Galvanic cell and Nernst equation

Galvanic cell and

Nernst equation

Galvanic cell

• Some times called Voltaic cell

• Spontaneous reaction redox reaction is

used to provide a voltage and an electron

flow through some electrical circuit

• When metallic zinc is dipped into a

solution of copper sulfate: dark brown ,

spongy layer of metallic copper forms on

the zinc.

Galvanic cell

• The reaction can be written as:

Cu

(

aq

) + Zn(

s

)



Cu(

s

) + Zn

(

aq

)

The blue color of copper sulfate gradually

disappeared and replaced by the colorless

zinc sulfate.

• This reaction will not produce any flow of

electrons as long as it occurs at the

Galvanic cell

• The flow of the electrons can be

achieved by making the oxidation and

reduction

half-reactions occur in separate

compartments of a galvanic cell.

Galvanic cell

• Each compartment called

half-cell

.

• Complete isolation of the two species

would lead to an electrical

Galvanic cell

Explanation:

• on the left side Zn ions entering the solution give the sloution overall positive charge. This would prevent additional Zn ions to enter the solution. • On the right side, Cu ions leave the solution, the

SO4 ions left behind would give the solution negative charge and the electrode becomes positively charge. This would cause the

electrode to repel Cu ions.

Galvanic cell

• The flow of current accompanied by continues electrical activity therefore the solution around the electrodes must be kept electrically neutral (even Zn leave the electrode compartment or anions enter to it; and cations enter the

compartment of Cu to balance the charge of SO4 or SO4 ions must leave .

How this can be achieved???

Galvanic cell

• The salt bridge and the porous partition allow the slow mixing of the ions in the two solution

.

Galvanic cell

• Salt bridge usually a tube filled with an

electrolyte such as KNO3 or KCl in gelatin.

• Cations from the salt bridge can move into

one compartment to compensate for the

excess of negative charge while anions

from the salt bridge diffuse into the other

compartment to neutralize the excess of

positive charge.

The signs of the electrodes in

galvanic cells

• Oxidation occurs at the anode and

reduction occurs at the cathode in both

galvanic cell and electrolyte cell.

What about the sign of the electrode at the

anode and cathode??????

The signs of the electrodes in

galvanic cells

• In the galvanic cell: anode has a negative

charge and cathode has positive charge

this is the opposite to the electrolyte

Cell potentials

• The force that moves the electrons

through the cell is called

electromotive

force (emf)

and measured as

volts (V)

.

• Emf is the passage of 1 coulomb is able to

accomplish 1 joule of work.

• (1 volt = 1joule / coulomb)

• The emf in the galvanic cell is called cell

potential (E

)

Cell potentials

• The potential of the cell depends on the

concentrations of the ions, the temperature, and the partial pressure of any gases that might be involved.

• Standard cell potential (Eo

cell) is measured when

the concentrations of the ions are 1 M, the

temperature is 25 C and the partial pressures of the gases are 1 atm.

• The cell potential is measured by a potentiomenter

Cell diagram

This allow us the describe the galvaic cell to give informations:

1. The nature of the electrode materials

2. The nature of the solution in contact with the electrodes (including the concentrations of the ions

3. which of the half cell is anode and cathode

4. The reactants and the products in each of the half cells In the previous fig suppose the concentrations of both Zn

and cu ions are 1 M

Reduction potential

What is the origin of the cell potential??

• In the cell we have two solution; one contains Zn ion and the other Cu ions.

• Each of these ions (zn and Cu) has a certain

tendency to acquire electrons from its respective electrode and become reduced.

Zn2+(aq) + 2e-





Reduction potential

• The larger reduction potential the larger tendency to undergo reduction

• When cell reaction takes place there is “tug of war” as each species attempts to pull electrons from its electrode so as to become reduced.

• The species with greater tendency to acquire electrons (greater reduction potential) wins the war.

Reduction potential

• The potential that we measure arises from

the

differences

in the tendency of the two

ions to become reduced and is equal to:

E

= E

- E

There fore for the previous reaction

Reduction potential

• Experimentally we can measure the overall cell potentials but can be calculated.

• We use standard hydrogen electrode which (arbitrarily) sets its reduction potential of zero volts.

• It consists of a platinum wire encased in a glass sleeve through which hydrogen gas is passing at pressure of 1 atm.

• The platinum wire is attached to a platinum foil that is coated with a black velvety layer of finely divided platinum.

Reduction potential

Standard Hydrogen Electrode

The convention is to select a particular electrode and assign its

standard reduction potential the value of 0.0000V. This electrode is the Standard Hydrogen Electrode.

2H+_{(aq) + 2e}–  _H 2(g) a H₂ H+ Pt

The “standard” aspect to this cell is that the activity of H₂(g) and that of H+_{(aq) are both 1.}

This means that the pressure of H₂ is 1 atm and the concentration of H+ _{is 1M, given that}

Reduction potential

H+|H2 =0.00V, means any substance that is more

easily reduced than H+ _{has a positive value of E}o_.

• What could happened if hydrogen electrode is paired with another half-cell?

• If the reduction potential of the species is greater than that for hydrogen electrode



-Reduction potential

• If the reduction potential of the species is

less than that for hydrogen electrode



reduction at hydrogen electrode

Reduction potential

when connecting a voltmeter, connect the positive terminal to the positive electrode. If it reads a positive potential, you have correctly identified all the terminals. If you read a negative

potential, then you have misidentified the

reactions in the cells, and you have hooked it up backwards. Reverse your assignment of anode and cathode.

• in a galvanic cell the cathode is +ive

Reduction potential

Suppose we connect both Cu and H electrode: to obtain proper reading (+) terminal should

connected to Cu electrode and (-) terminal to H electrode.

Cu2+(aq) + 2e-



Eo

cell = EoCu2+|Cu - EoH+|H2

0.34 V = Eo_Cu2+|Cu – 0.00 V

Reduction potential

If both Zn and H electrode are connected then: Positive terminal is for H electrode

Negative terminal is for Zn electrode 2H+_{(aq) + 2e}-  _H 2 (g) (cathode) Zn(s) Zn2+_{(aq) + 2e}- _(anode) Eo cell = EoH+|H2 - Eozn2+|zn 0.76 V = 0.000 V - Eo_{zn2+|zn V} ٍEozn2+|zn = -0.76 V

Reduction potential

• To calculate E

cell

For Cu and Zn cell

E

= E

- E

E

=+0.34 V– (-0.76 V)

E

= +1.10 V

• This value is precisely the value that we

observe experimentally

• Bear in mind, E

cell

positive means

More Example

Fe2+ _{+ 2e}–  _{Fe -0.44}

V2+ _{+ 2e}–  _{V -1.19}

To get a final positive cell

potential, the more negative half-reaction (V) must act as the anode.

Fe2+ _{+ V}  _{Fe + V}2+

E_cell = -0.44 - (-1.19) = +0.75 V

Sn2+ _{+ 2e}–  _{Sn -0.14}

Ag+ _{+ e}–  _{Ag +0.80}

More negative potential reaction is the anode.

Multiply the Ag reaction by 2, but don’t modify the cell potential.

2 Ag+ _{+ Sn}  _{2 Ag + Sn}2+

Standard Potential Tables

All of the equilibrium electrochemical data is cast in Standard Reduction Potential tables.

F₂ + 2e–  2F– +2.87 Co3+ _{+ e}–  _Co2+ _+1.81 Au+ _{+ e}–  _{Au +1.69} Ce4+ _{+ e}–  _Ce3+ _+1.61 Br₂ + 2e–  2Br– +1.09 Ag+ _{+ e}–  _{Ag +0.80} Cu2+ _{+ 2e}–  _{Cu +0.34} AgCl + e–  Ag + Cl– +0.22 Sn4+ _{+ 2e}–  _Sn2+ _+0.15 2H+ _{+ 2e}–  _H 2 0.0000 Pb2+ _{+ 2e}–  _{Pb -0.13} Sn2+ _{+ 2e}–  _{Sn -0.14} In3+ _{+ 3e}–  _{In -0.34} Fe2+ _{+ 2e}–  _{Fe -0.44} Zn2+ _{+ 2e}–  _{Zn -0.76} V2+ _{+ 2e}–  _{V -1.19} Cs+ _{+ e}–  _{Cs -2.92} Li+ _{+ e}–  _{Li -3.05}

Oxidative Strength

Consider a substance on the left of one of these equations. It will react as a

reactant with something below it and on the right hand side.

• higher in the table means more likely to act in a reducing manner.

• when something is reduced, it induces oxidation in something else.

• it is an oxidizing agent or an oxidant.

• F₂ is a stronger oxidant than Ag+_.

• Cu2+ _{is a weaker oxidant than Ce}4+_.

F₂ + 2e–  2F– +2.87 Co3+ _{+ e}–  _Co2+ _+1.81 Au+ _{+ e}–  _{Au +1.69} Ce4+ _{+ e}–  _Ce3+ _+1.61 Br₂ + 2e–  2Br– +1.09 Ag+ _{+ e}–  _{Ag +0.80} Cu2+ _{+ 2e}–  _{Cu +0.34} AgCl + e–  Ag + Cl– +0.22 Sn4+ _{+ 2e}–  _Sn2+ _+0.15

Reductive Strength

Substances on the right hand side of the equations will react so as to be oxidized.

• LOWER in the table means a greater tendency to be oxidized.

• when oxidized, it induces reduction in something else. It is a reducing agent or reductant.

• Ag is a stronger reductant than Au.

• Co2+ _{is a weaker reductant than}

Sn2+ F₂ + 2e–  2F– +2.87 Co3+ _{+ e}–  _Co2+ _+1.81 Au+ _{+ e}–  _{Au +1.69} Ce4+ _{+ e}–  _Ce3+ _+1.61 Br₂ + 2e–  2Br– +1.09 Ag+ _{+ e}–  _{Ag +0.80} Cu2+ _{+ 2e}–  _{Cu +0.34} AgCl + e–  Ag + Cl– +0.22 Sn4+ _{+ 2e}–  _Sn2+ _+0.15

Gibbs and the Cell Potential

• Here we can easily see how this Gibbs function relates to a potential. • By convention, we identify work which is negative with work which is

being done by the system on the surroundings. And negative free energy change is identified as defining a spontaneous process.

G_T,P  w_electrical  n F E

• Note how a measurement of a cell potential directly calculates the Gibbs free energy change for the process.

Gibbs and the Cell Potential

• Consider reaction with

n (number of electron) =2

F (number of coulombs per mole

electrons=96.500 C/mole e

₎

E (cell potential) = +1.1 V

Calculation of thermodynamic constants from standard cell potential

• ∆ Go _{= -RT ln K}

R=8.314 J/mole

• ∆ Go _{=- 2.303 RT log K}

• ∆ Go _{= -nFE}o_{= =- 2.303 RT log K}

• Eo _{= (2.303 RT/nF) log K}

• At T = 25, the quantity 2.303 RT/F is constant And equal to (2.303x8.314x298)/96.500 =0.0592V

Calculation of thermodynamic constants

from standard cell potential

• Finally we can get

Nernst Equation

• ∆ G = ∆ G

_{+ RT ln Q}

Q is the reaction quotient

• ∆ G = ∆ G

_{+ 2.303RT log Q}

• -nFE = -nFE

_{+ 2.303RT log Q}

Rearrangement

E = E

– (2.303RT/nF) log Q

Nernst Equation

• Consider the reaction of both Cu and Zn ar

discussed earlier :

• [Zn

_{] = o.4M and [Cu}

_{] = 0.02M, what is}

E?

E = E

– (0.059/n) log Q

Q = [Zn

] / [Cu

] and E

= 1.1V

Then