Answer please and thank you

Answer:

Based on the given reaction, it is evident that the reaction is endothermic as indicated by a positive sign of enthalpy of reaction. Thus, it can be stated that the favoring of the forward reaction will take place by upsurging the temperature of the reaction mixture.  

Apart from this, based on Le Chatelier’s principle, any modification in the quantity of any species is performed at equilibrium and the reaction will move in such an orientation so that the effect of the change gets minimized. Therefore, a slight enhancement in the concentration of the reactant will accelerate the reaction in the forward direction and hence more formation of the product takes place.  

Explanation:

This reaction is in equilibrium and would hence obey lechatelier's principle. This principle states that whenever a system at equilibrium undergoes a change, it would react in way so as to annul that change.

Since it is an endothermic reaction, increasing the temperature would cause the reaction to shift towards the right.

This means that it favours product formation and more of the product would be formed.

Answer:

[tex]m_{MgO}=0.239gMgO[/tex]

Explanation:

Hello,

In this case, the chemical reaction between magnesium and oxygen to yield magnesium oxide is:

[tex]2Mg+O_2\rightarrow 2MgO[/tex]

In such a way, for 0.144 g of magnesium reacting with sufficient oxygen, the mass of magnesium oxide, whose molar mass is 40.3 g/mol (2:2 mole ratio) turns out:

[tex]m_{MgO}=0.144gMg*\frac{1molMg}{24.3gMg} *\frac{2molMgO}{2molMg}* \frac{40.3gMgO}{1molMgO} \\\\m_{MgO}=0.239gMgO[/tex]

Best regards.

Answer:

D- poor conductor

Explanation:

metallic properties decrease as we go on the right of the periodic table. Carbon is a non metal hence it is dull and a poor conductor.

it has a low density and is ductile.

Answer: Poor conductor

Explanation:

Answer:

The correct answer is 8.10

Explanation:

Given:

A(g) + 2B(g) ↔ AB₂(g)   Kc = 59 ---- Eq. 1

A(g) + 3B(g) ↔ AB₃(g)   Kc = 478 ----- Eq. 2

We have to rearrange the chemical equations in order to obtain:

AB₂(g) + B(g) ↔ AB₃(g) Kc = ?

AB₂(g) is a reactant, so we have to use the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant is the same: Kc= 478.  The following is the sum of rearranged chemical equations, and the compounds in bold and italic are canceled:

 AB₂(g)       ↔   A(g) + 2B(g)          Kc₁= 1/59

A(g) + 3B(g) ↔   AB₃(g)                  Kc₂= 478

-----------------------------------------

AB₂(g) + B(g) ↔ AB₃(g)

If we add reactions at equilibrium, the equilibrium constants Kc are mutiplied as follows:

Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10

The value of the missing equilibrium constant is 8.10.

The value of the missing equilibrium constant is 8.10

Since

A(g) + 2B(g) ↔ AB₂(g)   Kc = 59 ---- Eq. 1

A(g) + 3B(g) ↔ AB₃(g)   Kc = 478 ----- Eq. 2

Now we have to rearrange the chemical equations in order to obtain:

AB₂(g) + B(g) ↔ AB₃(g) Kc = ?

Here AB₂(g) represents a reactant, so we have to applied the reverse reaction of Eq. 1, in this case Kc= 1/59. Since AB₃(g) is a product, we use the forward reaction of Eq.2, and the constant should be the same: Kc= 478.  

The following is the sum of rearranged chemical equations, and the compounds in bold and italic should be canceled:

AB₂(g)       ↔   A(g) + 2B(g)          Kc₁= 1/59

A(g) + 3B(g) ↔   AB₃(g)                  Kc₂= 478

-----------------------------------------

AB₂(g) + B(g) ↔ AB₃(g)

In the case when we add reactions at equilibrium, the equilibrium constants Kc are multiplied as follows:

Kc = Kc₁ x Kc₂ = 1/59 x 478 = 478/59 = 8.10

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The cell can obtain more sugar when the sugar undergo facilitated diffusion through a channel protein, and no energy is required.

FACILITATED DIFFUSION:

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Answer:

The possible solution is to balance the rate of water removal from the well to the rate of natural recharge of the well from its underground aquifer.

Explanation:

A well is an excavation in the earth, made with the aim of extracting water from the aquifers. The water from a well can be drawn up by the means of a pump, containers, such as buckets, or by hand. Aquifers can also be recharged through a well.

Well draw down occurs when water from the well is drained faster than it is naturally recharged from the aquifer. This can be as a result of over pumping, extended drought, among other factors. The use of the high-powered motor in this case, for pumping, might be the possible cause of the well drying up. The situation might have resulted from the pump drawing out water from the well at a rate tat exceeds the rate at which it is recharged naturally, causing the well water to start drying up. There's also a possibility that the well is pumped indiscriminately, possibly leading to wastage of water.

The solution to this problem is to give the well a time duration for it to recharge itself. Then, the rate of recharges should be calculated and determined by an hydrologist. When all these is done, a pump with a motor power that does not exceed the calculated recharge rate should be used in place of the high-powered motor. Also, water usage should be brought to the minimum level to prevent unnecessary pumping due to excessive, wasteful use of water.

Answer:

[tex][Cl^-]=232.3\frac{mgCl^-}{L}[/tex]

Explanation:

Hello,

In this case, we can represent the chemical reaction as:

[tex]Cl^-(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NO_3^-(aq)[/tex]

In such a way, since the mass of the obtained silver chloride is 93.9 mg, we can compute the chloride ions in the ground water by using the following stoichiometric procedure whereas the molar mass of chloride ions and silver chloride are 35.45 g/mol and 143.32 g/mol respectively:

[tex]m_{Cl^-}=93.3mgAgCl*\frac{1mmolAgCl}{143.32mgAgCl}*\frac{1mmolCl^-}{1mmolAgCl} *\frac{35.45mgCl^-}{1mmolCl^-} =23.23mgCl^-[/tex]

Finally, for the given volume of water in liters (0.100L), we compute the required concentration:

[tex][Cl^-]=\frac{23.2mgCl^-}{0.100L}\\[/tex]

[tex][Cl^-]=232.3\frac{mgCl^-}{L}[/tex]

Best regards.

The concentration of chloride ions in the groundwater sample is 230 mg/L.

Based on the given information,

Now the number of moles of AgCl precipitate can be calculated as,

n = Given mass/Molar mass

Now putting the values we get,

[tex]n = \frac{0.939 g}{143.32 g/mol} \\n = 6.5 * 10^{-4}[/tex]

Thus, 6.5 ×  10⁻⁴ moles of AgCl comprises 6.5 × 10⁻⁴ chloride ions. Therefore, 6.5 × 10⁻⁴ of chloride ions are present in the sample of 100 ml.

Now the molar mass of chloride ion is 35.453 g/mol, the mass of chloride ion will be,

Mass = Mole × Molar mass

Mass = 6.5 × 10⁻⁴ moles  × 35.453 g/mol

Mass = 0.0230 g or 23 mg

The volume of the groundwater sample is 100 ml or 0.1000 L.

Now the concentration of the chloride ions in the sample given is,

C = 23 mg/0.1000 L

C = 230 mg/L

Thus, the concentration of chloride ions in the groundwater sample is 230 grams per liter.

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Answer:

Using energy.

Explanation:

Answer:

Kc = 166.7

[Fe³⁺] =  0.18 M

[SCN⁻] = 2×10⁻⁴ M

Explanation:

In the reaction of Fe³⁺ and SCN⁻, it is formed a complex:

Fe³⁺  +  SCN⁻  ⇄  FeSCN²⁺             Kc

Let's make the expression for Kc →  [FeSCN²⁺] / [Fe³⁺] . [SCN⁻]

5.7×10⁻⁵ / 9.5×10⁻⁴. 3.6×10⁻⁴  = 166.7

We determine the mmoles, we add from each reactant:

18 ml . 0.2M = 3.6 mmoles of Fe³⁺

2 ml . 0.002M = 4×10⁻³ mmoles of SCN⁻

General form of the dilution equation is:

Concentrated [C] . Concentrated Volume = Diluted [C] . Diluted Volume

Total volume = 20mL

[Fe³⁺]: 3.6 mmoles /20mL = 0.18 M

[SCN⁻]: 4×10⁻³ mmoles /20mL = 2×10⁻⁴ M

The value should be 1.67 x 10^2

The initial concentration should be 0.18 M and 2.0 x 10^(-4) M

The value is

= 5.7 x 10^(-5)/(9.5 x 10^(-4) x 3.6 x 10^(-4))

= 167

= 1.67 x 10^2

we know that

Initial moles  = volume x concentration

So,

= 18/1000 x 0.200

= 0.0036 mol

Now

Initial moles  = volume x concentration

= 2/1000 x 0.0020

= 4.0 x 10^(-6) mol

So,

Total volume should be

= 18 + 2

= 20 mL

= 0.02 L

Now

Initial concentration   

= moles /total volume

= 0.0036/0.02

= 0.18 M

Now

Initial concentration

= moles  /total volume

= 4.0 x 10^(-6)/0.02

= 2.0 x 10^(-4) M

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Answer:

0.1110 mol

Explanation:

Mass = 2g

Molar mass = 18.016 g/mol

moles = ?

These quantities are realted by the following equation;

Moles = Mass / Molar mass

Substituting the values of the quantities and solving for moles, we have;

Moles = 2 / 18.016 = 0.1110 mol

Answer:

No, because sulfur does not typically form negative ions or oxidation states less than 2−. The binary compound formed by sulfur and sodium is Na2S

Explanation:

Sulphur is a member of group 16. The oxidation states expected for sulphur in group 16 are -1, -2, +1, +2,+3,+4,+5 or +6. The elements of group 16 usually form negative ions with oxidation number of -2. They do not typically form negative ions with oxidation state less than -2.

The implication of this is that we actually do not expect the existence of a compound in which sulphur forms an S^4- anion. In reality, such an anion does not exist. Rather a binary compound of sulphur and sodium will have the formula Na2S because it contains the S^2- anion.

Answer:

empirical formula = C3H7

molecular formula = C6H14

Answer:

Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰

Explanation:

Before proceeding to write out the electron configuration of Cd2+, we have to obtain the electron configuration of Cadmium (Cd),

Cadmium has an atomic number of 48, this means that a neutral cadmium atom will have a total of  48  electrons surrounding its nucleus.

The electronic configuration of Cadmium is;

Cd: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10

The shorthand notation is given as;

Cd: [Kr] 4d¹⁰5s²

Cd2+ means that it has two less electrons, hence it's electron configuration is given as;

Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰

Answer:

The voltage is  [tex]E_{cell} = 0.944 \ V[/tex]

Explanation:

Generally the half reaction for Zn, Zn2 half-cell is mathematically represented as

    [tex]Zn_{(s)}[/tex]   ⇔   [tex]Zn^{2+}_{ (aq)} + 2e^-[/tex]    (reference study academy)

and the electric potential for this is a constant value

     [tex]E_{zn } = -0.7618 \ V[/tex]

Generally the half reaction for Al, Al3 half-cell is mathematically represented as

      [tex]Al^{3+} _{(aq)} + 3e^-[/tex]  ⇔  [tex]Al_{(s)}[/tex]

and the electric potential for this is constant value  

       [tex]E_{Al } = -1.662 \ V[/tex]

Therefore the cell potential for an electrochemical cell  is mathematically represented as

          [tex]E_{cell} = E_{zn } - E_{Al }[/tex]

substituting values

         [tex]E_{cell} = -0.718 - (-1.662)[/tex]

        [tex]E_{cell} = 0.944 \ V[/tex]

Answer:

[tex]\Delta G_{rxn}=42.7\frac{kJ}{mol}[/tex]

Explanation:

In this case, for the dissociation of hypochlorous acid, we know that the acid dissociation constant (Ka) is 2.9x10⁻⁸, which is related with the Gibbs free energy as shown below:

[tex]\Delta G_{rxn}=-RTln(K)[/tex]

But in this case K is just Ka, therefore, at 296 K, it turns out:

[tex]\Delta G_{rxn}=-8.314\frac{J}{mol*K}*296K*ln(2.9x10^{-8})\\\\\Delta G_{rxn}=42.7\frac{kJ}{mol}[/tex]

Such result, means that the reaction is nonspontaneous at the given temperature, it means it is not favorable (not easily occurring).

Best regards.

Answer:

[tex]m_{CO_2}=2.8gCO_2[/tex]

Explanation:

Hello,

In this case, the combustion of octane is chemically expressed by:

[tex]C_8H_{18}+\frac{25}{2} O_2\rightarrow 8CO_2+9H_2O[/tex]

In such a way, due to the 25/2:8 molar ratio between oxygen and carbon dioxide, we can compute the yielded grams of carbon dioxide (molar mass 44 g/mol) as shown below:

[tex]m_{CO_2}=3.2gO_2*\frac{1molO_2}{32gO_2} *\frac{8molCO_2}{\frac{25}{2}molO_2 } *\frac{44gCO_2}{1molCO_2}\\ \\m_{CO_2}=2.8gCO_2[/tex]

Best regards.

Answer:

C3H7OH → C3H6 + H20

Explanation:

If we look at the reactant and the product we will realize that the reactant is an alcohol while the product is an alkene. The reaction involves acid catalysed elimination of water from an alcohol.

Water is a good leaving group, hence an important synthetic route to alkenes is the acid catalysed elimination of water from alcohols. Hence the conversion represented by C3H7OH → C3H6 + H20 is an elimination reaction in which water is the leaving group.

Answer:

B and C. Just finished my lesson on Edge.

Answer:

The hydrogen can be gotten from the added Acid or water during "workup".

Explanation:

Basically we can say that the reaction describe in this question is a Reduction reaction because of the chemical compound called sodium borohydride. In the reaction described above we can see that there is a Reduction of ketone to alcohol by the compound; sodium borohydride.

For the reduction Reaction to occur the C-O bond must break so as to enable the formation of O-H bond and C-H bond.

So, "the reaction mixture is treated with water and H2SO4 to produce the desired alcohol", thus, the oxygen will definitely pick up the hydrogen from H2SO4 or H2O.

Answer:

The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.

Explanation:

The decomposition of the compound has an exponential behavior and process can be represented by this linear first-order differential equation:

[tex]\frac{dc}{dt} = -\frac{1}{\tau}\cdot c(t)[/tex]

Where:

[tex]\tau[/tex] - Time constant, measured in seconds.

[tex]c(t)[/tex] - Concentration of the compound as a function of time.

The solution of the differential equation is:

[tex]c(t) = c_{o} \cdot e^{-\frac{t}{\tau} }[/tex]

Where [tex]c_{o}[/tex] is the initial concentration of the compound.

The time is now cleared in the result obtained previously:

[tex]\ln \frac{c(t)}{c_{o}} = -\frac{t}{\tau}[/tex]

[tex]t = -\tau \cdot \ln \frac{c(t)}{c_{o}}[/tex]

Time constant as a function of half-life is:

[tex]\tau = \frac{t_{1/2}}{\ln 2}[/tex]

Where [tex]t_{1/2}[/tex] is the half-life of the composite decomposition, measured in seconds.

If [tex]t_{1/2} = 8\,s[/tex], then:

[tex]\tau = \frac{8\,s}{\ln 2}[/tex]

[tex]\tau \approx 11.542\,s[/tex]

And lastly, given that [tex]\frac{c(t)}{c_{o}} = \frac{1}{9}[/tex] and [tex]\tau \approx 11.542\,s[/tex], the time taken for the concentration to decrease to one-ninth of its initial value is:

[tex]t = -(11.542\,s)\cdot \ln\frac{1}{9}[/tex]

[tex]t \approx 25.360\,s[/tex]

The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.

Answer:

0.619 M to 3 significant figures.

Explanation:

1 mole of [tex]NiI_{2}[/tex] - 312.5 g

? mole of [tex]NiI_{2}[/tex] - 2.9 g

= 2.9/312.5

= 0.0928 moles.

Concentration = no. of moles/vol in litres = [tex]\frac{0.0928}{0.150L}[/tex]

= 0.619 M

The representative elements are the elements in the tall columns of the periodic table.

The representative elements that can also be referred to as the main group elements. They can be used to represent the chemistry of the group to which they belong.

Hence, the representative elements are the elements in the tall columns of the periodic table.

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Answer:

a. Strong acid, pH = 1.69

b. Strong base, pH = 10

c. Strong base, pH = 11

d. Weak acid, pH = 4.90

e. Strong base, pH ≅ 7 (pH should be higher than 7, but the base is so diluted)

f. Weak base, pH = 10.96

g. Acidic salt, pH = 5.12

h. Basic salt, pH = 8.38

i. Neutral salt, pH = 7

j. Acidic salt, pH < 7

Explanation:

a. HBr →  H⁺  +  Br⁻

Hydrobromic acid is a strong acid.

pH = - log [H⁺]

- log 0.02 = 1.69

b. NaOH → Na⁺  +  OH⁻

Sodium hydroxide is a strong base.

pH = 14 - pOH

pOH = - log [OH⁻]

pH = 14 - (-log 0.0001) = 10

c. Ba(OH)₂ → Ba²⁺  +  2OH⁻

Barium hydroxide is a strong base

[OH⁻] = 2 . 0.0015 = 0.003M

pH = 14 - (-log 0.003) = 11

d. HCN + H₂O ⇄  H₃O⁺  + CN⁻

This is a weak acid, it reacts in water to make an equilibrium between the given protons and cyanide anion.

To calculate the [H₃O⁺] we must apply, the Ka

Ka = [H₃O⁺] . [CN⁻] / [HCN]

6.2×10⁻¹⁰ = x² / 0.25-x

As Ka is really small, we can not consider the x in the divisor, so we avoid the quadratic formula.

[H₃O⁺] = √(6.2×10⁻¹⁰ . 0.25) = 1.24×10⁻⁵

-log 1.24×10⁻⁵ = 4.90 → pH

e.  KOH →  K⁺  +  OH⁻

2×10⁻¹⁰ M → It is a very diluted concentration, so we must consider the OH⁻ which are given, by water.

In this case, we propose the mass and charges balances equations.

Analytic concentration of base = 2×10⁻¹⁰ M = K⁺

[OH⁻] = K⁺ + H⁺ → Charges balance

The solution's hydroxides are given by water and the strong base.

Remember that Kw = H⁺ . OH⁻, so H⁺ = Kw/OH⁻

[OH⁻] = K⁺ + Kw/OH⁻. Let's solve the quadratic equation.

[OH⁻] = 2×10⁻¹⁰ + 1×10⁻¹⁴ /OH⁻

OH⁻² = 2×10⁻¹⁰. OH⁻ + 1×10⁻¹⁴

2×10⁻¹⁰. OH⁻ + 1×10⁻¹⁴ - OH⁻²

We finally arrived at the answer [OH⁻] = 1.001ₓ10⁻⁷

pH = 14 - (- log1.001ₓ10⁻⁷) = 7

The strong base is soo diluted, that water makes the pH be a neutral value.

Be careful, if you determine the [OH⁻] as - log 2×10⁻¹⁰, because you will obtain as pOH 9.69, so the pH would be 4.31. It is not possible, KOH is a strong base and 4.30 is an acid pH.

f. Ammonia is a weak base.

NH₃ +  H₂O  ⇄  NH₄⁺  + OH⁻

Kb = OH⁻  .  NH₄⁺  /  NH₃

1.74×10⁻⁵ = x² / 0.05 - x

We can avoid the x from the divisor, so:

[OH⁻] = √(1.74×10⁻⁵ . 0.05) = 9.32×10⁻⁴

pH = 14 - (-log 9.32×10⁻⁴ ) = 10.96

g. NH₄Cl, an acid salt. We dissociate the compound:

NH₄Cl →  NH₄⁺  +  Cl⁻.  We analyse the ions:

Cl⁻ does not make hydrolisis to water. In the opposide, the ammonium can react given OH⁻ to medium, that's why the salt is acid, and pH sould be lower than 7

NH₄⁺  +  H₂O  ⇄  NH₃  +  H₃O⁺   Ka

Ka = NH₃  .   H₃O⁺ / NH₄⁺

5.70×10⁻¹⁰ = x² / 0.1 -x

[H₃O⁺] = √ (5.70×10⁻¹⁰  . 0.1) = 7.55×10⁻⁶

pH = - log 7.55×10⁻⁶ = 5.12

As Ka is so small, we avoid the x from the divisor.

h. CaF₂  →  Ca²⁺  +  2F⁻

This is a basic salt.

The Ca²⁺ does not react to water. F⁻ can make hydrolisis because, the anion is the strong conjugate base, of a weak acid.

F⁻  +  H₂O  ⇄  HF  +  OH⁻          Kb

Kb = x² / 2 . 0.2 - x

Remember that, in the original salt we have an stoichiometry of 1:2, so 1 mol of calcium flouride may have 2 moles of flourides.

As Kb is small, we avoid the x, so:

[OH⁻] = √(1.47×10⁻¹¹ . 2 . 0.2) = 2.42×10⁻⁵

14 - (-log 2.42×10⁻⁵) = pH → 8.38

i . Neutral salt

BaNO₃₂  →   Ba²⁺  +  2NO₃⁻

Ba²⁺ comes from a strong base, so it is the conjugate weak acid and it does not react to water. The same situation to the nitrate anion. (The conjugate weak base, from a strong acid, HNO₃)

pH = 7

j.  Al(NO₃)₃, this is an acid salt.

Al(NO₃)₃  →  Al³⁺  +  3NO₃⁻

The nitrate anion is the conjugate weak base, from a strong acid, HNO₃ so it does not make hydrolisis. The Al³⁺ comes from the Al(OH)₃ which is an amphoterous compound (it can react as an acid or a base) but the cation has an acidic power.

Al·(H₂O)₆³⁺  + H₂O ⇄  Al·(H₂O)₅(OH)²⁺  + H₃O⁺

Answer:

3.383x10⁻³ micromoles of HgCl

Explanation:

The chemist adds 170mL of a 1.99x10⁻⁵mmol/L Mercury (I) chloride, HgCl.

The solution contains 1.99x10⁻⁵milimoles of HgCl in 1L. That means in 170mL = 0.170L there are:

0.170L × (1.99x10⁻⁵milimoles HgCl / L) = 3.383x10⁻⁶ milimoles of HgCl.

Now, in 1milimole you have 1000 micromoles. That means in 3.383x10⁻⁶ milimoles of HgCl you have:

3.383x10⁻⁶ milimoles of HgCl ₓ (1000micromoles / 1milimole) =

Answer:

The heat absorbed by the sample of water is 3,294.9 J

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case:

Replacing:

Q= 4.184 [tex]\frac{J}{g*C}[/tex] * 45 g* 17.5 C

Solving:

Q=3,294.9 J

The heat absorbed by the sample of water is 3,294.9 J

A.S OLOS kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkll

Answer:

OH⁻ is the Bronsted-Lowry base.

Explanation:

A Bronsted-Lowry base is a substance that accepts protons. In the chemical equation, OH⁻ accepts a proton from H₂PO₄⁻ to become H₂O. H₂PO₄⁻ would be a Bronsted-Lowry acid because it donates a proton to OH⁻ and becomes HPO₄²⁻.

Hope that helps.

Answer:

The answer is 6 because the p sublevel holds 3 orbitals and since each orbital can hold 2 electrons, the answer is 3 * 2 = 6.

Answer:

6 electrons

Explanation:

Each principal energy level above the first contains one s orbital and three p orbitals. A set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. So the answer is 6 electrons.

Answer:

6

Explanation:

Alkanes undergo substitution reaction so the number of replacement reaction hydrogen is 6

Answer: Thus the cell potential of an electrochemical cell is +0.28 V

Explanation:

The calculation of cell potential is done by :

[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]

Where both [tex]E^0[/tex] are standard reduction potentials.

[tex]E^0_{[Fe^{2+}/Fe]}= -0.41V[/tex]

[tex]E^0_{[Pb^{2+}/Pb]}=-0.13V[/tex]

As Reduction takes place easily if the standard reduction potential is higher(positive) and oxidation takes place easily if the standard reduction potential is less(more negative). Thus iron acts as anode and lead acts as cathode.

[tex]E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Fe^{2+}/Fe]}[/tex]

[tex]E^0=-0.13- (-0.41V)=0.28V[/tex]

Thus the cell potential of an electrochemical cell is +0.28 V