identical, the standard half cell potentials are numerically equal and opposite in sign. This makes E cell =0. However, because the ion concentrations differ, there is a potential difference between the two half-cells. Spontaneous change in a concentration cell always occurs in the direction that produces a more dilute solution


Constructing and using a hydrogen electrode is difficult. A better approach is to replace the SHE by a different reference electrode and the other hydrogen electrode by a

glass electrode

. A glass electrode is a thin glass membrane enclosing HCl(aq) and a silver wire coated with AgCl(s) The potential of the glass electrode depends on the hydrogen ion concentration of the solution being tested. The difference in potential between the glass electrode and the reference electrode is converted to a pH reading by the meter


A device that stores chemical energy for later release as electricity is called a battery

Primary batteries:

The cell reaction is not reversible. When the reactants have been mostly converted to products, no more electricity is produced and the battery is dead.

Secondary batteries:

The cell reaction can be reversed by passing electricity through the battery(charging). This means that a battery can be used several times by charging.

Flow batteries:

Materials(Reactants, products, electrolytes) pass through the battery, which is the conversion of chemical energy into electrical energy


In this cell oxidation occurs at a zinc anode and reduction at an inert carbon cathode. The electrolyte is a moist paste of MnO 2 , ZnCl 2 , NH 4 Cl and carbon black.

Oxidation: Reduction:

2 OH (aq) Zn(s) Zn 2+ (aq) +2 e 2 MnO 2 (s) +H 2 O + 2e Mn 2 O 3 (s) + An acid-base reaction occurs between NH 4 + (aq) + OH (aq) NH 3 (g) + H 2 O(l).

The Leclanche cell is a primary cell. It is cheap to make, but it has some drawbacks. When current is rapidly drawn from the cell, products build up on the electrodes(e.g NH 3 ) and this causes the voltage to drop. Also because the electrolyte medium is acidic, zinc metal slowly dissolves. In order to overcome this problem, an alkaline form of this battery can be produced . The advantage of the alkaline battery are that zinc does not dissolve in a base and the battery does a better job in maintaining the voltage.


The most common secondary battery is the automobile storage battery (see below) . The electrodes are lead-antimony alloy. The anodes are impregnated with lead metal and the cathodes with red brown leaddioxide. The electrolyte is dilute sulfuric acid. When the cell is allowed to discharge, following reactions occur:


Pb(s) + SO 4 2 (aq) PbSO 4 (s) + 2 e -


PbO 2 (s) + 4 H + (aq) + SO 4 2 (aq)+ 2 e H 2 O


Pb(s) + PbO 2 (s)+ 4 H + (aq)+SO 4 2 (aq) 2 PbSO PbSO 4 4 (s) + 2 (s)+2 H 2 O To recharge it,electrons are forced in the opposite direction by connecting the battery to an external electric source. The reverse of reactions occurs when the battery is recharged


The cell-diagram of a silver-zinc cell is Zn(s)/,ZnO(s)/ KOH(satd)/Ag 2 O(s),Ag(s) Oxidation: Zn(s) + 2 OH (aq) ZnO(s) +H 2 O+ 2 e Reduction: Ag 2 O(s)+H 2 O+ 2 e 2 Ag(s)+ 2 OH (aq) Net: Zn(s) + Ag 2 O(s) ZnO(s) + 2 Ag(s) Because no solution species is involved in the net reaction, the quantity of the electrode is very small, and the electrodes can be maintained very close together.

The storage capacity of a silver zinc cell is about six times as great as a lead-acid cell of the same size.

These batteries are used in watches, electronic calculators, hearing aids and cameras


The essential process in a fuel cell is fuel+oxygen oxidation products. Applied to methan, a fuel cell can be obtained as the picture to the next.

One of the simplest and most successful fuel cells involves the reaction between H 2 (g) and O 2 (g) to produce water.

Oxidation: 2 H 2 (g) + 4 OH (aq) 4 H 2 O + 4 e Reduction: O 2 (g)+ 2 H 2 O+4 e 4 OH (aq) Net: 2 H 2 (g) + O 2 (g) 2 H 2 O (l)


As long as fuel and O 2 are available, the cell will produce electricity. It does not have the limited capacity of a primary battery, but neither does have the storage capacity of a secondary battery. Fuel cells have had their most notable successes as energy sources in space vehicles.

A fuel cell reaction is often rated in terms of the

efficiency value, ε= ΔG/ΔH

. For the methane fuel cell ε= - 818/-890 = 0,92

Construction of a PEMFC(Proton Exchange Membrane Fuel Cell


Another kind of flow battery, because it uses oxygen from air, is known as air battery. The substance that is oxidized is typically a metal.

One heavily studied battery system is the aluminium-air battery. In this battery oxidation occurs at an aluminium anode and reduction at a carbon-air cathode. The electrolyte circulated through the battery is NaOH(aq).The half reactions and the net reaction are:


4 { Al(s) + 4 OH (aq) [Al(OH) 4 ] (aq) + 3 e }

Reduction: Net:

3{O 2 (g) + 2 H 2 O + 4 e 4 OH (aq) 4 Al(s) + 3 O 2 (g)+ 6 H 2 O + 4 OH (aq) 4 [Al(OH) 4 ] (aq) Al-air batteries have one of the highest energy densities of the all bateries, but they are not widely used because of previous problems with cost,shelf life, start up time, by-product removal which have restricted their use to mainly military applications



is the wearing away of metals due to a chemical reaction. If an iron nail is embedded in a gel of agar(acid base indicator Phenolphtalein and the potassiumferricyanide) in water, it is observed that at the head and tip of the nail a deep blue precipitate forms. This shows us that oxidation occurs.

Oxidation : 2 Fe(s) 2 Fe 2+ (aq) +4 e Reduction: O 2 + 2 H 2 O+4 e 4 OH (aq) Some metals such as Aluminium form corrosion products that adhere tightly to the underlying metal and protect it from further corrosion. Iron oxide flakes off and constantly exposes fresh surface. The difference in corrosion behaviour explains why cans made of iron deteriorate rapidly in the environment, whereas aluminium cans have an almost unlimited lifetime. The simplest method of protecting a metal from corrosion is to cover it with a paint.

Rust, the most familiar example of corrosion


Another method of protecting an iron surface is to plate it with a thin layer of a second metal. iron can be plated with copper by electroplating or with tin by dipping the iron into the molten metal. In either case the underlying metal is protected only as long as the coating medium remains intact.

When iron is coated with zinc(galvanized iron) the situation is different. Zinc is more active than iron . If a break occurs in the zink plating, the iron is still protected.Zinc is oxidized instead of iron.(see below)


a sacrifical anode protection in the hull of a ship

Cathodic protection

is a technique is to control the corrosion of a metal surface by making the surface the cathode of an electrochemical cell. This is a useful protection method used with large iron and steel objects in contact with water,moist soils-ships,storage tanks, pipelines, etc.

This involves connecting a chunk of magnesium, aluminium, zinc or some active metal to the object either directly through a wire. Oxidation occurs at the active metal. The iron behaves like a cathode and supports a reduction half reaction. As long as some active metal remains, the iron is protected. The tive metal is called

sacrificial anode.


When the cell functions spontaneously, electrons flow from the zinc to the copper and the net chemical change in the voltaic cell is Zn(s) + Cu 2+ Zn 2+ (aq) + Cu(s) E cell = + 1,100 V Now suppose we connect the same cell to an external energy source of voltage greater than 1,100 V. That is the connection is made so that the electrons are forced into the zinc electrode(cathode) and removed from the copper electode(the anode) Oxidation Cu(s) Cu 2+ (aq) - E Cu 2+ /Cu ˚= - 0,337 V Reduction Zn 2+ (aq) + 2 e Zn(s) E Zn 2+ /Zn = - 0,763 V Net Cu(s) + Zn 2+ (aq) Cu 2+ (aq)+ Zn(s) E cell = - 1,100 V We conclude that when the cell reaction in a voltaic cell is reversed by reversing the direction of the electron flow , the voltaic cell is changed to an

electrolytic cell



Generally, we do not measure electric charge directly, we measure electric current. One ampere(A) of electric current represents the passage of 1 coulomb of charge per second (s).

1 mol e = 96485 C number of mol e = current(C/s) x time(s) x 1 mol e 96485 C