A chemist designs a galvanic cell that uses these two half-reactions:O2 (g) + 4H+(aq) + 4e− → 2H2O (l) Eo =+1.23V
Zn+2 (aq) + 2e− → Zn(s) Eo=−0.763V

Answer the following questions about this cell.

Write a balanced equation for the half-reaction that happens at the cathode.
Write a balanced equation for the half-reaction that happens at the anode.
Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written. Do you have enough information to calculate the cell voltage under standard conditions

Answers

Answer 1

Answer:

Answer: The reaction is spontaneous and there is not enough information to calculate the cell voltage.

Explanation:

The substance having highest positive $E^o$ reduction potential will always get reduced and will undergo reduction reaction.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

For a:

The half reactions for the cell occurring at cathode follows:

$O_2(g)+4H^+(aq)+4e^-\rightarrow H_2O(l);E^o_(cathode)=+1.23V$

For b:

The half reactions for the cell occurring at anode follows:

$Zn(s)\rightarrow Zn^(2+)+2e^-;E^o_(anode)=-0.763V$ ( × 2)

For c:

The balanced equation for the overall reaction of the cell follows:

$O_2(g)+4H^+(aq)+2Zn(s)\rightarrow H_2O(l)+2Zn^(2+)(aq)$

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_(cell)$

For a reaction to be spontaneous, the standard electrode potential must be positive.

To calculate the $E^o_(cell)$ of the reaction, we use the equation:

$E^o_(cell)=E^o_(cathode)-E^o_(anode)$

Putting values in above equation, we get:

$E^o_(cell)=1.23-(-0.763)=1.993V$

As, the standard electrode potential of the cell is coming out to be positive, the reaction is spontaneous in nature.

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_(cell)=E^o_(cell)-(0.059)/(n)\log ([Zn^(2+)]^2)/([H^(+)]^4* p_(O_2))$

As, the concentrations and partial pressures are not given. So, there is not enough information to calculate the cell voltage.

Hence, the reaction is spontaneous and there is not enough information to calculate the cell voltage.

Related Questions

Glucose, C 6 H 12 O 6 , is used as an energy source by the human body. The overall reaction in the body is described by the equation C 6 H 12 O 6 ( aq ) + 6 O 2 ( g ) ⟶ 6 CO 2 ( g ) + 6 H 2 O ( l ) Calculate the number of grams of oxygen required to convert 58.0 g of glucose to CO 2 and H 2 O . mass of O 2 : 61.76 g Calculate the number of grams of CO 2 produced.
Calculate the percent mass per volume, % (m/v), of a dextrose solution containing 7.00 g of dextrose in 2.00×102 mL of solution. Note that mass is not technically the same as weight, but the abbreviation % (w/v) is often used interchangeably with % (m/v).
QUESTION 2: The molar mass for this compound is 1.11g / m * o * l The molecular formula for this compound is
What is the mass of ammonium nitrate in 250 mL of a 75% by mass solution (density = 1.725 g/mL)?
Isopropyl methyl ether is slightly soluble with water because the oxygen atom of ethers with three or fewer carbon atoms can form a few hydrogen bonds with water.(A) True(B) False

For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomassie Blue followed by the addition of 50 ml of 85% v/v phosphoric acid. This entire mixture is then diluted to 1 liter with water. What is the final concentration of phosphoric acid?

Answers

Answer:

4,25% v/v H3PO4

Explanation:

The concentration of phosphoric acid (H3PO4) is expressed as a volume / volume percentage, which means:

%v/v H3PO4 = (mL of pure H3PO4/mL of solution)*100%

In other words, we are only interested in the final volume of the solution to which the phosphoric acid was diluted, regardless of its composition. Which in this case is 1 L (1000 mL).

We can then apply the following equation, commonly used to calculate the initial or final concentration (or volume) of a substance when it is diluted:

Ci*Vi=Cf*Vf

Where:

Ci, is the initial concentration of the substance.

Vi, the initial volume of the substance

Cf, the final concentration reached after dilution

Vf, the final volume of the solution at which the substance was diluted

In this case, the incognite would be the final concentration of H3PO4 reached after dilution, that is, Cf. Therefore, we proceed to clear Cf from the previous equation and replace our data:

Cf = (Ci*Vi)/Vf = (85% v/v * 50 mL)/1000 mL = 4,25 % v/v

Note that being up and down in the division, the mL unit is canceled to result in% v / v.

You have a sample of gold that contains 0.2 moles of gold (Au). How many gold atoms are present in the sample. HINT: Gold atoms represents the number of particles."1.20 x 10^23 atoms

3.3 x 10^-25 atoms

1.01 x 10^-3 moles

Answers

Given:

n = 0.20 mol

Required:

Solution:

N = n × 6.02 × 10²³ atoms/mol

N = 0.2 mol × 6.02 × 10²³ atoms/mol

N = 1.20 × 10²³ atoms

Therefore, there are 1.20 × 10²³ gold atoms in 0.2 mol of a gold sample.

#ILoveChemistry

#ILoveYouShaina

BEST ANSWER IS

have a great summer

1.01 x 10^-3 moles

An alloy is a mixture of metals to obtain a more durable and strongersubstance.
True
False

Answers

Answer: False

Explanation: Alloys are harder and stronger because the different-sized atoms of the mixed metals make the atomic layers less regular, so they cannot slide as easily.

A sample of phosgene gas at an initial concentration of 0.500 m is heated at 527 °c in a reaction vessel. at equilibrium, the concentration of co (g) was found to be 0.046 m. calculate the equilibrium constant for the reaction at 527 °c.

Answers

Answer : The equilibrium constant will be - 0.454

Explanation : The reaction is given below;

$COCl_(2) \ \textless \ ----\ \textgreater \ CO + Cl_(2)$

We need to find Equilibrium constant - $K_(c)$ ;

$K_(c)$ = [CO] [ $Cl_(2)$ ] / $[COCl_(2) ]$

So, $K_(c)$ = [0.046] X [0.046] / [0.5 - 0.046]

Therefore, $K_(c)$ = 0.454

If the enthalpy change of a reaction in a flask is δh = +67 kj, what can be said about the reaction?

Answers

Answer : Normally in any chemical reaction, if the enthalpy change i.e. ΔH is positive which means it is greater than zero then it can be called as an Endothermic Reaction.

Whereas, the system under study is absorbing heat that is produced during the reaction. So if ΔH is found to be positive then it can be called as endothermic reaction.

The enthalpy change of the reaction indicates that it is an endothermic process.

FURTHER EXPLANATION

Enthalpy (ΔH) is the amount of heat absorbed or released in a reaction. It is based on the amount of energy needed to break the bonds and the energy released during bond formation. Enthalpy change is the difference in the enthalpy of the reactants and the products. The positive or negative sign for an enthalpy value indicates the direction of the heat flow: a positive ΔH indicates that the reaction is endothermic while a negative value for ΔH means that the reaction is exothermic.

Endothermic Reactions

Endothermic reactions are reactions that absorb heat from the surroundings to the system. This is the case when more energy is absorbed to break the bonds than is released to form the bonds. Endothermic reactions can be identified in the lab by observing if the reaction vessel becomes cooler as the reaction proceeds.

Exothermic Reactions

When the amount of energy released during bond formation is greater than the amount of energy absorbed during bond breaking, a net release of energy to the surroundings takes place and the reaction is exothermic. Exothermic reactions can be identified when the reaction vessel becomes hot as the reaction progresses.

LEARN MORE

Heat capacity brainly.com/question/12976104

Equilibrium brainly.com/question/12983923

Gibbs Free Energy brainly.com/question/12979420

Keywords: Endothermic, Exothermic, Enthalpy

A student is heating a chemical in a beaker with a Bunsen burner.In a paragraph of at least 150 words, identify the safety equipment that should be used and the purpose of it for the given scenario.

Answers

When a student is warming a chemical in a container using a special burner, it is very important to focus on safety by using the right safety tools.

What is the safety equipment

First, the student needs to wear the right safety clothes like a lab coat, gloves, and goggles to protect themselves from getting splashed or hurt by chemicals. A lab coat stops chemicals from touching the skin, gloves keep the hands safe, and safety goggles protect the eyes from chemicals

and hot things.

Furthermore, using a fume hood is necessary to make sure there is enough fresh air circulating and to remove any dangerous fumes or gases that might be released while heating things up.

Read more about safety equipment here:

brainly.com/question/28389789

#SPJ3

Answer:The student should be wearing a lab coat or maybe an apron to prevent chemicals from spilling or exploding onto their clothes, I do recommend a lab coat better though because it can protect your skin better. Next, make sure while messing with chemicals you are always wearing goggles, if you are not wearing them there is a chance that after touching chemicals you could touch your eyes. And that brings me to washing your hands straight away after messing with chemicals. You could also wear gloves and just take them off when you're done but if you don't have clean hands afterward you could always put the chemicals all over your skin. But in case you do touch your eyes there is always an emergency eyewash station somewhere in the lab room. And if you are to get Chemicals on your skin, in your hair, on your clothes, or to be on fire, there shall be a shower somewhere to get rid of that. But if you read the instructions or listen closely to the teacher you shall have no problem.

Explanation:

I kinda got off topic

Other Questions

Find the mass of oxygen in grams produced by the decomposition of 100.0 g of CO2

Which expression correctly describes energy using SI units? A. 1 J=1kg•m^2/s^2 B. 1 J= 1kg•m/s^2 C. 1 J= 1kg• m/s D. 1 J= 1kg•m^2/s

In a calorimetry experiment, it was determined that a 92.0 gram piece of copper metal released 1860 J of heat to the surrounding water in the calorimeter (qcopper = −1860 J). If the final temperature of the copper metal-water mixture was 25.00°C, what was the initial temperature of the copper metal? The specific heat of copper is 0.377 J/(g°C). Group of answer choices Tinitial = 28.6°C Tinitial = −28.6°C Tinitial = 78.6°C Tinitial = 92.6°C Tinitial = 53.6°C

A student dissolves 12.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.01/gmL . The student notices that the volume of the solvent does not change when the sucrose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits. molarity = molality =