In the previous videos, we looked at electrochemical cells with reactive metal electrodes and...

In the previous videos, we looked at electrochemical cells with reactive metal electrodes and solutions containing their cations. However, some electrochemical cells have half-reactions involving gases. We’ll look at one of those here.

Electrochemical Cells

Involving Gases

Example 1

We’re given the diagram for an electrochemical cell

Sn

V

1 M HCl1 M

Sn(NO3)2

1 M KNO3

Pt

H2(g)

Given the electrochemical cell shown in the diagram:a) Write the half-reaction at the

anode along with its E°.b) Write the half-reaction at the

cathode along with its E°.c) Write the equation for the

overall redox reaction and its E°.

d) What is the initial voltage of this cell?

e) As the cell operates electrons will travel toward which electrode?

f) As the cell operates, what will happen to the pH of the solution near the Pt electrode?

g) As the cell operates, what will happen to the mass of Platinum?

The left beaker has a platinum electrode

Sn

V

1 M HCl1 M

Sn(NO3)2

1 M KNO3

Pt

H2(g)

platinum electrode

Which is surrounded by hydrogen gas

Sn

V

1 M HCl1 M

Sn(NO3)2

1 M KNO3

Pt

H2(g)

hydrogen gas

And dipped into a solution of 1 molar hydrochloric acid

Sn

V

1 M HCl1 M

Sn(NO3)2

1 M KNO3

Pt

H2(g)

1M HCl

Hydrochloric acid contains aqueous hydrogen ions and chloride ions.

Sn

V

1 M HCl1 M

Sn(NO3)2

1 M KNO3

Pt

H2(g)

H+(aq) and Cl–

(aq)H+

Cl–

The beaker on the right has a tin electrode

Sn

V

1 M HCl1 M

Sn(NO3)2

1 M KNO3

Pt

H2(g)

tin electrode

H+

Cl–

Immersed in a solution of 1 molar tin(II) nitrate.

Sn

V

1 M HCl1 M

Sn(NO3)2

1 M KNO3

Pt

H2(g)

H+

Cl–

tin(II) nitrate

Which contains aqueous Sn (2 plus) ions, and nitrate ions.

Sn

V

1 M HCl1 M

Sn(NO3)2

1 M KNO3

Pt

H2(g)

H+

Cl–

Sn2+(aq) NO3

–(aq)

Sn2+

NO3–

Looking back at the half-cell on the left, it’s important to know that (click) platinum is an inert metal. The platinum electrode does not react in this cell. It only provides a surface upon which the half-reaction can take place.

Sn

V

1 M HCl1 M

Sn(NO3)2

1 M KNO3

Pt

H2(g)

H+

Cl–

The Platinum Electrode is Inert

Inert metals like platinum are often used in half-cells where gases are involved. Gases and aqueous ions need a solid surface to react with each other on.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

Used in half-cells

where gases are involved

This particular half-cell is very important. (click) It’s called the standard half-cell.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

The Standard Half-

Cell

The reactive species in the standard half-cell are H2 gas and aqueous H+ ions.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

The StandardHalf-Cell

Reactive species are

H2(g) andH+ ions.

The chloride ions, Cl minus, are spectators in this particular half-cell.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–


Reactive species are

H2(g) andH+ ions.

The chloride ions, Cl–, are spectators in this half-cell

The half-reaction for the standard half-cell is the shaded half-reaction right in the middle of the reduction table

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–


The Standard Half-

Cell

Here it is enlarged .

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–


The Half-reaction for the Standard

Half-Cell

This half-cell is assigned a reduction potential of 0.00 Volts under standard conditions.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–


E° = 0.00 V

Temperature = 25°CPressure = 1 atm (101.3

kPa)[H+] = 1 M

Standard Conditions

The little “naught” on the E naught means standard conditions.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–


E° = 0.00 V


kPa)[H+] = 1 M

Standard Conditions

At standard conditions (click) the temperature is 25°C.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–


E° = 0.00 V


kPa)[H+] = 1 M

Standard Conditions

The pressure is 1 atmosphere, or 101.3 kilopascals.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–


E° = 0.00 V


kPa)[H+] = 1 M

Standard Conditions

And the concentration of H+ is 1 Molar

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–


E° = 0.00 V


kPa)[H+] = 1 M

Standard Conditions

The double arrow here reminds us that this half-reaction can occur either as a reduction and proceed to the right or as an oxidation and proceed to the left. The direction it goes depends on what other half-cell it is connected to.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–


Can occur either as a Reduction or an

Oxidation

Now that we know the half-cell on the left is the standard half-cell, we’ll look at the overall cell and go through a series of questions.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)









H+

Cl–

1 M KNO3

The “a” part of the question asks us to write the half-reaction at the anode, along with its E naught value.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)









H+

Cl–

1 M KNO3

We find the half-reactions on the reduction table.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3

a) Write the half-reaction at the anode along with its E°.

The higher half-reaction on the table, the hydrogen half-cell, will act as the cathode

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


cathode

And the lower one, the tin half-cell will act as the anode.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


cathode

anode

Because the tin half-reaction is the anode, oxidation is taking place, so the equation must be reversed (click) so its written as Sn solid gives Sn (2+) plus 2 electrons.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Anode

Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

anode

Also, because it the equation was reversed, the sign on the E naught must be switched, so (click) the E naught for this half-reaction is positive 0.14 Volts. This is the oxidation potential of tin metal

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Anode

Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

anode

The “b” part of the question asks us to write the half-reaction at the cathode, along with its E naught value.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)









H+

Cl–

1 M KNO3

ANODE

The cathode half-reaction is the hydrogen half-cell.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3

a) Write the half-reaction at the cathode along with its E°.

cathode

Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

anode

And The cathode half-reaction is not reversed, so its (click) 2H+ plus 2 electrons gives H2(gas)

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

anode

cathode

And Its E naught value is equal to zero Volts

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

anode

cathode

The “c” part of the question asks us to write the equation for the overall redox reaction, along with its E naught value.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)









H+

Cl–

1 M KNO3

ANODE

Cathode

To write the equation for the overall redox reaction, we add up the half-reactions the way they are written.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3

c) Write the equation for the overall redox reaction and its E°.

Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

anode

cathode

Electrons gained are equal to electrons lost, so we don’t need to multiply any of the half-reactions and we can (click) cancel out the electrons.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V Electrons Gained = Electrons

Lost

ANODE

Cathode

anode

cathode

On the left side we have Sn (solid) and 2H (plus)

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

And on the right side, we have Sn (2 plus) and H2 (gas)

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

anode

cathode

To find the overall E naught, we add up positive 0.14 and zero, giving us (click) positive 0.14 volts

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

anode

cathode

Remember the E naught value for an overall redox equation at standard conditions, is called (click) the standard cell potential

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V

Standard Cell

Potential

ANODE

Cathode

anode

cathode

The “d” part of the question asks us to state the initial voltage of this cell.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)









H+

Cl–

1 M KNO3

ANODE

Cathode

Remember that standard cell potential is the same as the initial voltage of a cell.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V

Standard Cell

Potential

Initial Voltage

ANODE

Cathode

anode

cathode

Initial Voltage

So the initial voltage of this cell is positive 0.14 volts.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

anode

cathode

We’ll make a note of this up here by the voltmeter.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V

V = +0.14 V

ANODE

Cathode

anode

cathode

The “e” part of the question asks which way electrons are travelling as this cell operates. The cell will operate if we replace the voltmeter with a light bulb.

Sn

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)









H+

Cl–

1 M KNO3

ANODE

Cathode

Remember that electrons always travel from the anode to the cathode through the wires.

Sn

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3

d) As the cell operates electrons will travel toward which electrode?

Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

Electrons travel from the Anode

toward the Cathode in the Wires

anode

cathode

So they are travelling from the tin electrode toward the platinum electrode.

Sn

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3

d) As the cell operates electrons will travel toward which electrode?

Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

Electrons travel toward the Platinum

Electrode

e–

anode

cathode

The “f” part of the question asks what will happen to the pH around the platinum electrode as this cell operates.

Sn

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)









H+

Cl–

1 M KNO3

ANODE

Cathode

To answer this question, we focus on the half-reaction taking place on the platinum electrode.

Sn

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3

d) As the cell operates what will happen to the pH of the solution near the Pt electrode?

Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

We see that H+ is on the left side of this half-reaction, so H+ is (click) being consumed as it is reduced to hydrogen gas.

Sn

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

H+ is being consumed as it is reduced to

H2(g)

anode

cathode

Remember from the Acid-base unit, that (click) pH is the NEGATIVE log of the hydrogen ion concentration.

Sn

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

pH = –log[H+]So as [H+]

decreases, the pH increases

anode

cathode

So as H+ is used up and the concentration of H+, or acidity decreases, (click) the pH will gradually increase near the platinum electrode.

Sn

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

pH = –log[H+]So as [H+]

decreases, the pH increases

anode

cathode

The “g” part of the question asks what will happen to the mass of the platinum electrode as this cell operates.

Sn

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)









H+

Cl–

1 M KNO3

ANODE

Cathode

Remember the half-reaction taking place on the platinum electrode is the reduction of H+ ions to hydrogen gas.

Sn

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3

d) As the cell operates, what will happen to the mass of Platinum?

Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

The platinum electrode itself is inert. It doesn’t undergo any reaction.

Sn

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

The Platinum Electrode is Inert

It simply provides a solid surface for this reaction to take place on.

Sn

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

The platinum electrode provides a surface for this half-reaction to take place

on

So as the cell operates (click), the mass of the Platinum electrode will NOT change in any way.

Sn

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3


Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

The mass of the platinum electrode will NOT change

The platinum electrode provides a surface for this half-reaction to take place

on

anode

cathode

To review, by using a cell diagram, the table of reduction potentials, and the principals we learned, we were able to answer a number of questions about this electrochemical cell.

Sn

V

1 M HCl1 M

Sn(NO3)2

Pt

H2(g)

H+

Cl–

1 M KNO3

Sn(s) Sn2+ + 2e– E°= +0.14 V2H+ + 2e– H2(g) E°= +0.00 V

.

Sn(s) + 2H+ Sn2+ + H2(g) E°= +0.14 V A

NODE

Cathode

V = +0.14 V

anode

cathode

In the previous videos, we looked at electrochemical cells with reactive metal electrodes and...

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Transcript of In the previous videos, we looked at electrochemical cells with reactive metal electrodes and...