CHEM1612 - Pharmacy
Week 9: Nernst Equation
Dr. Siegbert Schmid
School of Chemistry, Rm 223
Phone: 9351 4196
E-mail: siegbert.schmid@sydney.edu.au
Unless otherwise stated, all images in this file have been reproduced from:
Blackman, Bottle, Schmid, Mocerino and Wille,
Chemistry, John Wiley & Sons Australia, Ltd. 2008
ISBN: 9 78047081 0866
Electrochemistry
n
Blackman, Bottle, Schmid, Mocerino & Wille:
Chapter 12, Sections 4.8 and 4.9
Key chemical concepts:
n  Redox and half reactions
n  Cell potential
n  Voltaic and electrolytic cells
n  Concentration cells
Key Calculations:
n  Calculating cell potential
n  Calculating amount of product for given current
n  Using the Nernst equation for concentration cells
Lecture 25-3
Recap: Standard cell potential
n
The measured voltage across the cell under standard conditions is the
standard cell potential E0cell (also called emf).
Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)
Zn(s) + Cu2+(aq) à Zn2+(aq) + Cu(s)
E0cell = E0cathode – E0anode = 0.34 - ( - 0.76)= 0.34 +0.76 = 1.10 V
Lecture 26 - 4
Tricks to memorise anode/cathode
n
1. Anode and Oxidation begin with a vowels,
Cathode and Reduction with consonants.
n
2. Alphabetically, the A in anode comes before the C in cathode, as
the O in oxidation comes before the R in reduction.
n
3. Think of this picture:
AN OX and a RED CAT
(anode oxidation reduction cathode)
Lecture 26 - 5
Standard cell potential and free energy
n
For a spontaneous reaction, E0cell > 0
For a non-spontaneous reaction, E0cell <0
n
So there is a proportionality between E0 and -ΔG0.
n
You also know that the maximum work done on the surroundings is
n
and also ΔG0 < 0
and also ΔG0 > 0
-wmax = ΔG
n
Electrical work done by the cell is
w = Ecell × charge
Lecture 26 - 6
Standard cell potential and free energy
n
The emf E0cell is related to the change in free energy of a reaction:
∆G0 = –nFE0cell
∆G0 = Standard change in free energy
n = number of electrons exchanged
F = 96485 C/mol e- (Faraday constant)
Also, away from standard conditions:
∆G = –nFEcell
v  But what is Ecell?
Lecture 26 - 7
Example
Calculate ∆G0 for a cell reaction:
Cu2+(aq) + Fe (s) D Cu(s) + Fe2+ (aq)
Is this a spontaneous reaction?
Cu2+ + 2e– D Cu
E0 = 0.34 V
Fe2+ + 2e– D Fe
E0 = –0.44 V
E0cell = 0.34 - (-0.44) = 0.78 V
∆G0 = –nFE0cell
∆G0 = – 2 · 96485 · 0.78 = – 1.5 · 105 J
This process is spontaneous as indicated by the negative sign of ΔG0
and the positive sign for E0cell.
Lecture 26 - 8
Example: a Dental Galvanic Cell
Al(s)|Al3+(aq) || O2(aq), H+(aq), H2O(aq)|Ag,Sn,Hg
Al is very easily oxidised,
Al3+ + 3e− à Al Eo = -1.66V.
The filling is an inactive
cathode for the
reduction of oxygen,
O2 + 4H+ + 4e− à 2H2O
and saliva is an electrolyte.
Put the three together (biting on a piece of foil) results in generation
of a current and possible pain.
Lecture 26 - 9
Example: a Dental Galvanic Cell
Al(s)|Al3+(aq) || O2(aq), H+(aq), H2O(aq)|Ag,Sn,Hg
O2 + 4H+ + 4e– à 2H2O E0 = 1.23 V
Al3+ + 3e– à Al
E0 = –1.66 V
12H+ + 3O2 + 4Al à 6H2O + 4Al3+
E0cell = 2.89 V
∆G0 = –nFE0cell=
–12 · 96485 · 2.89 = –3346 kJmol–1
Lecture 26 - 10
v  But what is Ecell?
Recall ΔG = ΔG0 + RT ln(Q)
Since
Image fromnobelprize.org
Nernst Equation
(1)
ΔG0 = -nFE0cell and ΔG = -nFE
Equation (1) becomes
Walther Nernst
Nobel Prize 1920
-nFE = -nFE0cell + RTln(Q)
Q = [products] / [reactants]
dividing both sides by –nF gives:
E = E0 – RT ln(Q)
nF
Lecture 26 - 11
Nernst Equation
Ecell = E0 – RT ln(Q)
nF
Since ln (x) = 2.303 log (x)
E = E0 – 2.303 · RT log(Q)
nF
E = cell potential
E0 = Standard cell potential
R = Real Gas Constant
= 8.314 JK-1mol-1
T =Temperature (K)
n = no. of e- transferred
F = Faraday constant
= 96485 C mol-1
Q = Reaction quotient
(Q = K at equilibrium)
At 25°C, (2.303·R·298)/96485 = 0.0592
Ecell
0.0592
=E −
log(Q)
n
0
Nernst equation more commonly
written like this (note: only at 25°C)
Lecture 26 - 12
Example calculation 1
Calculate the expected potential for the following cell:
Zn(s) + Cu2+(aq) D Zn2+(aq) + Cu(s)
E0 = 1.1 V
i) [Cu2+] = 1.0 M; [Zn2+] = 10-5M
ii) [Cu2+] = 10-5M; [Zn2+] = 1.0 M
Ecell
0.0592
=E −
log(Q)
n
0
Firstly, work out the value of n :
e-
2 mol transferred per
mole of reaction: n = 2
Cu2+ + 2e- D Cu
Zn
D Zn2+ + 2e-
Zn(s) + Cu2+(aq) D Zn2+(aq) + Cu(s)
Lecture 26 - 13
Example calculation
Zn(s) + Cu2+(aq) D Zn2+(aq) + Cu(s)
Ecell
0.0592
0
=E −
log(Q)
n
[Zn2+] (M)
[Cu2+] (M)
1.0
1.0
10-5
1.0
(n=2)
[Zn 2+ ]
Q=
[Cu 2+ ]
Q
log(Q)
Ecell (V)
1.0
0.0
1.10
1.0
10-5
-5.0
1.25
10-5
105
5.0
0.95
Lecture 26 - 14
Demo: The effect of concentration
0.00 V
Cu|Cu2+||Cu2+|Cu
Both compartments of the voltaic cell are identical.
E0cell = E0copper – E0copper = 0 (in standard conditions, 1M concentrations)
What happens when the concentration of one cell is changed?
Lecture 26 - 15
Demo: Cu Concentration Cell
Low [Cu2+]
Cu D Cu2++ 2e-
High [Cu2+]
Cu2+ + 2e- D Cu
Cu |Cu2+||Cu2+|Cu
E0 same for both half-reactions, so E0cell= 0.
However, we have reduced the
concentration of Cu2+ in one cell = nonstandard conditions.
Electrical energy is generated until the
concentrations in each half-cell become
equal (equilibrium is attained).
Can we explain this?…
Add Na2S –precipitate forms.
Lecture 26 - 16
Cu Concentration Cell
Cu|Cu2+||Cu2+|Cu
Low [Cu2+]
Cu D Cu2++ 2en
n
n
n
High [Cu2+]
Cu2+ + 2e- D Cu
E0cell is same on both sides, but the Cu concentrations are different.
More charge carriers in one half-cell.
If we poured the two solutions together, we would expect
spontaneous mixing of two solutions of different concentrations to
give one of equal concentration.
The electrical connection allows electrons to pour from one half-cell
to the other.
Lecture 26 - 17
Concentration cells
n
“Voltage”
The measured cell potential in our experiment was “Voltage”.
Let’s work out what the voltage should be:
Cu à Cu2+ (0.01 M) + 2eCu2+ (0.1M) + 2e- à Cu
Cu2+ (0.1M)
Ecell
à Cu2+ (0.01 M)
Ecell = “Voltage”
0.0592
y ⎞
⎛0.01
=
0
.
0
−
0
.
0296
⋅
log
=E −
log(Q)
⎜
⎟ = “Voltage”
n
⎝ 0.1⎠
0
Solve for “Voltage”: “Voltage” = 0.0296 V
Lecture 26 - 18
Concentration cells
The cell potential depends on the concentration of reactants.
Corollary: It is useful to define a standard concentration, which is 1 M.
Implication: We need to specify concentration when referring to the cell
potential.
The overall potential for the Cu/Cu2+ concentration cell is:
E = E0cell – 0.0592/2 · log [Cu2+]dil / [Cu2+]conc
Lecture 26 - 19
Reference Electrodes
1.
Standard Hydrogen Electrode (SHE)
2.
Metal-Insoluble Salt Electrode: Standard Calomel
Electrode (SCE) and Silver Electrode
3.
Ion-Specific: pH electrode
Lecture 26 - 20
Standard Hydrogen Electrode (SHE)
Finely divided surface Pt
electrode.
HCl solution with [H+] =1,
H2 p= 1 atm bubbling over the
electrode.
H2 absorbs on the Pt, forming the
equivalent of a 'solid hydrogen‘
electrode in equilibrium with H+.
Platinum – gas electrode
H2 electrode
Metal – Metal ion Electrode
H2 D 2H+ + 2 e- Eo = 0.00 V
Cu2+ + 2e- D Cu Eo = 0.34 V
Pt|H2(g)|H+(aq)||Cu2+(aq)|Cu(s) Eocell = 0.34 – 0 = 0.34 V
anode
cathode
Lecture 26 - 21
Metal-insoluble salt electrodes
n
The Standard Hydrogen Electrode (SHE) isn't convenient to use in
practice (can be contaminated easily by O2 or organic substances).
n
There are more practical choices, like metal - insoluble salt electrodes.
The potential of these electrodes depends on the concentration of the
anion X- in solution.
n
In practice 2 interfaces:
X-
+
1. M / MX insoluble salt: MX (s) + e- à M (s) + X-(s) C
X-
C+
M
2. coating/solution: X- (s) à X- (aq)
C+
X-
Overall: MX (s) + e-(metal) à M (s) + X- (aq)
C+
MX
X
-
Lecture 26 - 22
Metal-insoluble salt electrodes
MX
n
The concentration of anions in solution is controlled
by the salt's solubility:
[Ag+] [X-] = Ksp
X-
C+
C+
XM
C+
XC+
n
Normal calomel electrode,
n
Saturated calomel electrode Pt | Hg | Hg2Cl2 | KCl(sat.) E0 = 0.24 V
n
Silver/Silver chloride,
Ag | AgCl | Cl- (1M) E 0 = 0.22 V
(used in pH meters)
X-
Pt | Hg | Hg2Cl2 | KCl (1M) E 0 = 0.28 V
Lecture 26 - 23
Saturated Calomel Electrode
n
n
The ‘saturated calomel electrode’ (SCE) features the reduction halfreaction:
Hg+ + e– D Hg
Hg2Cl2 D 2Hg+ + 2Cl–
Overall:
Hg2Cl2 (s) + 2e– à 2 Hg (s) + 2Cl– (sat)
5M
Pt | Hg | Hg2Cl2 | KCl ||
n
Standard cell potential of
E0 = 0.24 V.
2
Lecture 26 - 24
Cell Potentials 1
Q: The standard reduction potential of Zn2+/Zn is - 0.76 V. What
would be the observed cell potential for the Zn/Zn2+ couple when
measured using the SCE as a reference?
Calomel: Hg2Cl2(s) + 2e- à 2Hg(l) + 2Cl-(aq) E0 = 0.24V
Zn à Zn2+ + 2eE0 = + 0.76V
(reversed because it is written as an oxidation)
0.24 Hg+/Hg
0.0V H2/H+
Ans: The Zn will be oxidised (lower reduction potential), so
E (cell) = 0.76 + 0.24 = 1.00 V respect to the SCE
So to get the oxidation half-reaction E0 using the SCE as
cathode, subtract 0.24 V from the volt meter reading.
-0.76 Zn2+/Zn
Lecture 26 - 25
Summary
CONCEPTS
q  Concentration cells
q  Nernst equation
CALCULATIONS
q  Work out cell potential from reduction potentials;
q  Work out cell potential for any concentration (Nernst equation)
Lecture 26 - 26
Silver electrode
Ag+ + e– D Ag
AgCl D Ag+ + Cl–
Overall: AgCl + e– D Ag + Cl–
E0 = 0.22 V
AgCl (s) + e– à Ag (s) + Cl– (sat)
A thin coating of AgCl is deposited
on the pure metal surface.
Ag | AgCl | Cl–
Lecture 26 - 27
Cell Potentials 2
n
A Fe3+/Fe2+ cell with [Fe3+]=[Fe2+] =1 M has a potential of
0.55V respect to the Ag/AgCl electrode (E0= 0.22 V). What
is the potential of this electrode with respect to the SHE?
?
0.22 Ag+/Ag
Answer.
The reactions that occur in the two half-cells are:
Fe3+ + e- → Fe2+
at the cathode E = 0.55 V; E0 = ?
Ag + Cl- +e-→ AgCl
at the anode
E0= 0.22 V
0.0V H2/H+
The potential of this electrode with respect to the SHE is the
difference of the two electrode potentials:
E = E0 - 0.22
E0Fe3+/Fe2+ = 0.55 + 0.22 = 0.77 V
Lecture 26 - 28
Measurement of pH
We could construct a concentration cell,
using the standard hydrogen electrode (SHE)
and a hydrogen electrode:
unknown
H2(g, 1 atm) à 2H+(aq, unknown) + 2e-
anode
2H+ (aq, 1M) + 2e- à H2(g, 1 atm)
cathode
2H+ (1M) à 2H+ (unknown)
Ecell = ?
Using Nernst equation:
2
RT
[H+ ]un
RT
0
Ecell = E −
ln + 2 = 2 ln [H+ ]un = 0.0592 ⋅ pH
nF
2F
[H ]ref
at 25°C,
i.e. the measurement of the cell potential provides pH directly!
Lecture 26 - 29
pH electrode
The pH electrode potential is typically measured
versus a fixed reference calomel electrode.
Eglass electrode= E’ + RT/2.303F log [H+]
E’ is the sum of the constant offset potentials of
the inner glass surface/solution, the Ag/AgCl
electrode, and the calomel electrode.
•  Based on a thin glass membrane: a modified glass enriched in H+
and resulting in a hydration layer a few micrometers thick.
•  Inside the membrane is a 'reference solution' of known [H+] (1M HCl).
•  The potential difference relevant to pH measurement builds up across
the outside glass/solution interface marked ||.
Lecture 26 - 30
Nernst Equation
The Nernst equation describes the effect of concentration on cell potential.
Ecell = E 0 −
Q=
0.0592
log(Q)
n
[products]
[reactants]
Zn(s) + Cu2+(aq) D Zn2+(aq) + Cu(s)
When Q < 1, [reactants] >[products] and the cell can do more work.
When Q = 1, Ecell = E0cell (standard conditions [x] = 1 M).
When Q > 1 , [products] > [reactants] and Ecell is lower.
Lecture 27 - 31
Difference between Q and K
n
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
Breakdown of N2O4 to NO2:
N2O4 (g, colourless) → 2 NO2 (g, brown)
Q=
[ NO2 ]2
[ N2O4 ]
Q =K
Lecture 27 - 32
Potential of an electrochemical cell
Zn(s) + Cu2+(aq) D Zn2+(aq) + Cu(s)
Ecell (Volts)
Ecell
0.0592
=E log( Q )
n
0
K
[Zn 2+ ]
Q=
[Cu 2+ ]
0V
~1037
Ecell decreases as the reaction proceeds, until at equilibrium Ecell =0 and
E0 =
0.0592
log(K ) .
n
Lecture 27 - 33
Recap: Examining Q To K Ratios
Ecell = E 0 −
0.0592
log(Q)
n
If Q/K < 1
n  Ecell is positive for the reaction as written. The smaller the Q/K ratio,
the greater the value of Ecell and the more electrical work the cell
can do.
If Q/K = 1,
n  Ecell = 0. The cell is at equilibrium and can no longer do work.
If Q/K > 1,
n  Ecell is negative for the reaction as written. The cell will operate in
reverse – the reverse reaction will take place and do work until Q/K
= 1 at equilibrium.
Lecture 27 - 34
Link between E 0 and K
Q: What happens if the reaction proceeds until equilibrium is reached?
A: The reaction stops, therefore the voltage, or electrical potential, is zero
(the battery is flat). In mathematical terms:
Ecell
0.0592
=E −
log(Q) = 0
n
0
At equilibrium Q=K
E0 =
0.0592
log(K )
n
So the equilibrium constant determines the cell potential.
Large K _ products favoured _ large standard cell potential, E0
Lecture 27 - 35
Redox reactions are special
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
§  For redox reactions
there is a direct
experimental method to
measure K and ΔG°.
Lecture 27 - 36
Relation between
0
E
and K
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
0.0592
E =
log(K )
n
0
K is plotted on a
logarithmic scale to
give a straight line.
Lecture 27 - 37
Example question 2
Q: A voltaic cell consisting of a Ni/Ni2+ half-cell and Co/Co2+ half-cell is
constructed with the following initial concentrations: [Ni2+] = 0.80 M;
[Co2+]=0.2 M.
a) What is the initial Ecell?
b) What is the [Ni2+] when the voltage reaches 0.025 V?
c) What are the equilibrium concentrations of the ions?
Given: E 0 Ni2+/Ni = -0.25 V; E 0 Co2+/Co = -0.28 V
Ni2+ + 2e- → Ni
E 0 = -0.25V
Co → Co2+ + 2e-
E 0 = +0.28V
Co(s) + Ni2+(aq) D Co2+(aq) + Ni(s) E 0 = 0.03 V
Lecture 27 - 38
Example question 2a
a) What is the initial Ecell?
Co(s) + Ni2+(aq) D Co2+(aq) + Ni(s)
E 0 = 0.03 V
[Co 2+ ] 0.2
Q=
=
= 0.25
2+
[Ni ] 0.8
Ecell
0.0592
0.0592
=E −
log(Q) = 0.03 −
log(0.25)
n
2
0
= 0.03 − 0.0296 × (−0.602) = 0.048 V
Lecture 27 - 39
Example question 2b
b) What is the [Ni2+] when the voltage reaches 0.025V?
Ecell = E 0 −
0.0592
log(Q)
n
0.025 = 0.03 − 0.0296 × log(Q)
0.005
log(Q) =
= 0.169
0.0296
Co(s) + Ni2+(aq) D Co2+(aq) + Ni(s)
0.80 - x
Q = 1.47
0.20+x
[Co 2+ ] (0.2 + x)
Q = 1.47 =
=
2+
[Ni ] (0.8 − x)
[Co2+] = 0.60 M
1.176 − 1.47 x = 0.2 + x
2.47 x = 0.976
So when Ecell = 0.025 V
x = 0.40
[Ni2+] = 0.40 M
Lecture 27 - 40
Example question 2c
c) What are the equilibrium concentrations of the ions?
0.03 = 0.0296 × log(K )
0.03
log(K ) =
= 1.014
0.0296
0.0592
0=E −
log(K )
n
0
Co(s) + Ni2+(aq) D Co2+(aq) + Ni(s)
0.80-x
K = 10.24
0.20+x
[Co 2+ ] (0.2 + x)
K = 10.24 =
=
2+
[Ni ] (0.8 − x)
8.192 − 10.24 x = 0.2 + x
11.24 x = 7.986
So at equilibrium,
[Co2+] = 0.91 M
x = 0.71
[Ni2+] = 0.09 M
Lecture 27 - 41
Concentration Cells in Nature
Concentration cells are present all around us, e.g.
n
nerve signalling: concentration gradients produce electrical current
n
ion pumps across cell membranes: Na+ / K+ pump, Ca2+ pump
n
energy production and storage in cells: ATP
Lecture 27 - 42
Nerve cells
Energy from ATP hydrolysis is used by ion pumps,
so that across the nerve cell membrane
concentration gradients are maintained:
Figure from Silberberg, “Chemistry”,
Ion Concentration Gradient:
McGraw Hill, 2006.
Inside
Outside
K+
High
Low
Na+
Low
High
The membrane potential is more positive outside than inside the cell.
n  On nerve stimulation, Na+ enters cell, the inside cell membrane
becomes > +ive, then K+ ions leave cell to re-equilibrate the outside.
n  These rapid (ms) changes in charge across the membrane stimulate
the neighbouring region and the electrical impulse moves down the
length of the cell.
Lecture 27 - 43
Nerve cells
n
Nernst Equation gives the membrane potential generated by the
differing extracellular versus intracellular concentrations of each ion:
Eion =
2.303RT
[ion]outside
log
nF
[ion]inside
Substituting n = 1, T =
Eion = 61.5 log
37oC:
[ion]outside
[ion]inside
(Eo = 0)
Consider K+:
( in mV)
1 mV = 10-3 V
EK +
[K+]outside = 3 mM,
[K+]inside = 135 mM
[K + ]outside
= 61.5 log +
[K ]inside
⎛ 3 ⎞
EK + = 61.5 log⎜
⎟
⎝ 135 ⎠
= 61.5 ⋅ (− 1.65 ) mV
= −102 mV
Lecture 27 - 44
Cellular Electrochemistry
n
Biological cells apply the principles
of electrochemical cells to generate
energy in a complex multistep -O
process.
O
P
NH2
adenosine
triphosphate
O
O
O-
P
N
O
O
O-
P
O
O-
Bond energy in food is used to
generate an electrochemical
potential.
N
N
ATP
O
H
n
N
H
H
OH
H
OH
H2O
NH 2
n
The potential is used to create the
bond energy of the high-energy
molecule adenosine triphosphate
(ATP) (energy currency for the cell).
adenosine
diphosphate
O
-O
P
O
ATP4- + H2O → ADP3- + HPO42- + H+
ΔG °’ = -30.5 kJ mol-1
N
O
O
-
N
P
O
O
-
O
H
H
H
OH
H
OH
N
N
ADP
+ HPO42- + H+
ΔGo’ (solution at pH 7 and at human body temperature 37oC.)
Lecture 27 - 45
Bond Energy to Electrochemical Potential
n
Inside mitochondria, redox reactions are performed by a series of
proteins that form the electron-transport chain (ETC) which contain
redox couples, such as Fe2+/Fe3+.
Large potential differences provide enough energy to convert ADP
into ATP.
Lecture 27 - 46
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
n
Bond Energy to Electrochemical Potential
In nature, the most important reducing agent is a complex molecule
named nicotinamide adenine dinucleotide, abbreviated NADH, which
functions as a hydride donor (H-).
= NADH = biological reducing
agent
NAD+ = biological oxidising
agent
Lecture 27 - 47
Electron Transport Chain (ETC)
ETC consists of three main steps, each of which
has a high enough potential difference to produce
one ATP molecule.
The reaction that ultimately powers ETC is the
reduction of oxygen in the presence
of NADH:
2H+ + 2e- + ½ O2 → H2O
Eo’ = +0.82
NADH + H+ → NAD+ + 2H+ + 2e- Eo’ = -0.32
NADH(aq) + H+(aq) + ½O2(aq) → NAD+(aq) + H2O(l)
Eo’overall = 1.14 V
ΔGo’ = -nFEo’ = -2 · 96485 C mol-1· 1.14 V = - 220 kJ mol-1
Substantial
energy release!
Lecture 27 - 48
ATP Synthesis
§  In short: e- are transported along the
chain, while protons are forced into the
intermembrane space.
§  This creates a H+ concentration cell
across the membrane.
§  In this step, the cell acts as an
electrolytic cell, i.e. uses a spontaneous
process to drive a non-spontaneous
process.
Figure from Silberberg, “Chemistry”,
McGraw Hill, 2006.
§  When [H+]intermembrane/[H+]matrix ~ 2.5 a
trigger allows protons to flow back
across membrane, and ATP is formed.
§  It’s not simple: Noble prize in 1997 to Boyer and Walker for elucidating this.
Lecture 27 - 49
Summary
CONCEPTS
q  Concentration cells
q  Nernst equation
q  E 0 and K
q  Link between E , Q and K
q  Applications of concentration cells
CALCULATIONS
q  Work out cell potential from reduction potentials;
q  Work out cell potential for any concentration (Nernst equation)
q  Work out K from E 0
q  Work out pH from concentration cell
Lecture 27 - 50
Pop Quiz 1
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Balance the following reaction in basic solution:
MnO4- + CN- à MnO2 + CNO-
Answer: H2O + 2 MnO4- + 3 CN- --> 2 MnO2 + 3 CNO- + 2 OH-
Lecture 26 - 51
Pop Quiz 2
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Balance the following reaction in basic solutions:
NO3- + Zn à Zn2+ + NH3
Answer: NO3- + 4 Zn + 6 H2O → 4 Zn2+ + NH3 + 9 OH-
Lecture 26 - 52