The given question is incomplete, the complete question is:
Balance the following chemical equations: A) P4+ O2 → P406 B) _P406+ LO2 → P4010 a) A reaction container holds 5.33 g of P4 and 3.77 g of O2 and reaction A occurs. If enough oxygen is available then the P406 reacts further to undergo reaction B. What is the limiting reactant for the formation of P406? b) What mass of P406 is produced (theoretical yield in grams)? c) If 7.12g of P406 were obtained what is the percent yield? d) Will reaction B occur? Why or why not? What mass of excess reactant is left in the reaction container?
Answer:
The balanced reaction will be,
A) P₄ + 3O₂ ⇒ P₄O₆
B) P₄O₆ + 2O₂ ⇒ P₄O₆
a) Based on the given information, the reaction container holds 5.33 grams of P₄ and 3.77 grams of oxygen. Thus, the moles of P₄ will be,
Moles = mass of P₄/Molar mass of P₄ = 5.33 grams/124 g/mole = 0.043 mole
Now the moles of O₂ will be,
Moles = mass of O₂/Molar mass of O₂ = 3.77 grams/32 g/mol = 0.112 mole
Now the moles of P₄O₆ formed when 0.043 moles of P₄ react completely will be = 1/1 × 0.043 = 0.043 mole of P₄O₆
Similarly, the moles of P₄O₆ formed, when 0.112 moles of O₂ react completely will be = 1/3 × 0.112 = 0.0373 mole of P₄O₆
Thus, from the analysis, the maximum moles of P₄O₆ formed will be 0.0373 moles. Therefore, oxygen will be the limiting reagent, which will react completely in the reaction.
b) From the above findings, the maximum moles of P₄O₆ produced is 0.0373 mole. Thus, the theoretical yield of P₄O₆ produced will be,
= Moles of P₄O₆ × Molar mass of P₄O₆
Theoretical yield = 0.0373 mole × 220 g/mole = 8.206 grams
c) Based on the given information, the actual mass of P₄O₆ produced is 7.12 grams.
Hence, percent yield = Actual yield/Theoretical yield * 100
= 7.12/8.206 × 100 = 86.77 %
d) In the given case, reaction B will not take place. This is due to the fact that oxygen is not left for reaction B, which was the limiting regent for reaction A. Here P₄ is the excess reactant, which was left in the reaction.
The initial moles of P₄ is 0.043, O₂ is 0.112, and P₄O₆ is O. The final moles of P₄ is 0.043 -1/3 × 0.112 = 0.0057 mole, O₂ is 0, and P₄O₆ is 0.0373 mole.
Thus, moles of P₄ left is 0.0057 mole. Hence, the mass of P₄ left will be,
= 0.0057 mole × Molar mass of P₄
= 0.0057 mole × 124 g/mole = 0.7068 grams.
What does a decrease in temperature indicate about the molecules in a sample of liquid water? (3 points) a Decreased chemical energy of the molecules b Decreased kinetic energy of the molecules c Increased chemical energy of the molecules d Increased kinetic energy of the molecules
Answer:
option b is correct
Explanation:
when the temperature of liquid solid and gases increases the kinetic energy of their molecules increases but when the temperature of liquid solid or gases decreases then kinetic energy of the molecule decreases
Answer:
B. Decreased kinetic energy of the molecules
Explanation:
The average kinetic energy of molecules determines the temperature of a substance.
Hotter objects
have faster-moving particles and higher temperatures
Temperature measures the averages because moving particles can absorb/release heat with each collision as heat is converted to kinetic energy and back.
Chemistry V22 flvs 6.01
Express 7,376,000,000 (population of the world) in scientific notation
Answer:
7.376 × 10^9 people
Explanation:
yes
Answer the following questions in brief.
Why does silver get tarnished on exposure to air?
Answer: When silver is exposed to sulfur-containing gases in the air, it discolours and then darkens as it reacts with the gas to form a surface layer of tarnish. This process is called tarnishing.
Hope this helped :)
What is the diameter of the coin to the nearest
millimeter?
36 mm
37 mm
38 mm
39 mm
Answer:
Explanation:
Diameter in the coin is the distance from side to side.
The numbers in the ruler are showing the centimeters. Each line is a subdivision in ten parts between centimeter and centimeter (Each line is equal to 1mm).
As you can see in the ruler, in the edge of the coin, the ruler is showing 3.8cm
As 1cm = 10mm:
3.8cm * (10mm / 1cm) =
38mm is the diameter of the coin to the nearest milimeter
Answer:
38m
Explanation:
1. Determine whether or not the equation below is balanced. If it isn’t balanced, write the balanced form. Also, identify the reactant(s) and product(s) in this equation. Finally, label this as one of the five types of reactions: combination, decomposition, substitution, double replacement, or reversible.
Answer:
- Not balanced.
- Reactants: Zn and HCI .
- Products: ZnCl₂ and H₂.
- Substitution reaction.
Explanation:
Hello,
In this case, for the given reaction:
Zn + HCI → ZnCl₂ + H₂
We can see that it is not balanced due to the fact that at the left side we have one hydrogen atom whereas at the right side two, taking into account the number must be same as well as chlorine. Thus, in order to balance we write:
Zn + 2HCI → ZnCl₂ + H₂
And that is enough. Moreover, we can see that the chemical species at the left side of the equation are the reactants and those at the right side the products, thus we have:
Reactants: Zn and HCI .
Products: ZnCl₂ and H₂.
Finally, since we can see that the chlorine is at the reactants with hydrogen, but at the end with the zinc, and the initial zinc is alone as well as the yielded hydrogen we can infer this is a substitution reaction.
Best regards.
Answer:
The equation isn't balanced because there aren't equal numbers of hydrogen or chlorine atoms on both sides of the equation. To correct this, a coefficient of 2 must be placed in front of HCl on the left-hand side of the equation.
Explanation:
What is the limiting reactant for the reaction below given that you start with 10.0 grams of Al (molar mass 26.98 g mol-1) and 19.0 grams of O2 (molar mass 32.00 g mol-1)? Equation: 4Al + 3O2 → 2Al2O3
Answer:
Al
Explanation:
4 Al + 3 O₂ → 2 Al₂O₃
You need to figure out which one has the smaller mole ratio. Convert both substances from grams to moles.
(10.0 g Al)/(26.98 g/mol) = 0.3706 mol Al
(19.0 g O₂)/(32.00 g/mol) = 0.5938 mol O₂
Now, use the mole ratios of reactant to product to see which substance produces the least amount of product.
(0.3706 mol Al) × (2 mol Al₂O₃/4 mol Al) = 0.1853 mol Al₂O₃
(0.5938 mol O₂) × (2 mol Al₂O₃/3 mol O₂) = 0.3958 mol Al₂O₃
Since aluminum produces the least amount of product, this is the limiting reagent.
Al is a limiting reagent. A further explanation is below.
Given equation,
[tex]4Al+3O_2 \rightarrow 2Al_2O_3[/tex]So,
4 moles of Al reacts with 3 moles of O₂ to produce 2 moles of Al₂O₃.
Mass of Al = 10 gMolar mass = 26.98 g/molthen,
Number of moles of Al,
= [tex]\frac{10 \ g}{26.98 \ g/mol}[/tex]
= [tex]3.706\times 10^{-1} \ moles[/tex]
Now,
Mass of O₂ = 19 gMolar mass = 32 g/molthen,
Number of moles of O₂,
= [tex]\frac{19 \ g}{32 \ g/mol}[/tex]
= [tex]5.937\times 10^{-1} \ moles[/tex]
Now,
[tex]5.937\times 10^{-1} \ moles[/tex] of O₂ can react with,
= [tex]4\times \frac{5.937\times 10^{-1}}{3}[/tex]
= [tex]7.916\times 10^{-1} \ moles \ of \ Al[/tex]
But there is only [tex]3.706\times 10^{-1}[/tex] moles of Al available.
Thus the above approach is right.
Learn more:
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How will temperature affect the spontaneity of a reaction with positive H and S?
A. It will be spontaneous only at T = H/S.
B.Changing the temperature will not affect spontaneity.
C.A low temperature will make it spontaneous.
D.A high temperature will make it spontaneous.
Answer:
D.A high temperature will make it spontaneous.
Explanation:
we know that
Δ G = Δ H - TΔS
For spontaneous reaction
Δ G should be negative . If Δ G = 0
Δ H = TΔS
T = Δ H / Δ S
At temperature above this T , Δ G becomes negative if Δ S and ΔH are positive .
So at a high temperature will make it spontaneous.
Answer:
D. A high temperature will make it spontaneous.
Explanation:
A P E X
Determine the standard enthalpy of formation in kJ/mol for NO given the following information about the formation of NO2 under standard conditions, and (NO2) = + 33.2 kJ/mol. Keep one decimal point. 2 NO(g) + O2(g) → 2 NO2(g) ∆Hrxn = –114.2 kJ
Answer:
90.3 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ
We can find the standard enthalpy of formation for NO using the following expression.
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol
∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))
ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol
ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol
ΔH°f(NO(g)) = 90.3 kJ/mol
How many lbs. of O2 are required to fully convert 1-lb of glucose (C6H12O6) to carbon dioxide and water
Answer:
The correct answer is : 1.07 lbs
Explanation:
solution:
molar mass glucose (C6H12O6) = 180 g/mol
molar mass of oxygen molecule (O2) = 32 g/mol (as we know molar mass of O = 16 g/mol)
the balanced reaction of conversion of water and oxygen to glucose is:
[tex]C_{6}H_{12}O_{6} + 6O_{2} \rightarrow 6CO_{2} + 6H_{2}O ...(A)[/tex]
1 mol of C6H12O6 = 6 mol of O2 (from reaction A)
so, 180 g C6H12O6 = 192 g O2
that is, 0.396 lb of C6H12O6 = 0.423 lb of O2 ( 1 g = 0.00220462 lb )
so,
1 lb C6H12O6 =[tex]\frac{1(lb) \times 0.4233 (lb)}{0.3968 (lb)}[/tex] = 1.07 lb O2
therefore, the correct answer is : 1.07 lbs
1. While searching through a an old trunk you found a treasure map that says a treasure chest was burried 27
chains due west of the US Capitol Building. The treasure chest has 50 talents of gold and is buried 3 chains deep
under ground. Through your research you have discovered that 1 talent is 30.2 kg and 1 chain is 20.1 meters. How
many meters from the US Capitol should you dig, and how deep will you have to dig to find the treasure chest?
Answer:
See the answer below
Explanation:
From the illustration;
1 chain = 20.1 meters
1 talent = 30.2 kg
The map says the treasure chest was buried 27 chains due west and 3 chains deep under the ground.
27 x 20.1 = 542.7 meters.
3 x 20.1 = 60.3 meters
This means that one would need to travel 542.7 meters due west from the US Capitol and dig 60.3 meters under the ground in order to locate the treasure chest.
The treasure chest has 50 talents of gold:
50 x 30.2 = 1,510 kg of gold.
The product of the reaction between Aluminium Oxide (Al2O3) and Carbon (C) is Aluminium (Al) and carbon monoxide (CO). The reaction of 60.0 g of Al2O3 with 30.0 g of C produced 22.5 g of Al. What is the percent yield for this reaction?
Answer:
Percent yield = 70.8%
Explanation:
Equation of the reaction:
Al2O3 + 3C ----> 2Al + 3CO
From the reaction equation, 1 mole of Aluminum oxide produces 2 moles of Aluminum when reacted with 3 moles of carbon.
Molar mass of Aluminum oxide = 102 g; molar mass of Aluminum = 27 g; molar mass of C = 12
3 moles of C = 36 g
2 moles Al = 54 g
102 g of Al2O3 reacts with 36 g of C
60 g of Al2O3 will react with (36/102) × 60 g of C = 21.2 g of C
Therefore, Al2O3 is the limiting reactant while C is in excess.
102 g of Al2O3 produces 54 g of Al
60 g of Al2O3 will produce (54/102) × 60 of Al = 31.8 g of Al
Percent yield = (actual yield/ theoretical yield) × 100%
Percent yield = (22.5 g / 31.8 g) × 100%
Percent yield = 70.8 %
5.195 kg + 10.95 kg + 73.2 kg
Answer: 89.345
Explanation:
What are three ways that a solid, liquid, gas and plasma are different?
Answer:
like, plasmas have no fixed shape or volume, and are less dense then solid or liquids but unlike ordinary gases, plasma are made up of atoms in which some or all of the electrons have been stripped away and positively charged nuclei, called ions, roem freely
Which direction will the following reaction (in a 5.0 L flask) proceed if the pressure of CO_2(g) is 1.0 atm? CaCO_3(s) rightarrow CaO(s) + C02(g) Kp = 1.9 times 10^-23
a. To the right because Q > K_p
b. To the right because Q < K_p
c. To the left because Q < K_p
d. To the left because Q > K_p
Answer:
d. To the left because Q > K_p
Explanation:
Hello,
In this case, for the given reaction:
[tex]CaCO_3(s) \rightarrow CaO(s) + CO_2(g)[/tex]
The pressure-based equilibrium expression is:
[tex]Kp=p_{CO_2}[/tex]
In such a way, since Kp is given we rather compute the reaction quotient at the specificed pressure of carbon dioxide as shown below:
[tex]Q=p_{CO2}=1.0[/tex]
Therefore, since Q>Kp we can see that there are more products than reactants, which means that the reaction must shift leftwards towards the reactants in order to reestablish equilibrium, thus, answer is d. To the left because Q > Kp.
Regards.
A number should be rounded up if________? it is the first digit, the number after it is between 0 and 4, the number after is between 5 and 9, it is the last digit
Answer:
a number should be rounded up it is 5 or higher
Determine if the following compounds will be soluble or insoluble in water?
a) CrPO4
b) Na2S
c) PbBr2
d) Ag2SO4
e) Ca(ClO3)2
f) K3PO4
Answer:
a) Insoluble
b) Soluble
c) Insoluble
d) Insoluble
e) Soluble
f) Soluble
Explanation:
A sample of serum of mass 25 g is cooled from 290 K to 275 K at constant pressure by the extraction of 1.2 kJ of energy as heat. Calcuate q and ∆H and estimate the heat capacity of the sample.
Which is an example of the practical pursuit of alchemy
Answer:
A practical pursuit of alchemy was the development of metallurgy practices.
Explanation:
As alchemists always tried to turn various metals into other things, metalworking techniques were developed even though their experiments were often unsuccessful.
Answer: developing metalworking techniques
Explanation:
trust it!!
i need to know the measurements of this to the appropriate amount of significant figures
Answer:
[See Below]
Explanation:
13 cm.
For a voltaic cell consisting of Al(s) in Al(NO3)3(aq) and Cu(s) in Cu(NO3)2(aq), what is Ecell, given [Al3 ]
Answer:
2.0 V
Explanation:
For the oxidation half cell;
Al(s) -------> Al^3+(aq) + 3e.
For reduction half cell;
Cu^2+(aq) +2e ------> Cu(s).
E°cell = E°cathode - E°anode
But;
E°cathode= 0.34 V
E°anode = -1.66 V
E°cell= 0.34 -(-1.66)
E°cell= 2.0 V
The public is not yet able to ourchase cars powered by hydrogen fuel cells because engineers have to determine how the cars perform based on which scenario
Answer:
in real-world conditions
Explanation:
) Why is the Lewis structure for CO2 usually written as O = C = O rather than
O ≡ C - O ?
Answer:
The Lewis structure for CO2 is O = C = O rather than O ≡ C - O because of the covalent nature of the bonds that makes CO2 a polar molecule and to become stable it is important to fill valence shell of both atoms equally.
CO2 can not be written as O ≡ C - O because then there will be one non-paired electron in the oxygen atom that can disturb the stability of CO2.
Temperature is a measure of the amount of heat in a substance. 1) True 2) False
Answer:
False
Explanation:
Temperature is the measure of the amount of kinetic energy in a substance.
It is commonly assumed that temperature measures the amount of heat, because it is commonly associated with the weather or checking for fevers.
However, temperature is truly a measure of the average kinetic energy of a substance.
Answer:
Temperature is the measure of average kinetic energy in a substance.but heat is also a form of kinetic energy so i think its also true to say that temperature is a measure of heat in a substance.
Explanation:
A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure of 783 Torr . What is the final volume of the gas? g
Answer:
the final volume of the gas is [tex]V_2[/tex] = 1311.5 mL
Explanation:
Given that:
a sample gas has an initial volume of 61.5 mL
The workdone = 130.1 J
Pressure = 783 torr
The objective is to determine the final volume of the gas.
Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.
Converting the external pressure to atm ; we have
External Pressure [tex]P_{ext}[/tex]:
[tex]P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}[/tex]
[tex]P_{ext} = 1.03 \ atm[/tex]
The workdone W = [tex]P_{ext}[/tex]V
The change in volume ΔV= [tex]\dfrac{W}{P_{ext}}[/tex]
ΔV = [tex]\dfrac{130.1 \ J \times \dfrac{1 \ L \ atm}{ 101.325 \ J} }{1.03 \ atm }[/tex]
ΔV = [tex]\dfrac{1.28398717 }{1.03 }[/tex]
ΔV = 1.25 L
ΔV = 1250 mL
Recall that the initial volume = 61.5 mL
The change in volume V is [tex]\Delta V = V_2 -V_1[/tex]
[tex]- V_2= - \Delta V -V_1[/tex]
multiply through by (-), we have:
[tex]V_2= \Delta V+V_1[/tex]
[tex]V_2[/tex] = 1250 mL + 61.5 mL
[tex]V_2[/tex] = 1311.5 mL
∴ the final volume of the gas is [tex]V_2[/tex] = 1311.5 mL
The final volume of the gas when the volume of 61.5 mL should be 1311.5 mL.
Calculation of the final volume:Since
a sample gas has an initial volume of 61.5 mL
The workdone = 130.1 J
Pressure = 783 torr
So, here the external pressure should be
= 783 * 1 / 760
= 1.03 atm
Now the change in volume is
= 130.1 J * 1 / 101.325 J / 1.03
= 1250 mL
Now the change in volume should be
= 1250 + 61.5
= 1311.5 mL
hence, The final volume of the gas when the volume of 61.5 mL should be 1311.5 mL.
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A graduated cylinder contains 4.50 mL of water. After a piece of zinc is dropped into the cylinder, the water level rises to 9.24 mL. What is the volume of the piece of zinc?
Answer: Well, the volume of the copper is (63.4 - 40.0) * mL = 23.4 * mL
Explanation:
Do you agree? The copper displaces the given volume of water.
Now ρ Cu = 8.90 ⋅ g ⋅ c m 3 OR ρ Cu = 8.90 ⋅ g ⋅ m L − 1, i.e. 1 ⋅ m L ≡ 1 ⋅ c m 3
But by definition, ρ density = mass volume
And thus mass = ρ × volume = 8.90 ⋅ g ⋅ m L − 1 × 23.4 ⋅ m L
= 208.3 * G
What is the partial pressure of nitrogen in a container that contains 3.96mol of oxygen, 7.49 mol of nitrogen and 1.19 mol of carbon dioxide when the total pressure is 563 mmHg?
Answer:
333.6 atm
Explanation:
The following data were obtained from the question:
Mole of O2 (nO2) = 3.96 moles
Mole of N2 (nN2) = 7.49 moles
Mole of CO2 (nCO2) = 1.19 moles
Total pressure = 563 mmHg
Partial pressure of N2 =..?
Next, we shall determine the total number of mole in the container.
This can be obtained as follow:
Mole of O2 (nO2) = 3.96 moles
Mole of N2 (nN2) = 7.49 moles
Mole of CO2 (nCO2) = 1.19 moles
Total mole =?
Total mole = nO2 + nN2 + nCO2
Total mole = 3.96 + 7.49 + 1.19
Total mole = 12.64 moles
Next, we shall determine the mole fraction of N2.
This can be obtained as follow:
Mole fraction = mole of substance/total mole
Mole of N2 (nN2) = 7.49 moles
Total mole = 12.64 moles
Mole fraction of N2 =?
Mole fraction of N2 = 7.49/12.64
Finally, we shall determine the partial pressure of N2.
This can be obtained as follow:
Mole fraction of N2 = 7.49/12.64
Total pressure = 563 mmHg
Partial pressure of N2 =..?
Partial pressure = mole fraction x total pressure
Partial pressure of N2 = 7.49/12.64 x 563
Partial pressure of N2 = 333.6 atm
The, the partial pressure of nitrogen, N2 is 333.6 atm
A 500.0-mL buffer solution is 0.10 M in benzoic acid and 0.10 M in sodium benzoate and has an initial pH of 4.19. What is the pH of the buffer upon addition of 0.010 mol of NaOH?
Answer:
pH after the addition of NaOH is 4.37
Explanation:
When the amount of the weak acid = Amount of the conjugate base (As in the problem) pH = pKa. That means pKa of benzoic buffer is 4.19.
Now, to solve the pH of the buffer we need to use H-H equation for benzoic buffer:
pH = pKa + log [Benzoate] / [Benzoic acid]
pH = 4.19+ log [Benzoate] / [Benzoic acid]
You can take [] concentrations as the moles of both species.
When you add NaOH to the buffer, it reacts with benzoic acid producing more sodium benzoate and water, thus:
NaOH + Benzoic Acid → Sodium benzoate + Water.
Before the reaction, moles of benzoic acid and sodium benzoate were:
500.0mL = 0.500L × (0.10mol / L) = 0.050 moles
After the reaction, 0.010 moles of Benzoic acid are consumed and the same 0.010 moles of sodium benzoate are produced. That means moles of both species after reaction are:
[Benzoate] = 0.050 moles + 0.010 moles = 0.060 moles
[Benzoic acid] = 0.050 moles - 0.010 moles = 0.040 moles
Replacing in H-H equation:
pH = 4.19+ log [Benzoate] / [Benzoic acid]
pH = 4.19+ log [0.060mol] / [0.040mol]
pH = 4.37
pH after the addition of NaOH is 4.37
Fill in the [?]:
82 ul = [?]x10^[?] L
Answer:
8.2×10¯⁵ L
Explanation:
82 μL.
We can convert 82 μL to L by doing the following:
To convert from microlitre (μL) to litre(L), we need to know how many microlitre (μL) that makes up a litre (L)
Recall:
1 μL = 1×10¯⁶ L
Therefore,
82 μL = 82 μL × 1×10¯⁶ L/1 μL
82 μL = 8.2×10¯⁵ L
Therefore, 82 μL is equivalent to 8.2×10¯⁵ L
Both hydrogen sulfide (H2S) and ammonia (NH3) have strong, unpleasant odors. Which gas has the higher effusion rate? If you opened a container of each gas in a corner of a large room, which odor would you detect first on the other side of the room? Assume the temperature is constant.
The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster.
What is Effusion rate?Effusion rate is inversely proportional to molar mass. NH3 will have a higher average molecule velocity, so it will diffuse faster and will reach the other side of the room more quickly.
The concentration gradient, or the rise or fall in concentration from one site to another, the amount of surface area available for diffusion, and the distance the gas particles must travel all affect the diffusion rate.
The effusion of gas molecules into a vacuum through a small hole, as a pinhole in a balloon, is a process that involves movement of gaseous species comparable to diffusion.
Therefore, The molar mass of H2S is greater than the molar mass of NH3, making the velocity and effusion rate of NH3 particles faster.
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The standard cell potential for a reaction in an electrolytic cell is always:_____
a. Positive
b. Negative
c. Zero
d. Impossible to determine
Answer:
The correct answer is b. Negative
Explanation:
An electrolytic cell is based on a reduction- oxidation reaction which is non spontaneous. That means that the standard cell potencial (Eº) is negative. For this reason, an electrical potential must be applied in order to force the reaction. Conversely, a galvanic cell is based on a spontaneous redox reaction, so the galvanic cell produces electrical energy.