20.1 The electrochemical series

Question 1

What is the basis of how batteries work

Answer 1

. If you connect two different metals together in a salt solution, an electrical current flows

. The current flows from the more reactive metal to the less reactive metal

Question 2

How is an equilibrium set up in electrochemical cells

What does this have to do with half cells

Answer 2

When you dip a rod of metal into a solution of its own ions, an equilibrium is set up

Eg dip zinc into zinc sulphate solution sets up equilibrium:
Zn(s) <—> Zn2+ + 2e-

This set up is called an electrode, or a half cell.

Question 3

What is a half cell
How is an electrical cell made

Answer 3

. It is an electrode
. Two half cells join together to make an electrical cell

Question 4

If you could measure the potential of a half cell, eg
Zn <—> Zn2+ + 2e-
What would it tell us

Answer 4

It would tell us how readily the electrons are released by the metal, so how good a reducing agent it is

Question 5

Since you can’t measure the electrical potential directly, only Pd, how can you find the potential

Answer 5

. Connect together two different electrodes and measure the potential difference between them with a voltmeter for copper and zinc electrodes

Question 6

What is a salt bridge

Why is it used instead of a piece of wire

Answer 6

It connects two electrodes
Eg it can be filter paper soaked in a solution of salt eg potassium nitrate KNO3

And it transfers ions

A piece of wire only transfers electrons, whilst a salt bridge transfers ions

Question 7

When the two electrodes are zinc and copper, what does the reading of the voltmeter tell us

Answer 7

. When they are connected to voltmeter it gives reading 1.1 V
(if solutions are 1moldm^-3 and temp 298K)
. So this shows that the zinc is more negative so loses its electrons more readily than copper does
. So zinc is a better reducing agent

Question 8

If the voltmeter was removed from the connected electrodes zinc and copper, and the electrons were allowed to flow from zinc to copper, what changes would take place

What would the half equations for this be

How is electricity produced from this

Answer 8

. Zinc would dissolve to form Zn2+, which would increase the concentration of Zn2+(aq)
. The electrons would flow through the wire to the copper rod where they combine with Cu2+ ions from the copper sulphate solution
so will deposit fresh copper on the rod and decrease the concentration of Cu2+

Zn(s) –> Zn2+ + 2e-
Cu2+ + 2e- —> Cu

So combine to give overall equation
Zn(s) + Cu2+(aq) —> Zn2+(aq) + Cu(s)

So it is a redox reaction with zinc being oxidised and copper ions being reduced.
So if these half cells are connected they generate electricity.

This electrical cell is called the Daniell cell

Question 9

What is the standard hydrogen electrode used for

Answer 9

. To compare the tendency of different metals to release electrons, you can connect a standard electrode to another half cell to compare them

Question 10

How does the standard hydrogen electrode work

What is the metal used for electrical contact

Answer 10

. Hydrogen gas is bubbled into a solution of H+ ions

. Since hydrogen doesn’t conduct, electrical contact is made with a piece of un-reactive platinum metal (coated with finely divided platinum to increase surface area)

. The electrode is used under standard conditions:
298K, 100kPa,
and 1moldm^-3 conc of H+ ions

Question 11

What is the emf

What is Eº

What happens if electrodes have negative values of Eº

Answer 11

. The potential of SHE is defined as zero so if it is connected to another electrode, the measured voltage called the emf, is the electrode potential of that cell.

If the second cell is under standard conditions, the emf is given symbol Eº

So electrodes with negative values of Eº are better at releasing electrons so are better reducing agents than hydrogen.

Question 12

Is it possible to change electrode potential?

Answer 12

Yes, changing conditions such as temperature and concentration of ions

Question 13

What is the electrochemical series

Answer 13

. A list of Eº values for metal or metal ion standard electrodes
. The equilibria are always written as reduction reactions with electrons on the left of the arrow

So they are arranged with the most negative values at the top, eg
Li+ + e- –> Li which is -3.03

Question 14

If you connect an
Al3+(aq)/Al(s) standard electrode with Eº: -1.66
to a Cu2+(aq)/Cu(s) electrode
Eº : +0.34

How do you get the voltage

Answer 14

Voltage is obtaned by connectin gthw two standard electrodes together and finding the difference between the two Eº values

So 0.34- -1.66 equals 2.00V

The Al3+(aq)/Al(s) electrode will be negative

Question 15

If you connect an Al3+(aq)/Al(s) standard electrode with Eº: -1.66 and a Pb2+(aq)/Pb(s) standard electrode
Eº : -0.13

What will the emf be

Answer 15

The emf is the difference so
-0.13- -1.66 equals 1.53V

And the Al one is the negative electrode of the cell

Question 16

What is the way to write down the cell formed by connecting two electrodes

Answer 16

. A vertical solid line indicates a phase boundary, so eg between a solid and aqueous solution
. Double vertical line shows salt bridge
. The species with the highest oxidation state is written next to the salt bridge

When giving value of emf, state the polarity ( if it is positive or negative,) of the right hand electrode, as the cell representation is written .
So the side that is more positive (if it is connected to positive terminal of voltmeter) means that electrons flow from the more negative one to the more positive one

Question 17

How do you write the two cells aluminium and copper together

Using dashes instead of vertical lines

Answer 17

Al(s)/Al2+(aq)//Cu2+(aq)/Cu(s)
and Eº is +2V

Or
Cu(s)/Cu2+(aq)//Al2+(aq)/Al(s)
so Eº is-2V

Because electrons flowing from aluminium to copper as the polarity of the right hand electrode is always given

So emf= Eº(R) - Eº(L)
Where the R is the right electrode and the L is the left