2-5-2011

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2-5-2011

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Galvanic Cell: Electrochemical cell in which chemicalreactions are used to create spontaneous current (electron) flow

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Galvanic Cells

spontaneousredox reaction

anodeoxidation

cathodereduction

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Galvanic Cells

The difference in electrical potential between the anode and cathode is called:

• cell voltage

• electromotive force (emf)

• cell potential Cell Diagram

Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)

[Cu2+] = 1 M & [Zn2+] = 1 M

Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)

anode cathode

Cell Representation

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For example, look at the following cell representation: 

Cu(s) | CuSO4 (0.100 M) || ZnCl2 (0.200 M) | Zn(s) This cell is read as follows: a copper electrode is

immersed in a 0.100 M CuSO4 solution (this is the first half-cell, anode), 0.200 M ZnCl2 solution in which a Zn(s) electrode is immersed (this is the second half-cell, cathode).

If the Ecell is a positive value, the reaction is spontaneous (galvanic cell) and if the value is negative, the reaction is nonspontaneous (electrolytic cell) in the direction written and will be spontaneous in the reverse direction.

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Standard Reduction Potentials and the Standard hydrogen electrode (SHE)

Standard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm (standard state).

E0 = 0.00 V

2e- + 2H+ (1 M) H2 (1 atm)

Reduction Reaction

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Calculating the Cell Potential

The process of calculating the cell potential is simple and involves calculation of the potential of each electrode separately, then the overall cell potential can be determined from the simple equation:

Ecell = Ecathode - Eanode

Sometimes, this relation is written as: 

Ecell = Eright – Eleft

Ecell = Ereduction half cell - Eoxidation half cell

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Standard Reduction Potentials

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

2e- + 2H+ (1 M) H2 (1 atm)

Zn (s) Zn2+ (1 M) + 2e-Anode (oxidation):

Cathode (reduction):

Zn (s) + 2H+ (1 M) Zn2+ + H2 (1 atm)

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E0 = 0.76 Vcell

Standard emf (E0 )cell

0.76 V = 0 - EZn /Zn 0

2+

EZn /Zn = -0.76 V02+

Zn2+ (1 M) + 2e- Zn E0 = -0.76 V

E0 = EH /H - EZn /Zn cell0 0

+ 2+2

Standard Reduction Potentials

E0 = Ecathode - Eanodecell0 0

Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)

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Standard Reduction Potentials

Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)

2e- + Cu2+ (1 M) Cu (s)

H2 (1 atm) 2H+ (1 M) + 2e-Anode (oxidation):

Cathode (reduction):

H2 (1 atm) + Cu2+ (1 M) Cu (s) + 2H+ (1 M)

E0 = Ecathode - Eanodecell0 0

E0 = 0.34 Vcell

Ecell = ECu /Cu – EH /H 2+ +2

0 0 0

0.34 = ECu /Cu - 00 2+

ECu /Cu = 0.34 V2+0

11These electrodes are Active (they take part in the reaction)

Combining the two half reactions:

12 Electrodes are passive (not involved in the reaction)

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Standard Electrode Potential

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• E0 is for the reaction as written

• The more positive E0 the greater the tendency for the substance to be reduced

• The half-cell reactions are reversible

• The sign of E0 changes when the reaction is reversed

• Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

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Diagonal Rule

It is possible to determine whether an electrochemical reaction would occur spontaneously or not by considering the values of the standard electrode potentials of the two half reactions.

Steps:1. Write down the two half reactions placing the one

with the larger Eo up.2. Draw a diagonl from up left pointing to down right.3. The species in the up left will react with the

species in the products of the down right.

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Predict what will happen if Br2 is added to a solution containing NaCl and NaI. Assume all species are in their standard states.

Solution:

From Eo Table we have:

Cl2 + 2e 2Cl- Eo = 1.36V

Br2(l) + 2e 2Br- Eo = 1.07V

I2(s) + 2e 2I- Eo = 0.53V

Applying the diagonal rule suggests that Br2 will not be able to react with Cl-, but will react with I-.

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Br2(l) + 2e 2Br- Eo = 1.07V

2I- I2(s) + 2e Eo = 0.53V

Br2(l) + 2I- I2(s) + 2Br-

Eorxn = Eo

cathode - Eoanode

Eorxn = 1.07 – 0.53

Eorxn = 0.54V

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Can Sn(s) reduce Zn2+ under standard state conditions? Sn(s) + Zn2+ Sn2+ + Zn(s) ?

To answer this question, we should write the two half reactions starting with the one with larger Eo

Sn2+ + 2e Sn(s) Eo = -0.14V

Zn2+ + 2e Zn(s) Eo = -0.76V

Applying the diagonal rule, we conclude that Sn(s) will not be able to reduce Zn2+.

The spontaneous reaction will be:

Sn2+ + Zn(s) Zn2+ + Sn(s)

This means that Zn(s) will reduce Sn2+.

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Calculation of the standard emf of an electrochemical cell

The procedure is simple:

1. Arrange the two half reactions placing the one with the larger Eo up (the cathode).

2. The half reaction with the lower Eo is placed down (the anode).

3. Eocell = Eo

cathode - Eoanode