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)  2AlCl3(aq) + 3 Cu(s)
This reaction can be re-written as two separate HALF-REACTIONS.
OXIDATION
OXIDATION HALF-REACTION
REDUCTION
REDUCTION HALF-REACTION
-2When a substance becomes OXIDIZED:
Ex.
When a substance becomes REDUCED:
Ex.
REDUCTION-OXIDATION REACTION
Ex. Consider the following reaction:
2 Al + 3 Cu2+  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.
-3Ex. Are the following atoms being oxidized or reduced (and would that make them the oxidizing
agent or the reducing agent?)
a. I-  I3+ + 4eb. Au3+ + 3e-  Au
c. Cu+  Cu2+ + ed. F2 + 2e-  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  Zn + Mg2+
2 Na + Br2  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
1.
2.
i.
-4ii.
iii.
3.
4.
5.
6.
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.
-5Consider the following
ClO3-  ClO4H2O2  H2O
Cr3+  CrO42NO2  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 FAg+ + 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)
1.
2.
3.
-6Ex.
4.
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.
-7Try: 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.
-8-
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:
-9There are three possibilities:
A.
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:
i.
Ex. Assume the reactants in a vessel are Cu2+ and Zn.
ii.
Ex. Assume the reactants are Zn2+ and Cu.
-10+
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.
-11IV. BALANCING HALF-REACTIONS
A half-reaction must be balanced, just as other chemical reactions must be balanced, however halfreactions 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:
1.
2.
3.
4.
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:
-12Ex. Balance the half-reaction:
Ex. Balance the half-reaction:
RuO2 ↔ Ru
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 ↔ HPbO21. Balance the major atoms
2. Balance the oxygen atoms
3. Balance the hydrogen atoms
4. Balance the charge.
-13Now, 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  Cl- + IO3- in acidic solution.
1.
-14-
2.
3.
4.
Try: Balance MnO4- + C2O42-  MnO2 + CO2 in basic solution.
-15DISPROPORTIONATION
REACTION
Ex. Balance ClO2-  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 halfreaction, so the following two statements are true:
Ex. Balance the following redox reaction
1.
2.
ClO4- + I2  Cl- + IO3-
-163.
4.
Ex. Balance P4  H2PO2- + PH3 in acidic solution
Try: Balance Zn + As2O3  AsH3 + Zn2+ in basic solution.
-17Try: Balance S
2-
+ ClO3  Cl + S (basic)
-
-
Try: Balance CN- + IO3-  I- + CNO- (acidic)
Try: Balance As4 + NaOCl + H2O  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:
-18It 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 halfreaction:
Titrations involving I- generally involve two consecutive steps:
1.
2.
-19-
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-  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  S4O62- + 2 IWhat is the [OCl-] in the bleach?
-20-
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
-21CATHODE
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:
-224. Overall, the electron flow is:
5. The salt bridge:
6.As Cu2+ ions are formed:
7.As the [Ag+] is depleted around the cathode:
-23Ex. Assume two half-cells consisting of Pb(s) in a Pb(NO3)2 solution and Zn(s) in a ZnCl2 solution
are connected to make an electrochemical cell. Draw and label the parts of the cell, write the
equations for the individual half reaction and overall reaction, and indicate the direction in which
the ions and electrons move.
-24VIII. STANDARD REDUCTION POTENTIALS
VOLTAGE
Since electrons cannot flow in an isolated half-cell, the voltage of an individual half-cell cannot be
determined. However:
A ZERO-POINT is arbitrarily defined on the voltage scale. Specifically the voltage for the
HYDROGEN HALF-CELL:
An electrochemical cell is said to be at STANDARD STATE if:
All voltages listed in the table of STANDARD REDUCTION POTENTIALS are determined at
standard state and are compared to the standard reduction potential of the hydrogen half-cell.
Ex. Cu2+ + 2e- ↔ Cu
……… Eo = + 0.34
Ex. Zn2+ + 2e- ↔ Zn ……… Eo= -0.76
-25Since the voltage is a measure of work done (so work in either done or is being done), reversing a
reduction reaction such as
Zn2+ + 2e- ↔ Zn ……… Eo= -0.76
produces:
When two half reactions are combined,
Ex.
Ex. Calculate the potential of the cell Ni2+ + Fe  Ni + Fe2+
-26-
Ex. Calculate the potential of the cell: Mn + Mg2+  Mn2+ + Mg
Ex. Calculate the potential of the cell: 3 Ag+ + Al  3Ag + Al3+
Try: Calculate the potential for the cell and state if it would be expected to be spontaneous:
2 H2O2  2 H2O + O2
Although Eo can be used to predict if a reaction is spontaneous, it has no correlation to the rate of the
reaction. Recall that the ___________________________ of a reaction determines the rate at which
it will proceed.
-27COMMENT ON WATER
If a reaction occurs in a neutral solution, the reduction of neutral water (at – 0.41 V) may be a possible
reaction and must be considered along with any other reductions possible.
If a reaction occurs in an acidic solution, the reduction of H+ (at 0.00 V) may be a possible reaction
and must also be considered along with any other reductions possible.
(you won’t be asked to deal with basic solutions)
SURFACE AREA OF ELECTRODES
HALF-REACTIONS NOT AT STANDARD STATE
If a half-cell is not at standard conditions, there will be a change in the potential.
Ex.
-28According to Le Chatalier’s principle, if the [Cu ] is increased so it is > 1.0 M, the
equilibrium will shift
2+
And if the [Cu2+] is decreased so it is < 1.0 M, the equilibrium will shift:
Notice that the ‘o’ is dropped from the above equations as they are NOT at standard state.
CELLS THAT REACH EQUILIBRIUM
Operating chemical cells ARE NOT AT EQUILIBRIUM. The reaction arrow is in one direction.
Ex.
THE REDUCTION REACTION
THE OXIDATION REACTION
Overall, the following occurs as cell go towards equilibrium:
INITALLY
AS RXN PROCEEDS
(TIME PASSES)
EVENTUALLY
-29-
IX. SELECTING PREFERRED REACTIONS
When a cell contains a mixture of substances, several different half-reactions (and therefore overall
reactions) may appear to be possible.
Consider the following cell:
The species present are:
WHEN SEVERAL DIFFERENT REDUCTION HALF-REACTIONS CAN OCCUR
WHEN SEVERAL DIFFERENT OXIDATION HALF-REACTIONS CAN OCCUR
-30Ex. A strip of iron metal is placed in a mixture of Br2(aq) and I2(aq). What is the preferred reaction
that will occur?
SPECTATOR IONS
Any ion capable of being reduced will be a spectator ion if:
Any ion capable of being oxidized will be a spectator ion if:
Some ions are more commonly used in making electrochemical cells because they have such a low
tendency to oxidize or reduce that they rarely come into play in a reaction.
Common Spectator Ions:
X.CORROSION OF METALS
CORROSION
When a drop of water rests on an iron surface, a spontaneous reaction occurs:
At the oxygen-poor region in the center of the drop, the iron oxidizes.
-31Once the Fe is formed, it tends to migrate away from the anode (random movement from higher to
lower concentrations). As they move away, more Fe(s) is exposed underneath the drop.
2+
At the same time, the reaction:
Is occurring at the oxygen-rich outer surface of the drop. When the Fe2+ reaches this region, it
encounters the OH- and precipitates as Fe(OH)2(s).
The Fe(OH)2 is eventually oxidized to a complex mixture of Fe2O3 and H2O by the O2 in the air. Rust
is Fe2O3.XH2O where ‘X’ can change. Rust can have numerous different colours (red, brown, yellow,
black) since differing numbers of water molecules attached to the iron(III) oxide will change the
colour of the compound.
A metal can corrode if it touches a different type of metal in the presence of an electrolyte solution
exposed to oxygen. For example, if iron touches copper wire and the spot where they touch gets wet,
then:
The copper conducts the electrons away from the Fe and makes them available to the oxygen/water
touching the wire.
PREVENTING CORROSION
There are several ways to stop or at least slow down corrosion, all of which fall into two main
categories.
1. ISOLATING THE METAL FROM ITS ENVIRONMENT
a.
b.
2. ELECTROCHEMICAL METHODS
a. Cathodic protection
-32Ex. Both Mg and Zn are ________________________ than iron, so if pieces of
magnesium or zinc are attached to the surface of iron, the Mg or Zn will:
Ex.
Strips of zinc are often bolted to the iron-hull of ships below the water line. The zinc is
‘sacrificed’ (and eventually needs to be replaced) to keep the iron of the ship from
oxidizing and corroding. Some ships even pass a low voltage electric current into the
hull from an electric generator. This forces electrons into the metal and prevents it
from being oxidized.
Ex.
Galvanized iron simply has a zinc coating. The zinc reacts preferentially with air and
water, forming a zinc oxide layer which protects the iron as it adheres strongly and
prevents the exposure of the underlying metal to air and water.
Ex.
Some buried gas and oil tanks made of steel have a thick braided wire connected to
them. The wire comes to the surface and is attached to a post in the ground that is
made of an easily oxidized metal such as magnesium or zinc. Because the post will
oxidize first, the buried tank is protected.
b. Change the conditions of the surroundings
When iron is placed in contact with water containing oxygen, the following halfreaction will oxidize the iron:
If oxygen is removed from the system, the tendency for it to reduce is drastically
reduced. (Hydrogen will be reduced first as it is higher on the table).
Another method is to lower the [H+] ions by adding OH- ions. A piece of iron will not
rust (or a very little amount until all the oxygen has reacted) in a basic solution.
XI. ELECTROLYSIS
ELECTROLYSIS
-33ELECTROLYSIS OF A MOLTEN BINARY SALT
BINARY
SALT
When such a salt is melted, the ions are free to move in the liquid form (molten NaCl is NaCl(l), as
only Na+ and Cl- ions are present. Do not confuse it with NaCl(aq) which is a solution containing
Na+, Cl-, and H2O.) Also note that there is no need for a salt bridge to keep the reactant separated as
no spontaneous reaction will occur between the reactants.
The only reactants present are Na+ and Cl-
The anode reaction is:
The cathode reaction is:
The overall reaction is:
In order for the above cell to operate:
Since the half cells are not at standard state, the reduction potentials will be different than those listed
on the table.
-34ELECTROLYSIS OF AN AQUEOUS SOLUTION
Consider the electrolysis of aqueous sodium iodide, NaI(aq). This involves another consideration –
now water is also present in the system. Inert electrodes are used and the cell is set up as follows:
There are two possible reducing agents:
There are two possible oxidizing agents:
So, in order to determine which reaction will occur, think back to the definition of electrolysis –
electrical energy is applied to produce a chemical change. It makes sense that the reaction that:
Looking at the above example. The preferred reaction (requiring the least voltage input) involves the
higher of the two possible reductions (strongest oxidizing agent) and the lower of the two possible
oxidations (strongest reducing agent). The same situation applies to electrolysis as did for
spontaneous reactions:
-35The half-reactions and overall reaction for the electrolysis of NaI(aq) is:
Note – The concentrations of the materials in cells in not relevant – as long as there is sufficient
material in the cell you can assume that the reactions proceed as predicted.
IN REALITY
It is most often found that a higher potential than calculated must be applied to cause electrolysis.
This is due to:
The difference between actual potentials required for electrolysis and the calculated potentials are
termed_________________________. As a result of the overpotential effect, when dilute neutral
solutions ( <1.0 M) containing Cl- or Br- are electrolyzed:
-36Oxidation:
You would expect that O2 would be produced. In reality it is found that _____ or _____ are actually
produced.
When dilute solution of certain metals are electrolyzed:
Reduction:
You would expect that H2O is produced, but in practice _____________________ are produced.
Ex.
What products are formed at the anode and cathode and what is the overall reaction when a
solution containing NiSO4(aq) is electrolyzed using inert electrodes? Determine the minimum
voltage required.
Ex.
-37What is the overall reaction which occurs when a 1.0 M solution of HCl(aq) is electrolyzed
using carbon electrodes?
XII. ELECTROPLATING
ELECTROPLATING
The Cathode is:
The Electroplating Solution is:
The Anode is:
You will only be asked about electroplating in NEUTRAL solutions.
-38Ex. Design a cell to electroplate a copper medallion with nickel metal. Include in the design:
a. the ions present in the solution
b. the direction of ion flow
c. the substance used for the anode and cathode
d. the direction of electron flow when the cell is connected to a DC power source
ELECTROREFINING
At the anode the small amounts of Zn or Pb present is preferentially oxidized as it is exposed at the
surface. When any exposed Pb/Zn atoms have oxidized and gone into solution as ions, only the Cu
atoms are available to be oxidized. Any Au, Ag, or Pt atoms present cannot be oxidized because the
anode is mostly copper which is oxidized in preference to Au, Ag, or Pt which simply drop off and
accumulate on the bottom of the cell. This ‘anode sludge’ can be purified to obtain the valuable
metals.
-39XIII. APPLIED ELECTROCHEMISTRY
A. THE BREATHALYSER
When alcohol is consumed, it is absorbed into the blood stream from the stomach. Some of it passes
through the cell walls of blood capillaries into the alveoli (air sacs that make up the lungs), in a similar
manner that CO2(g) and O2(g) enter in and out of the lungs. The process of alcohol entering the
bloodstream and entering the lungs is at equilibrium so that the greater the concentration of alcohol in
the blood, the greater the concentration of alcohol in the lungs. When the lungs exhale, the ethanol in
the lungs is expelled.
Ethanol undergoes oxidation by an acidic solution of dichromate ions as follows:
3 C2H5OH + 2 K2Cr2O7 + 8 H2SO4  3 CH3COOH + 2 Cr2(SO4)3 + 2 K2SO4 + 11 H2O
(Cr2O72- is orange/yellow)
(Cr3+ is dark green)
The expelled air is blown into a breathalyzer which can measure and record the amount of green
colourization. The more green the more alcohol in the breath. The machine is calibrated to be able to
accurately calculate the blood alcohol content.
B. BATTERIES
1. The Lead-Acid Storage Battery
A car battery is of this type, consists of alternating pairs of plates of Pb(s) and PbO2(s)
immersed in dilute sulphuric acid.
The anode reaction is: Pb(s) + HSO4-(aq)  PbSO4(s) + H+(aq) + 2e- (or Pb  Pb2+ + 2e-)
The cathode reaction is: PbO2(s) + HSO4-(aq) +3H+ +2e-  PbSO4(s) + 2H2O
(or Pb4+ + 2e-  Pb2+)
The overall reaction (Pb + Pb4+  Pb2+ + Pb2+) occurs when the battery is discharging –
spontaneously reacting to produce electrical energy. This forms an insoluble PbSO4(s) layer
on the plates of the battery. When an external charge is applied, the reaction is driven
backwards. (Over time the PbSO4 tends to flake off the plates so less Pb and PbO2 can be
formed, so eventually the battery needs to be replaced.)
-402. The alkaline battery
Gets its name from the alkaline (basic) electrolyte (KOH) that is used.
The cathode reaction: 2 MnO2(s) + H2O(l) +2e-  Mn2O3(s) + 2 OH-(aq)
The anode reaction: Zn(s) + 2OH-(aq)  ZnO(s) + H2O(l) +2e-
Cheap to make, but cannot be recharged (or reversed), has a relatively short shelf life (reaction
continually occurs slowly whether the battery is being used or not).