Answer:
Au^3+(aq) +3e ------> Au(s). 1.50 V
Explanation:
When we construct the galvanic cell, our intention is to produce energy by spontaneous electrochemical reactions. In order to have a spontaneous electrochemical reaction, E°cell must be positive. The more positive the value of E°cell, the more spontaneous the reaction is.
E°cell= E°cathode - E°anode
If E°cathode= 1.50 V
E°anode= -0.25 V
E°cell= 1.50 -(-0.25)
E°cell= 1.75 V
Hence the process; Au^3+(aq) +3e ------> Au(s) yields the highest standard cell potential
Which of the following is the balanced reaction, given the rate relationships below.
a. rate = − 13 Δ[CH4] Δt = − 12 Δ[H2O] Δt = − Δ[CO2] Δt = 14 Δ[CH3OH] Δt
b. rate = − 12 Δ[N2O5] Δt = 12 Δ[N2] Δt = 15 Δ[O2] Δt
c. rate = − 12 Δ[H2] Δt = − 12 Δ[CO2] Δt = − Δ[O2] Δt = 12 Δ[H2CO3] Δt
Answer:
a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]
b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]
c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]
Explanation:
Hello,
In this case, since those rate relationships have the stoichiometric coefficient at the denominators of the fractions preceding each ratio and the negative terms account for reactants and positive for products, we have:
a. [tex]3CH_4+2H_2O+CO_2\rightarrow 4CH_3OH[/tex]
b. [tex]2N_2O_5\rightarrow 2N_2 + 5O_2[/tex]
c. [tex]2H_2+2CO_2+O_2\rightarrow 2H_2CO_3[/tex]
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How many unit cells share an atom that is located at the center of a cube edge of a unit cell?
Answer:
zero
Explanation:
In a unit cell, an atom that is located at the center of a cube edge is not involved in sharing unit cells because a central atom of a unit cell belongs to the entire cell and only to that unit cell of the lattice.
Hence, the center atom of a unit cell do not share any unit cell and the correct answer is "Zero".
Which solution has the greatest buffer capacity? Select the correct answer below: 1 mole of acid and 1 mole of base in a 1.0 L solution
Answer:
The answer is
Explanation:
1 mole of acid.
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A buffer that is 1 M in acid and base will have the greatest capacity of buffer, and therefore the greatest buffer capacity.
What do you mean by the buffer solution ?A weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base, are combined to form the buffer solution, a water-based solvent solution.
In a biological system, a buffer's keep intracellular and extracellular pH levels within a relatively small range and to withstand pH fluctuations brought on by both internal and external factors.
A buffer is a substance that can withstand a pH shift when acidic or basic substances are added. It may balance out little quantities of additional acid or base, keeping the pH stable.
Thus, 1 M in acid and base solution has the greatest buffer capacity.
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What is the pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid with 15.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive. K a = 1.8 ×× 10-5 for CH3CO2H.
Answer:
pH = 8.72
Explanation:
This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
As this compound acts like a base, we propose this equilibrium:
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰
Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M
So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
We can avoid the quadractic equation because Kb is so small
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH = 8.72
The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
Calculation of the pH of the solution:Since the following equation should be used.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
Now
(0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
So,
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
Now
Kw = Ka. Kb
Kb = Kw/Ka
And,
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵
= 5.55×10⁻¹⁰
Now
[CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
Now
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH
= 8.72
Hence, The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
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If the amount of radioactive iodine-123, used to treat thyroid cancer, in a sample decreases from 3.2 to 0.4 mg in 39.6 h, what is the half-life of iodine-123?
Answer:
Half life = 13.197 hour
Explanation:
Given:
Old amount (A₀) = 3.2
New amount (A) = 0.4
Radiation decay time (t) = 39.6 hour
Half life = T(1/2)
Find:
Half life = T(1/2) = T
Computation:
A = A₀[tex]e^{-(\frac{0.693t}{T} )}[/tex]
[tex]e^{-(\frac{0.693t}{T} )}[/tex] = 0.4 / 3.2
-[27.4428 / T] = In (0.125)
-[27.4428 / T] = -2.0794
[27.4428 / T] = 2.0794
T = 13.197
Half life = 13.197 hour
20. Stoichiometry is based on
A. molecular weight.
B. temperature.
C. conservation of matter.
D. pressure.
Answer:
The correct option is (c)
Answer:
the law of conservation of mass
what is the bond energy required to break one mole of carbon-carbon bonds
Answer:
100 kcal of bond energy
9. In order to determine the melting point of tetradecanol, it was important that _____. the thermometer was attached to the capillary tube the water was frozen the capillary tube was immersed in the water without the thermometer
To record the melting point of tetradecanol, it is important the temperature is increase by little degrees after each trial.
The thermometer attached to the capillary tube was immersed in water without thermometer.
Usually heating curve describes the temperature change for a sample which passes through different states of matter.
The temperature than decreases and the solution in beaker cools down. To determine the melting point, it is important to keep the temperature consistent.
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The second law of thermodynamics predicts that heat flow from a cooler object to a hotter object:________
a) will be spontaneous at high pressure
b) will be spontaneous at low pressure
c) will never be spontaneous at any pressure
d) will always be spontaneous
Answer:
c) will never be spontaneous at any pressure.
Explanation:
Hello,
In this case, we need to remember that the second law of thermodynamics states that heat flow is transferred from a hot object to a cold object only, never the other way around, therefore, the answer is c) will never be spontaneous at any pressure. This is supported by driving force that in this case is the temperature difference which must be negative for the hot object (it gets eventually cold) and positive for the cold object (it gets eventually hot) until they reach the equilibrium temperature.
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Pentanone was treated with excess sodium cyanide in HCl (aq) followed by hydrogen gas has over Pd. This produced:________
A. 2-amino-1-hexanol
B. 1-amino-2-methylpentan-2-ol
C. 1-cyano-1-pentanol
D. 2-aminomethylpentan-1-ol
Answer:
B. 1-amino-2-methylpentan-2-ol
Explanation:
In this case, the first step, we have the attack of the nucleophile cyanide ([tex] CN^-[/tex] produced by sodium cyanide to the carbon on the carbonyl group (C=O) producing a negative charge in the oxygen.
Then HCl protonates the molecule to produce a cyanohydrin. This cyanohydrin can be reduced by the action of hydrogen gas ([tex]H_2[/tex]) in the presence of a catalyst ([tex]Pd[/tex]), producing an amino group. With this in mind, the final molecule is: 1-amino-2-methylpentan-2-ol.
See figure 1 to further explanations
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. You have two solutions, both with a concentration of 0.1M. Solution A contains a weak acid with a pKa of 5. ThepH of solution A is 3. Solution B contains a weak acid with a pKa of 9. The pH of solution B is:
Answer:
pH of solution B is 5
Explanation:
A weak acid, HA, is in equilibrium with water as follows:
HA(aq) + H₂O(l) ⇄ A⁻(aq) + H₃O⁺(aq)
Where Ka (10^-pKa = 1x10⁻⁹) is:
Ka = 1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
Where concentrations of this species are equilibrium concentrations
As initial concentration of HA is 0.1M, the equilibrium concentrations of the species are:
[HA] = 0.1M - X
[A⁻] = X
[H₃O⁺] = X
Where X is the amount of HA that reacts until reach the equilibrium, X is reaction coordinate.
Replacing in Ka expression:
1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
1x10⁻⁹ = [X] [X] / [0.1 - X]
1x10⁻¹⁰ - 1x10⁻⁹X = X²
1x10⁻¹⁰ - 1x10⁻⁹X - X² = 0
Solving for X:
X = -0.00001 → False solution, there is no negative concentrations.
X = 1x10⁻⁵ → Right solution.
As [H₃O⁺] = X
[H₃O⁺] = 1x10⁻⁵M
And pH = -log[H₃O⁺]
pH = 5
pH of solution B is 5
Which option is an isotope with 6 protons and 8 neutrons? (1 point)
O oxygen-6
o carbon-14
o oxygen-14
o carbon-8
Answer:
Carbon-14
Explanation:
It is because it have 6 protons and 8 neutrons which are added together to form an atomic mass of 14.
The mass number of an isotope is the sum of the number of protons and number of neutrons hence the correct answer is carbon-14.
Isotopes are atoms of the same element that has the same number of protons but different number of neutrons.
The atomic number is the same as the number of protons in the atom. Since there are six protons in the atom, the element must be carbon.
There are eight neutrons meaning that the mass number is 8 + 6 = 14
Therefore, the isotope is carbon-14.
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What must happen to uranium before it can be used as a fuel source?
Answer: Uranium enrichment. Uranium is used to fuel nuclear reactors; however, uranium must be enriched before it can be used as fuel. Enriching uranium increases the amount of uranium-235 (U235) that can sustain the nuclear reaction needed to release energy and produce electricity at a nuclear power plant.
How many neutrons does Carbon- 14 and Carbon -15 have? *
Answer: 8 for both
Explanation:
Janet observes that bubbles rise inside water when water is heated. Which of the following best names and explains the change that causes bubbles to rise?
Answer:
Boiling
Explanation:
When a liquid is heated, the vapor pressure rises steadily. When water attains a temperature of 100°C or 212°F its vapor pressure is now equal to the atmospheric pressure at sea level, this is what we mean by boiling.
When this occurs, water continues to evaporate untill the vapor pressure inside the bubbles becomes high enough to stop water bubbles from collapsing again from the pressure of the water around it so the bubbles rise and break the surface.
g Which ONE of the following pure substances will exhibit hydrogen bonding? A) methyl fluoride, FCH3 B) dimethyl ether, CH3C–O–CH3 C) formaldehyde, H2C=O D) trimethylamine, N(CH3)3 E) hydrazine, H2N-NH2
Answer:
C) formaldehyde, H2C=O.
Explanation:
Hello,
In this case, given that the hydrogen bondings are known as partial intermolecular interactions between a lone pair on an electron rich donor atom, particularly oxygen, and the antibonding molecular orbital of a bond between hydrogen and a more electronegative atom or group. Thus, among the options, C) formaldehyde, H2C=O, will exhibit hydrogen bonding since the lone pair of electrons of the oxygen at the carbonyl group, are able to interact with hydrogen (in the form of water).
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How many moles of gold are equivalent to 1.204 × 1024 atoms? 0.2 0.5 2 5
Answer:
c
Explanation:
How many moles of gold are equivalent to 1.204 × 1024 atoms?
0.2
0.5
2
5
C) 2 Is the correct answer, I took the test and it was correct.
According to the concept of Avogadro's number, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.
What is Avogadro's number?
Avogadro's number is defined as a proportionality factor which relates number of constituent particles with the amount of substance which is present in the sample.
It has a SI unit of reciprocal mole whose numeric value is expressed in reciprocal mole which is a dimensionless number and is called as Avogadro's constant.It relates the volume of a substance with it's average volume occupied by one of it's particles .
According to the definitions, Avogadro's number depend on determined value of mass of one atom of those elements.It bridges the gap between macroscopic and microscopic world by relating amount of substance with number of particles.
Number of atoms can be calculated using Avogadro's number as follows: mass/molar mass×Avogadro's number.Number of moles=number of atoms/Avogadro's number=1.204×10²⁴ /6.023×10²³=1.999≅2
Thus, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.
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When cyclopentane undergo free-radical substitution with bromine (Br2 /Heat) the product:
a. Bromo cyclopentene
b. Bromo cyclopentane + HBr
c. Bromo cyclopentene + HBr
Explanation:
b. Bromo cyclopentane + HBr
Calculate the silver ion concentration in a saturated solution of silver(I) sulfate (K sp = 1.4 × 10 –5). 1.5 × 10–2 M 1.4 × 10–5 M 3.0 × 10–2 M 2.4 × 10–2 M None of the above.
Answer:
3.0x10⁻²M
Explanation:
Silver sulfate, Ag₂SO₄, has a product constant solubility equilbrium of:
Ag₂SO₄(s) ⇄ 2Ag⁺ + SO₄²⁻
When an excess of silver sulfate is added, some Ag₂SO₄ will react producing Ag⁺ and SO₄²⁻ until reach the equilbrium determined for the formula:
ksp = 1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]
Assuming the Ag₂SO₄ that react until reach equilibrium is X, we can replace in Ksp expression:
1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]
1.4x10⁻⁵ = [2X]² [X]
1.4x10⁻⁵ = 4X³
3.5x10⁻⁶ = X³
0.015 = X
As [Ag⁺] is 2X:
[Ag⁺] = 0.030 = 3.0x10⁻²M
The answer is:
3.0x10⁻²MThe substance formed on addition of water to an aldehyde or ketone is called a hydrate or a/an:_______
A) vicinal diol
B) geminal diol
C) acetal
D) ketal
Answer:
B) geminal diol
Explanation:
Hello,
In this case, considering the attached picture, you can see that the substance resulting from the hydrolysis of an aldehyde or a ketone is a geminal diol since the two hydroxyl groups are in the same carbon. Such hydrolysis could be carried out in either acidic or basic conditions depending upon the equilibrium constant.
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Provide two real world examples of habitats and niches within ecosystems? Explain the interdependence that exists within those ecosystems.
Answer:
An ecosystem encompasses living organisms and the nonliving elements of their environments. Hence, the components of an ecosystem include animals, plants, microorganisms, rocks, soil, minerals, atmosphere, and the surrounding water masses. An ecosystem can be huge, cutting across several nations, or it can be relatively small, such as the body of an animal, which is home to numerous microorganisms.
We can describe ecosystems under two headings – natural ecosystems and unnatural ecosystems. Unnatural ecosystems, which include agricultural and urban areas, are greatly modified and maintained by human activity. Conversely, natural ecosystems are self-sufficient, balanced ecological units, with a high proportion of native biodiversity and minimal human disruption.
The natural ecosystem is broad. It is divided into two major groups – terrestrial and aquatic ecosystems. These are further divided into many other smaller types of ecosystems as outlined in this article.
Explanation:
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You are performing an experiment that involves the electrolysis of gold (I) bromide, also know as AuBr. You know that gold is less reactive than hydrogen. Which of the following would be the product of the reaction?
A. Hydrogen gas
B. Gold bromide
C. Oxygen gas
D. Pure gold
Answer:
D. Pure gold
Explanation:
Hello,
In this case, since gold, as a heavy metal, is said to be less reactive than hydrogen, when it undergoes electrolysis process when forming a salt, due to the action of the electric current, we can appreciate the formation of a layer of gold on the surface of the cathode via a reduction half-reaction from gold (I) to metallic gold:
[tex]Au^++1e^-\rightarrow Au^0[/tex]
Thereby, D. Pure gold is formed as the product of the reaction.
In contrast, more reactive metals than hydrogen such as sodium or potassium, will remain in solution so the hydrogen converted to hydrogen gas.
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Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days. If you begin with 48.0 mg of this isotope, what mass remains after 47.9 days have passed?
Answer:
After 47.9 days, will remain 14.5mg of the isotope
Explanation:
The radioactive decay follows always first-order kinetics where its general law is:
Ln[A] = -Kt + ln[A]₀
Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.
We can find rate constant from half-life as follows:
[tex]t_{1/2} = \frac{ln2}{K}[/tex]
K = ln 2 / 27.7 days
K = 0.025 days⁻¹
Replacing, initial amount of isotope is 48.0mg = [A]₀ , K is 0.025 days⁻¹ and t = 47.9 days:
Ln[A] = -Kt + ln[A]₀
Ln[A] = -0.025 days⁻¹*47.9 days + ln (48.0mg)
ln [A] = 2.6726
[A] = e^ (2.6726)
[A] = 14.5mg
After 47.9 days, will remain 14.5mg of the isotope
For the following reaction, 22.8 grams of diphosphorus pentoxide are allowed to react with 13.5 grams of water . diphosphorus pentoxide(s) water(l) phosphoric acid(aq) What is the maximum mass of phosphoric acid that can be formed
Answer:
[tex]m_{H_3PO_4}=31.5gH_3PO_4[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]P_2O_5(s)+3H_2O(l)\rightarrow 2H_3PO_4(aq)[/tex]
Thus, since the diphosphorus pentoxide to water molar ratio is 1:3 and we are given the mass of both of them, for the calculation of the maximum mass phosphoric acid that is yielded, one could first identify the limiting reactant, for which we compute the available moles of diphosphorus pentoxide (molar mass 142 g/mol):
[tex]n_{P_2O_5}=22.8gP_2O_5*\frac{1molP_2O_5}{142gP_2O_5}=0.161molP_2O_5[/tex]
And the moles of diphosphorus pentoxide that are consumed by 13.5 g of water (molar mass 18 g/mol):
[tex]n_{P_2O_5}^{consumed}=13.5gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molP_2O_5}{3molH_2O} =0.25molP_2O_5[/tex]
Hence, since less moles of diphosphorus pentoxide are available, we sum up it is the limiting reactant, therefore, the maximum mass of phosphoric acid (molar mass 98 g/mol) is computed by considering the 1:2 molar ratio between them as follows:
[tex]m_{H_3PO_4}=0.161molP_2O_5*\frac{2molH_3PO_4}{1molP_2O_5} *\frac{98gH_3PO_4}{1molH_3PO_4} \\\\m_{H_3PO_4}=31.5gH_3PO_4[/tex]
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For the following set of volume/temperature data, calculate the missing quantity after the change is made. Assume that the pressure and the amount of gas remain constant.
V=2.91 L at 23.0 °C
V= 4.20 L at ? °C
Answer:
155 °C
Explanation:
Step 1: Given data
Initial volume (V₁): 2.91 LInitial temperature (T₁): 23.0°CFinal volume (V₂): 4.20 LFinal temperature (T₂): ?Step 2: Convert the initial temperature to Kelvin
We will use the following expression.
K = °C + 273.15 = 23.0°C + 273.15 = 296.2 K
Step 3: Calculate the final temperature
Assuming an ideal gas behavior, we can calculate the final temperature using Charles' law.
V₁/T₁ = V₂/T₂
T₂ = V₂ × T₁/V₁
T₂ = 4.20 L × 296.2 K/2.91 L
T₂ = 428 K
Step 4: Convert the final temperature to Celsius
We will use the following expression.
°C = K - 273.15 = 428 - 273.15 = 155 °C
What will be formed when 2,2,3-trimethylcyclohexanone reacts with hydroxylamine?
Answer:
Following are the solution to this equation:
Explanation:
In the given-question, an attachment file of the choices was missing, which can be attached in the question and its solution can be defined as follows:
In the given question "Option (iii)" is correct, which is defined in the attachment file.
When 2,2,3-trimethylcyclohexanone reacts with hydroxylamine it will produce the 2,2,3-trimethylcyclohexanoxime.
2) What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly 230 mL of solution?
If a salt is formed by combining NH3 (Kb=1.8×10−5) and CH3COOH (Ka=1.8×10−5), an aqueous solution of this salt would be:
Answer:
Neutral
Explanation:
pKa of acid = -log Ka
= -log (1.8 x 10^-5)
= 4.74
pKb of base = -log Kb
= 4.74
pKa of acid = pKb of base
salt pH formula : pH = 7 + 1/2 [pKa -pKb ]
here pKa = pKb
so pH = 7
the salt it is CH3COONH4 exactly neutral solution .
If a salt is formed by combining NH₃ (Kb=1.8×10⁻⁵) and CH₃COOH (Ka=1.8×10⁻⁵), an aqueous solution of this salt would be neutral.
What information does pH convey?pH of any solution tells about the acidity or basicity or neutral nature of the solution.
pH of any solution is directly proportional to the acid dissociation constant value (Ka) and base dissociation constant (Kb). In the question it is given that,
Value of Kb for NH₃ = 1.8×10⁻⁵
Value of Ka for CH₃COOH = 1.8×10⁻⁵
Ka & Kb values for the base and acid is same means it dissociates with same extent. So the aqueous solution of this acid and base is a neutral in nature as they have same number of acid and base ions in it.
Hence resultant solution will be a neutral solution .
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What is the maximum number of electrons in the following energy level?
n = 4
Answer: 32 electrons
Explanation:
Determine what product will be produced at the negative electrode for the following reaction:
2KCl(aq) + 2H20(1) -> H2(g) + Cl2(g) + 2KOH(aq)
A. H2
B. Cl2
с. КОН
D. K
Answer:
Choice A. [tex]\rm H_{2}[/tex] would be produced at the negative electrode.
Explanation:
Ionic equation for this reaction:
[tex]2\, {\rm K^{+}} + 2\, {\rm Cl^{-}} + {2\, \rm H_{2} O} \to {\rm H_{2}} + {\rm Cl_{2}} + 2\, {\rm K^{+}} + {\rm 2\, OH^{-}}[/tex].
Net ionic equation:
[tex]2\, {\rm Cl^{-}} + 2\, \rm H_{2} O} \to {\rm H_{2}} + {\rm Cl_{2}} + 2\, {\rm OH^{-}}[/tex].
Half-equations:
[tex]2\, {\rm Cl^{-}} \to {\rm Cl_{2}} + 2\, {e^{-}}[/tex].
(Electrons travel from the solution to an electrode.)
[tex]2\, {\rm \overset{+1}{H}_{2} O} + 2\, {e^{-}} \to \overset{0}{\rm H}_{2} + 2\, {\rm O\overset{+1}{H}\!^{-}}[/tex].
(An electrode supply electrons to the solution to reduce some of the [tex]\rm H[/tex] atoms from [tex]\rm H_{2}O[/tex].)
In a DC circuit, electrons always enter the circuit from the negative terminal of the power supply and return to the power supply at the positive terminal.
The negative electrode is connected to the negative terminal of the power supply. Electrons from the power supply would flow into the solution through this electrode.
This continuous supply of electrons at the negative electrode would drive a reduction half-reaction. In this question, that corresponds to the reduction of water: [tex]2\, {\rm \overset{+1}{H}_{2} O} + 2\, {\rm e^{-}} \to \overset{0}{\rm H}_{2} + 2\, {\rm O\overset{+1}{H}\!^{-}}[/tex]. Hence, [tex]\rm H_{2}[/tex] would be produced at the negative electrode.