REDOX REACTIONS CLASS XI CHEMISTRY

DR. TANUJA NAUTIYAL

SHARDA PUBLIC SCHOOL , ALMORA

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SYLLABUS

 Concept of oxidation and reduction,

 redox reactions,

 oxidation number,

 balancing redox reactions,

 applications of redox reactions.

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OXIDATION AND REDUCTION

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Oxidation is a process which involves either of the following :

• Addition of oxygen and removal of hydrogen.
• Addition of electronegative element and removal of

electropositive element.

• Increase in the valency of an electropositive element.
• Loss of one or more electron by an atom or an ion or molecule.

Reduction is the reverse process of oxidation.

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REDOX REACTION

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oxidation-reduction reaction An oxidizing chemical

change, where increased

(electron

loss), accompanied by

an element’s positive valence is

a

reduction of an associated element

simultaneous (electron gain

oxidation state The number of electrons to be added (or subtracted) from an atom in a combined state to convert it to elemental form. Also known as oxidation number.

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OXIDATION NUMBER

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Oxidation number (Oxidation state) is the charge that an atom appears to have in a given species when the bonding electrons are counted using following rules:

1. The oxidation number (O.N.) of an atom in its elemental state is zero, no matter how complicated the molecule is, e.g.,
• H in H₂= 0
• S in S₈= 0
• P in P₈= 0
1. O in O₂ or O₃= 0

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OXIDATION NUMBER

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• (b). F has oxidation number -1.

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• (c). Oxidation number of oxygen is -2 in all

compounds except in peroxides, super oxides and oxygen fluoride. The Oxidation Number of

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• Oxygen In O₂^2–is -1
• Oxygen in O^2- is -½
• Oxygen in OF₂ is +2
• Oxygen O₂F₂ is +1

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OXIDATION NUMBER

4. The oxidation number of
• Hydrogen is +1 in all its compounds
• Hydrogen is -1 in Metallic Hydrides

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• The oxidation number of
• Group I elements is +1
• Group 2 elements is +2.

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(f) For complex ion, the algebric sum of oxidation numbers of all

the atoms in equal to the net charge on the ion.

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TYPES OF REDOX REACTIONS :

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Combination reaction :

0 0

3Mg(s) + N₂(g) Δ

+2 -3

Δ Mg₃N₂ (s)

+1 +5 -2

3KClO₃(s)

Decomposition :

+1 -1

2KCl

0

+ Δ 3O₂(g)

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TYPES OF REDOX REACTIONS

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Metal displacement :

+2 +6 -2 0

3CuSO₄ + Zn(s)

+2 +6 -2

ZnSO₄ +

0

Cu

Non-metal displacement :

0 +1 -1

3Ca_(s) + 2H₂O(l)

+2 -2 +1

Ca (OH)_2(aq)

+

0

^H2(g)

Disproportionation :

0 +1 +1

Cl₂ + 2OH(aq) Cl⁻ (aq) + ClO ⁻

+ H₂O

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OXIDIZING AGENT AND REDUCING AGENT

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Oxidizing agents are those chemicals, which oxidizes other

chemicals but reduces itself.

Reducing agents are those substances, which reduces other but oxidizes itself.

Example:

In this chemical reaction:

• Oxidizing agent: 3CuSO₄
• Reducing agent: Zn

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OXIDIZING AGENT

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Two common oxidizing agents are potassium permanganate

(KMnO₄) and potassium dichromate (K₂Cr₂O₇).

MnO₂

purple

_Mn2+

light pink

Cr₂O₇^2-

orange

_Cr3+

green yellow

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BALANCING REDOX EQUATIONS

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Equations for redox reactions like the preceding one are

relatively easy to balance.

However, in the laboratory we often encounter more complex redox reactions involving oxo anions such as

• chromate (CrO4^2-),
• dichromate (Cr2O7^2-),
• permanganate(MnO4^-),
• nitrate (NO3^- ),
• sulfate (SO4^2-).

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BALANCING REDOX EQUATIONS

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In principle, we can balance any redox equation using the

procedure ,but there are some special techniques for handling redox reactions, techniques that also give us insight into

electron transfer processes.

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Here we will discuss one such procedure, called the ion-electron

method.

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In this approach, the overall reaction is divided into two half-

reactions,

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• one for oxidation half reaction
• one for reduction half reaction
• The equations for the two half-reactions are balanced separately
• then added together to give the overall balanced equation.

BALANCING REDOX EQUATIONS

ION-ELECTRON METHOD.

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ION ELECTRON METHOD

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Suppose we are asked to balance the equation showing the

oxidation of Fe²⁺ ions to Fe ³⁺ ions by dichromate ions(Cr ₂O₇^2- )

in an acidic medium. As a result, the Cr₂O₇^2- ions are reduced to Cr³⁺ ions. The following steps will help us balance the equation .

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• Step 1: Write the unbalanced equation for the reaction in ionic

form.

Fe²⁺ + Cr₂O₇^2- Cr³⁺ + Fe³⁺

• Step 2: Separate the equation into two half-reactions.

+ 2

 Oxidation half reaction: Fe²⁺

+ 6

 Reduction half reaction : Cr ₂O₇^2-

+ 3

_Fe3+

+ 3

_Cr3+

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ION ELECTRON METHOD

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 Step 3: Balance each half-reaction for number and type of

atoms and charges. For reactions in an acidic medium, add H2O to balance the O atoms and H1 to balance the H atoms.

Oxidation half-reaction: The atoms are already balanced. To

balance the charge, we add an electron to the right-hand side of the arrow:

_Fe2+ _Fe3+

+ e-

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ION ELECTRON METHOD

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Reduction half-reaction: Because the reaction takes place in an acidic medium, we add 7H₂O molecules to the right-hand side of the arrow to balance the O atoms:

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Cr₂O₇^2- 2Cr³⁺ + 7H₂O

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 To balance the H atoms, we add 14 H⁺ ions on the left-hand side:

14H ⁺ + Cr₂O₇^2- + 6e- 2Cr³⁺ + 7H₂O

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ION ELECTRON METHOD

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 Step 4: Add the two half-equations together and balance the

final equation by inspection. The electrons on both sides must cancel. If the oxidation and reduction half-reactions contain

different numbers of electrons, we need to multiply one or both half-reactions to equalize the number of electrons .

Here we have only one electron for the oxidation half-reaction and 6 electrons for the reduction half-reaction, so we need to multiply the oxidation half-reaction by 6 and write

14H ⁺

_Fe2+ _Fe3+

+ Cr₂O₇^2-+ 6e-

+ e-

2Cr³⁺ + 7H₂O

2-

6Fe²⁺ + 14H ⁺ + Cr₂O₇ + 6e-

6Fe²⁺ + 2Cr³⁺ + 7H₂O + 6e-

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ION ELECTRON METHOD

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• The electrons on both sides cancel, and we are left with the

balanced net ionic equation:

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6Fe²⁺ + 14H ⁺ + Cr₂O₇^2-+ 6e^- 6Fe²⁺ + 2Cr³⁺ + 7H₂O + 6e^-

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• Step 5: Verify that the equation contains the same type and numbers of atoms and the same charges on both sides of the equation.

• A final check shows that the resulting equation is “atomically” and “electrically” balanced.
• For reactions in a basic medium, we proceed through step 4 as if the reaction were carried out in a acidic medium. Then, for every

_H+

ion we add an equal number of OH^- ions to both sides of the

equation. Where H⁺ and OH^- ions appear on the same side of the equation,

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GALVANIC CELLS

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• When a piece of zinc metal is placed in a CuSO₄ solution,
• Zn is oxidized to Zn²⁺ ions while Cu²⁺ ions are reduced to metallic copper
• The electrons are transferred directly from the reducing agent (Zn) to the oxidizing agent (Cu²⁺) in solution. However, if we physically separate the oxidizing agent from the reducing agent, the transfer of electrons can take place via an external conducting medium (a metal wire). As the reaction progresses, it sets up a constant flow of electrons and hence generates electricity (that is, it produces electrical work such as driving an electric motor).

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GALVANIC CELLS

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 The experimental apparatus for generating electricity through

the use of a spontaneous reaction is called a galvanic cell or

voltaic cell, after the Italian scientists Luigi Galvani and

the

Alessandro Volta, who constructed early versions of device.

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GALVANIC CELL

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 In Galvanic cell, a zinc bar is immersed in a ZnSO₄ solution,

and a copper bar is immersed in a CuSO₄ solution. The cell

operates on the principle that the oxidation of Zn to Zn²⁺ and the reduction of Cu²⁺ to Cu can be made to take place

simultaneously in separate locations with the transfer of

electrons between them occurring through an external wire.

The zinc and copper bars are called electrodes. This particular

arrangement of electrodes (Zn and Cu) and solutions (ZnSO₄

and CuSO₄) is called the Daniel cell.

 By definition, the anode in a galvanic cell is the electrode at which oxidation occurs and the cathode is the electrode at which reduction occurs.

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GALVANIC CELL

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Zn (s) ZnSO₄(aq) CuSO₄ (aq) Cu(s)

Anode

(oxidation)

Cathode

(reduction)

voltameter

Zn(s) Zn²⁺(aq) + 2e–

2e– + Cu²⁺(aq) Cu(s)

Anode

Cathode

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GLAVANIC CELLS

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 Zn is oxidized to Zn ²⁺ at anode.

 Cu2+ is reduced to Cu at cathode.

 The net reaction:

Zn(s) Zn²⁺(aq) + 2e–

2e– + Cu²⁺(aq) Cu(s)

Zn(s) + Cu²⁺(aq) Zn²⁺(aq) + Cu(s)

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SIGNIFICANCE OF SALT BRIDGE

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Salt bridge and its significance :

• An inverted U-tube filled with concentrated solution of inert electrolyte like NH₄NO₃, KCl.
• (a) It connects the solution of two half-cells.
• (b) It prevents the accumulation of charges in anodic as well

as cathodic half-cells.

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ELECTRODE POTENTIAL

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• When the concentrations of the Cu²⁺and Zn²⁺ ions are both 1 .0 M, we find that the voltage or emf of the Daniel cell is 1 .10 V at 25°C. This voltage must be related directly to the redox reactions, but how?
• The overall cell reaction can be thought of as the sum of two half-cell reactions, the measured emf of the cell can be treated as the sum of the electrical potentials at the Zn and Cu electrodes. Knowing one of these electrode potentials, we could obtain the other by subtraction (from 1 .10 V). It is impossible to measure the potential of just a single electrode, but, if we arbitrarily set the potential value of a particular electrode at zero, we can use it to determine the relative potentials of other electrodes. The hydrogen electrode, serves as the reference for this purpose. Hydrogen gas is bubbled into a hydrochloric acid solution at 25°C.

STANDARD HYDROGEN ELECTRODE (SHE)

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• The platinum electrode has two functions:
• First, it provides a surface on which the dissociation of hydrogen

molecules can take place:

• Second, it serves as an electrical conductor to the external circuit.
• Under standard-state conditions (when the pressure of H2 is 1 atm and the concentration of the HCl solution is 1 M; the

potential for the reduction of H1 at 25°C is taken to be exactly

zero:

2H₁(1 M) + 2e^- → H₂(1 atm) E° =0 V

 E° is the standard reduction potential, or the voltage associated with a reduction reaction at an electrode, when all solutes are 1 M and all gases are at 1 atm. Thus, the standard reduction

potential of the hydrogen electrode is defined as zero.

The hydrogen electrode is called the standard hydrogen electrode (SHE).

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STANDARD HYDROGEN ELECTRODE (SHE)

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USE OF STANDARD HYDROGEN ELECTRODE (SHE)

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 When all the reactants are in their standard states (that is, H ₂ at 1 atm, H₁ and Zn⁺² ions at 1 M), the emf of the cell is 0.76 V at 25°C, We can write the half-cell reactions as follows:

Anode (oxidation): Zn_(s)→ Zn⁺² (1 atm) )+ 2e^- E° =? V

Cathode (reduction): 2H₁(1 M) + 2e^- →H₂(1 atm) E°=0V Overall: Zn_(s) + 2H₁(1 M) → Zn⁺² (1 atm) )+ H₂(1 atm)

E°cell = E°cathode - E°anode

E°cell = E°_H1/H2 - E°_{Zn /Zn}

+2

0.76 V = 0 V - E°_{Zn /Zn}

+2

E°_{Zn /Zn} = 0.76 V

+2

STANDARD REDUCTION POTENTIAL OF COPPER

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• The standard electrode potential of copper can be obtained in a similar fashion, by using a cell with a copper electrode and a SHE. In this case, the copper electrode is the cathode

because its mass increases during the operation of the cell, as is consistent with the reduction reaction:

 Cu²⁺ (aq) → 2e^- + Cu_(s)

• The cell diagram is

 Pt(s) H₂(1 atm) H⁺ (1 M) Cu⁺² (1 M) Cu(s)

the half-cell reactions are

Anode (oxidation): H₂(1 atm) →2H⁺ +(1 M) + 2e^- E°=0V Cathode (reduction): Cu⁺² (1 atm) + 2e^- → Cu_(s) E°=?V

Overall: Cu⁺² (1 atm) + H ₂(1 atm) → Cu_(s) + 2H⁺(1 M)

STANDARD REDUCTION POTENTIAL OF COPPER

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the half-cell reactions are

Anode (oxidation): H₂(1 atm) →2H⁺ +(1 M) + 2e^- E°=0V

Cathode (reduction): Cu⁺² (1 atm) + 2e^- → Cu_(s) E°=?V Overall: Cu⁺² (1 atm) + H₂(1 atm) → Cu_(s) + 2H⁺(1 M)

• Under standard-state conditions and at 25 °C, the emf of the cell is 0.34 V, so we write

E°cell = E°cathode - E°anode E°cell = E° _{Cu+2 /cu} - E°_H⁺_/H2 0.34 V = E°_{Cu+2 /cu} – 0 V

E° _{Cu+2 /cu} = 0.34 V

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• In this case, the standard reduction potential of copper, E°_{Cu+2 /cu}

is 0.34 V,

ELECTRODE POTENTIAL OF ELECTROCHEMICAL CELL

 By convention, the standard emf of the cell, E°cell , which is

composed of a contribution from the anode and a contribution

from the cathode, is given by

E°cell = E°cathode - E°anode

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where both E°cathode and E°anode are the standard reduction

potentials of the electrodes.

 For a cell

E°cell = E°cathode - E°anode E°cell = E°Cu²+/Cu - E°Zn²+/Zn

= 0. 34 V - 0.76 V

= 1.10 V

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STANDARD OXIDATION-REDUCTION POTENTIALS AT 25⁰C

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SIGNIFICANCE OF STANDARD OXIDATION- REDUCTION POTENTIALS

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Oxidizing and Reducing Strengths

Calculating the emf of the cell

Feasibility of a redox reaction

Predicting the Liberation of Hydrogen Gas from Acids by Metals

Comparison of Reactivity of Metals

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OXIDIZING AND REDUCING STRENGTHS

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• The electrochemical series helps to pick out substances that are good oxidizing agents and those which are good reducing agents. For example, a very high negative reduction potential of lithium electrode indicates that it is very difficult to reduce Li⁺ ions to Li atoms. Therefore, Li⁺ cannot accept electrons easily and so loses electrons to behave as a reducing agent. As the reduction potential increases (negative value decreases), the tendency of the electrode to behave as reducing agent decreases. Thus, all the substances appearing on the top of the series behave as good reducing agents. For example Li and K are good reducing agents while F^- and Au are the poorest reducing agents.
• Similarly, substances at the bottom of the table have high reduction potential and they can be easily reduced. Therefore,

they act as strong oxidizing agents. From the table we can

conclude that

_H+ _Zn2+

is a better oxidizing agent than while

_Cu2+

is a better oxidizing agent than H⁺; Fe is a better oxidizing

agent than Cl₂ and so on. All the substances appearing at the bottom of the table are good oxidizing agents.

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COMPARISON OF REACTIVITY OF METALS

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• The relative ease with which the various species of metals and ions may be oxidized or reduced is indicated by the reduction potential values. The metals with lower reduction potential are not reduced easily but are easily oxidized to their ions losing electrons. These electrons would reduce the other metals having higher reduction potentials. In other words, a metal having smaller reduction potential can displace metals having larger reduction potentials from the solution of their salt.
• For example, copper lies above silver in the electrochemical series, therefore, if copper metal is added to AgNO₃ solution, silver is displaced from the solution. In general a metal occupying higher position in the series can displace the metals lying below it from the solutions of their salts and so are more reactive in displacing the other metals. Thus, Li is the most electropositive element in solutions and fluorine is the most electronegative element.

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CALCULATION OF THE EMF OF THE CELL

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The following steps determine the reduction potential of the cathode and

anode:

Step I : The two half- cell reactions are written in such a way that the reaction taking place at the left hand electrode is written as an oxidation reaction and that taking place at the right electrode is written as reduction reaction.

Step II: The number of electrons in the two equations are made equal by multiplying one of the equations if necessary by a suitable number. However, electrode potential values (E°) are not multiplied.

Step III: The electrode potentials of both the electrodes are taken to be reduction potentials and so the EMF of the cell is equal to the difference between the standard potential of the right hand side and the left hand side electrode.

^Ecell ^{= E}R ^{- E}L

Step IV: If the EMF of the cell is + ve, the reaction is feasible in the given direction and the cell is correctly represented, i . e., oxidation occurs at left electrode ( anode) and reduction occurs at the right electrode ( cathode). If it is - ve, the cell reaction is not feasible in the given direction and the cell is wrongly represented. Thus, to get positive value for the EMF the electrodes must be reversed.

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• All metals having negative electrode potentials (negative E °

values) show greater tendency of losing electrons as compared to hydrogen. So, when such a metal is placed in an acid solution, the metal gets oxidized, and H⁺ (hydrogen) ions get reduced to

form hydrogen gas. Thus, the metals having negative E ° values liberate hydrogen from acids.

• metal having negative E° value
• For example, metals such as Mg (E (Mg ²+ Mg) = - 2.37 V),

 Zn (E (Zn²⁺ Zn) = - 0.76 V), Iron (E (Fe²⁺ Fe) = - 0.44 V) etc., can

displace hydrogen from acids such as HCl and HSO₄.

• But metals such as Copper, (E (Cu ²⁺ Cu) = + 0.34V), silver (E

(Ag⁺ Ag) = + 0.80V) and gold (E (Au ³⁺ Au) = +1 .42 V) cannot

displace hydrogen from acids because of their positive reduction potential value.

PREDICTING THE LIBERATION OF

HYDROGEN GAS FROM ACIDS BY METALS

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FEASIBILITY OF A REDOX REACTION

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• From the E° values of the two electrodes one can find out whether a given

redox reaction is feasible or not. A redox reaction is feasible only if the

species which has higher potential is reduced i. e., accepts the electrons and the species which has lower reduction potential is oxidized i. e., loses electrons.

• The electrochemical series gives the increasing order of electrode

potentials ( reduction) of different electrodes on moving down the table.

This means that the species, which accept the electrons ( reduced) must be

lower in the electrochemical series as compared to the other which is to lose electrons. ( oxidized). For example ,

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FEASIBILITY OF A REDOX REACTION

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• From the electrochemical series E° value of Cu = +0.34 V and

that of Ag = +0.80 V since the reduction potential of Ag is more than that of Cu, this means that silver has greater tendency to get reduced in comparison to copper. Thus, the reaction occurs more readily than the reaction

• The reduction potential of copper is less than that of Ag, this

means that copper will be oxidized or will go into solution as ions in comparison to Ag. Thus, the reaction, occurs more readily than Therefore, silver will be reduced and copper will be oxidized and

the above reaction is not feasible. Rather the reverse

reaction,can occur. Thus a metal will displace, any other metal, which occurs below it in the electrochemical series from its salt solution. When a metal having lower E° value is placed in a

solution, containing ions of another metal having higher E °

value, then the metal having lower E° value gets dissolved and the ions of the metal having higher E° value get precipitated.

PRESENTATION BY: DR. TANUJA NAUTIYAL

SHARDA PUBLIC SCHOOL ALMORA

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