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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
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–
The StandardHalf-Cell
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 StandardHalf-Cell
The Standard Half-
Cell
Here it is enlarged .
Sn
V
1 M HCl1 M
Sn(NO3)2
Pt
H2(g)
H+
Cl–
The StandardHalf-Cell
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–
The StandardHalf-Cell
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–
The StandardHalf-Cell
E° = 0.00 V
Temperature = 25°CPressure = 1 atm (101.3
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–
The StandardHalf-Cell
E° = 0.00 V
Temperature = 25°CPressure = 1 atm (101.3
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–
The StandardHalf-Cell
E° = 0.00 V
Temperature = 25°CPressure = 1 atm (101.3
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–
The StandardHalf-Cell
E° = 0.00 V
Temperature = 25°CPressure = 1 atm (101.3
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–
The StandardHalf-Cell
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)
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?
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)
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?
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
a) Write the half-reaction at the anode along with its E°.
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
a) Write the half-reaction at the anode along with its E°.
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
a) Write the half-reaction at the anode along with its E°.
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
a) Write the half-reaction at the anode along with its E°.
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)
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?
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
a) Write the half-reaction at the cathode along with 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
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
a) Write the half-reaction at the cathode along with 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
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)
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?
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
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 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
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
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
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
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
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
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
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
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)
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?
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
d) What is the initial voltage of this cell?
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
d) What is the initial voltage of this cell?
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
d) What is the initial voltage of this cell?
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)
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?
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)
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?
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
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
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
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
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
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
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)
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?
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
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 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
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 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
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 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