UNIT V ELECTROCHEMISTRY Ver. 1.2

Chemistry 12

UNIT V – ELECTROCHEMISTRY ver. 1.2

I. INTRODUCTION

Put a strip of copper into a concentrated solution of nitric acid and it will quickly begin to

bubble, turning the solution green and eating away the metal. Put an identical strip of copper

into a concentrated solution of hydrochloric acid, and nothing happens.

The reason is based on ELECTROCHEMISTRY

ELECTROCHEMISTRY

ELECTROCHEMICAL
REACTIONS

Consider the following reaction

2 Al(s) + 3 CuCl2(aq) à 2AlCl3(aq) + 3 Cu(s)

This reaction can be re-written as two separate HALF-REACTIONS.

OXIDATION

OXIDATION HALF-REACTION

REDUCTION

REDUCTION HALF-REACTION

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When a substance becomes OXIDIZED:

Ex.

When a substance becomes REDUCED:

Ex.

REDUCTION-OXIDATION REACTION

Ex. Consider the following reaction:

2 Al + 3 Cu2+ à 2 Al3+ + 3 Cu

Al:

Cu2+:

ANY TIME THAT YOU SEE THAT AN ATOM OR ION HAS CHANGED ITS CHARGE DURING A REACTION, YOU ARE DEALING WITH A REDOX REACTION.

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Ex. Are the following atoms being oxidized or reduced (and would that make them the oxidizing

agent or the reducing agent?)

a. I- à I3+ + 4e-

b. Au3+ + 3e- à Au

c. Cu+ à Cu2+ + e-

d. F2 + 2e- à 2F-

Ex. For the following reactions indicate which substance is: a. being oxidized, b. being reduced,

c. the oxidizing agent, d. the reducing agent.

Zn2+ + Mg à Zn + Mg2+

2 Na + Br2 à 2 Na+ + 2 Br-

II. OXIDATION NUMBERS

OXIDATION

NUMBER

You will need to be able to calculate the oxidation numbers of various atoms.

RULES FOR DETERMINING OXIDATION NUMBER

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ii.

iii.

Ex. Determine the oxidation number of each atom for the following:

a. Na c. N2O4

b. CCl4 d. PO43-

Try: Determine the oxidation number of each atom for the following:

a. P4 c. H4P2O7

b. PbSO4 d. NH4+

An atoms change in oxidation number indicates if it is being reduced or oxidized.

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Consider the following

ClO3- à ClO4-

H2O2 à H2O

Cr3+ à CrO42-

NO2 à N2O3

III. PREDICTING THE SPONTANEITY OF A REDOX REACTION

There is a table called the Table of Standard Reduction Potentials, which is partially shown below.

F2 + 2 e- ↔ 2 F-
Ag+ + e- ↔ Ag
Cu2+ + 2 e- ↔ Cu
Zn2+ + 2 e- ↔ Zn
Li+ + e- ↔ Li

The following observations can be made from the Table of Standard Reduction Potentials.

(These will help you locate a half-reaction much more quickly)

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Ex.

The Table of Standard Reduction Potentials is used in a similar way to the Table of Relative Strengths of Acids and Bases.

Ex. The half reaction for Zn and Zn2+ is:

If there was a piece of Zn(s) in a solution of Zn2+(aq) you can write either:

Note – When referring to an ISOLATED half-reaction, use equilibrium arrows to show that the

reaction can go forward or backwards.

If the half-reaction is made to UNDERGO EITHER REDUCTION OR OXIDATION AS A

RESULT OF BEING PART OF A REDOX REACTION, then use a one-way reaction arrow.

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Try: State if the following ions could undergo reduction, oxidation, or both:

a. Ni2+ c. NO e. Fe2+

b. Cl- d. Cu+ f. Al3+

Become familiar with the idea that reduction equations are those on the table in the forward direction and oxidation equations are those in the reverse direction.

Assume you have two different half-reactions in two different beakers. In one beaker there is Zn(s) in a solution of Zn2+, and in the other is Cu(s) in a solution of Cu2+. The two possible half-reactions are:

Of the two oxidizing agents _____ and _____ , the ______is higher on the left side of the table, and therefore has a greater tendency to become reduced. The reduction reaction will be:

Of the two reducing agents _____ and _____ , the _____ is lower on the right side of the table, and therefore has a greater tendency to be oxidized. The oxidation reaction will be:

Recall:

The overall reaction, which in this particular case will occur spontaneously, is found by adding together the two half reactions – ONE OXIDATION AND ONE REDUCTION.

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Try: Which is the stronger oxidizing agent in each of the following pairs:

a. Ag+ or Cu2+ b. Co2+ or Au3+

Try: Which is the stronger reducing agent in each of the following pairs:

a. H2O or H2O2 b. Sn2+ or Cu+

Ex.

If you are only given two species (or ions) rather than the four needed for two complete half-reactions as given above, you will need to be able to predict if that particular redox reaction will occur.

RULES FOR PREDICTING SPONTANEOUS REDOX REACTIONS

1. If only one species in a half reaction is present, don’t assume that the other species is also present.

You need to be explicitly told if the other species is present.

Ex.

2. If you are only given two potential reactants, rather than complete half-reactions, a reaction may or

may not occur. In order to determine if a reaction will occur, the first thing to do is:

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There are three possibilities:

Ex. Assume the only reactants are Zn and Cu.

Ex. Assume the only reactants are Br- and Cl-.

B. If one is on the left of the table and one is on the right of the table two different cases are

possible:

Ex. Assume the reactants in a vessel are Cu2+ and Zn.

ii.

Ex. Assume the reactants are Zn2+ and Cu.

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Try: Which of the following species will oxidize Cu+?

Ni2+ Ag+ Pb2+ I2 H2O

Try: Predict whether or not a reaction will occur when the following are mixed:

a. Cl2 and Br-

b. Sn and Mn

c. Ni2+ and Pb

d. Cl2 and K

COMMENT ON H+

Some half-reactions require H+ to occur.

Ex.

If H+ is present in a particular half-reaction, it must be treated like any other reactant. For example, if you are asked whether the SO42- in a solution of Na2SO4 will reduce:

Your answer should be ‘there is no reaction unless H+ is could is present also” (or likewise “only if the solution is acidic”), just as nothing would happen if SO42- wasn’t present – same with H+. There will be no reaction if it were not present.

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IV. BALANCING HALF-REACTIONS

A half-reaction must be balanced, just as other chemical reactions must be balanced, however half-reactions are balanced for mass and CHARGE. Balancing half-reactions is not overly complicated, but it is very easy to make mistakes if you are careless about writing the charges on ions.

BALANCING HALF-REACTION STEPS

Usually when you are required to balance a half-reaction, you will be given a ‘skeleton equation’ containing the major atoms. It is up to you to complete the balancing by applying other species as follows:

Note – NEVER vary the order of balancing, it will make it difficult to impossible if you don’t follow

the above order EXACTLY.

Memory aid:

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Ex. Balance the half-reaction: RuO2 ↔ Ru

Ex. Balance the half-reaction: Cr2O72- ↔ Cr3+

Try: Balance the half-reaction: MnO4- ↔ Mn2+

BALANCING BASIC HALF-REACTIONS

All the above solutions were assumed to be in acidic solutions. Sometimes you will be required to balance half-reactions in BASIC conditions.

First, balance as if it were in acidic conditions

Ex. Pb ↔ HPbO2-

1. Balance the major atoms

2. Balance the oxygen atoms

3. Balance the hydrogen atoms

4. Balance the charge.

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Now, CONVERT THE EQUATION TO BASIC CONDITIONS which is done by:

5. Adding the water equilibrium equation in such a way as to CANCEL OUT ALL THE H+

in the half reaction.

Try: Balance the following:

a. HC2H3O2 ↔ C2H5OH in acidic conditions

b. MnO4- ↔ MnO2 in basic conditions

V. BALANCING REDOX EQUATIONS USING HALF-REACTIONS

Note: There are two common methods for balancing redox equations: using half reaction and using

oxidation numbers. You are not required to know both methods (as they end up with the

same results) but it may be to your advantage to be familiar with both of them. Balancing using half-reactions is easier for more complicated redox equations, the oxidation number is easier for simpler redox equations.

STEPS IN BALANCING EQUATIONS USING HALF REACTIONS

Ex. Balance ClO4- + I2 à Cl- + IO3- in acidic solution.

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Try: Balance MnO4- + C2O42- à MnO2 + CO2 in basic solution.

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DISPROPORTIONATION

REACTION

Ex. Balance ClO2- à ClO3- + Cl- in basic solution.

STEPS IN BALANCING REDOX EQUATIONS USING OXIDATION NUMBERS

This method is somewhat of a shortcut, based on that fact that since the total number of electrons lost in an oxidation half-reaction must equal the total number of electrons gained in a reduction half-reaction, so the following two statements are true:

Ex. Balance the following redox reaction ClO4- + I2 à Cl- + IO3-

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Ex. Balance P4 à H2PO2- + PH3 in acidic solution

Try: Balance Zn + As2O3 à AsH3 + Zn2+ in basic solution.

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Try: Balance S2- + ClO3- à Cl- + S (basic)

Try: Balance CN- + IO3- à I- + CNO- (acidic)

Try: Balance As4 + NaOCl + H2O à AsO43- + NaCl

VI. REDOX TITRATIONS

Acid-base titrations are very useful as they allow an accurate determination of an unknown concentration of an acid or a base. In a similar manner, there are many occasions where you may need to know the concentration of a substance that is capable of undergoing an oxidation or reduction reaction.

A. OXIDIZING AGENTS

One of the most useful oxidizing agents that you will encounter is ______. The half reaction:

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It has such a strong tendency to reduce (note its position on the table - ______) that it is able to oxidize a large number of substances (the K+ in KMnO4 is left out as it is a spectator ion).

Ex. To find the [Fe2+] in an unknown solution, react it with acidic MnO4- as follows:

Recall that acid base titrations use an indicator to help see the equivalence point of the titration. The above redox titration also requires some way to identify the equivalence point. Another reason the KMnO4 is so commonly used in redox titrations is that:

Ex. A 100.0 mL sample containing FeCl2 is titrated with 0.100 M KMnO4 solution. If 29.15 mL of

KMnO4 was required to reach the endpoint, what was the [Fe2+]?

B. REDUCING AGENTS

Two commonly used reducing agents are ______and ______. A large number of substances can oxidize I- to I2 (as it is relatively low – about halfway down the table) according to the following half-reaction:

Titrations involving I- generally involve two consecutive steps:

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An example of a reaction involving I- is the reduction of laundry bleach, NaOCl. The reaction between I- and OCl- proceeds as follows:

No attempt is made to add exactly enough I- to react with the OCl-. Rather:

The above reaction between OCl- and I- is the ‘initial’ reaction because the actual redox titration involves a second reaction between the I2 produced, and another ion present in the titrating substance, the reducing agent sodium thiosulphate, Na2S2O3.

When the addition of S2O32- has reacted most of the I2 present, the brown colour of the I2 almost disappears (a diluted colour appearing pale yellow remains). Some starch solution is then added to the titration, which produces a dark blue colour (this is caused by the reaction between starch and the remaining I2 in solution). After adding the starch (which acts as a more noticeable and therefore more precise indicator), the last of the S2O32- is added, causing the blue colour of the starch-I2 mixture to fade – so that the last of the colour just disappears at the equivalence point between the I2 and the S2O32-.

Ex. A 25.00 mL sample of bleach is reacted with excess KI according to the following equation:

2 H+ + OCl- + 2 I- à Cl- + H2O + I2

The I2 produced requires exactly 46.84 mL of 0.7500 M Na2SO3 to bring the titration to the

endpoint using starch solution as an indicator, according to the following equation:

2 S2O32- + I2 à S4O62- + 2 I-

What is the [OCl-] in the bleach?

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VII. ELECTROCHEMICAL CELLS

Recall the half-reactions need to be somehow connected in order for both reactions to occur (a donating and accepting of electrons needs to occur).

ELECTROCHEMICAL

CELL

Consider the reaction:

A spontaneous reaction will occur when zinc metal is placed into a solution of CuSO4. However,

The exact same reaction can be used to produce electricity if:

ELECTRODE

ANODE

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CATHODE

Memory aid:

ELECTROCHEMICAL CELL DIAGRAM

1.The possible half-reactions are:

2. After the half-cells are connected:

3. As electrons are supplied to the Ag electrode:

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4. Overall, the electron flow is: