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2 years ago

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Calculate the standard cell emf for the reaction: ClO3−(aq)+3Cu(s)+6H+(aq)→Cl−(aq)+3Cu2+(aq)+3H2O(l) Pt is used as an inert elec

trode in contact with the ClO3− and Cl−. Calculate the standard emf using data in Appendix E and given the following: ClO3−(aq)+6H+(aq)+6e−→Cl−(aq)+3H2O(l); E∘=1.45 V Express the emf to three significant figures with the appropriate units. V

1 answer:

Rama09 [41]2 years ago

6 0

<u>Answer:</u> The standard EMF of the cell is 1.11 V

<u>Explanation:</u>

For the given chemical equation:

$ClO_3^-(aq)+3Cu(s)+6H^+(aq.)\rightarrow Cl^-(aq.)+3Cu^{2+}(aq.)+3H_2O(l)$

The half reaction follows:

<u>Oxidation half reaction:</u> $Cu(s)\rightarrow Cu^{2+}+2e^-;E^o_{Cu^{2+}/Cu}=0.34V$ ( × 3)

<u>Reduction half reaction:</u> $ClO_3^-(aq.)+6H^+(aq.)+6e^-\rightarrow Cl^-(aq.)+3H_2O(l);E^o=1.45V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Calculating the $E^o_{cell}$ using above equation, we get:

$E^o_{cell}=1.45-0.34=1.11V$

Hence, the standard EMF of the cell is 1.11 V

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How many pints of a 30% sugar solution must be added to a 5 pint of a 5% sugar solution to obtain a 20% sugar solution?

ser-zykov [4K]

You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.

You can use a modified dilution formula to calculate the volume of 30 % sugar.

<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3

Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt

(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar

30<em>x</em> + 25 = 20x + 100

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2 years ago

How would each of the following procedural errors affect the results to be expected in this experiment? Give your reasoning in e

Digiron [165]

Answer:

a) if the liquid is not vaporized completely, then the condensed vapor in the flask contains the air which is initially occupied before the liquid is heated. When calculating the molar mass of the vapor the moles of air which are initially present are not excluded, so that the molar mass of the vapor would be an increase in value.

b) While weighing the condensed vapor, the flask should be dried. If the weighing flask is not dried then the water which is layered on the surface of the flask is also added to the mass of the vapor. Therefore, the mass of the vapor that is calculated would be increase.

c) When condensing the vapor, the stopper should not be removed from the flask, because the vapor will escape from the flask and a small amount of vapor will condense in the flask. Therefore, the mass of the condensed vapor would be In small value.

d) If all the liquid is vaporized, when the flask is removed before the vapor had reached the temperature of boiling water, then the boiling

temperature of that liquid would be lower than that of the boiling temperature of the water.Therefore, the liquid may have more volatility.

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2 years ago

3. According to the label on a bottle of concentrated hydrochloric acid, the contents are 36.0% HCl by mass and have a density o

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Answer:

a) 11.64 M

b) 43 mL

c) 1.7 kg

Explanation:

a) Let's use a basis of the calculus of 1000 mL (1 L) of the concentrated solution. If the solution has 1.18 g/mL, it has:

1.18*1000 = 1180 g.

The mass of HCl will be then:

mHCl = 1180*0.36 = 424.8 g

The molar mass of HCl is 36.5 g/mol, so the number of moles is the mass divided by the molar mass:

nHCl = 424.8/36.5 = 11.64 mol

The molarity is the number of moles divided by the volume in L:

Molarity = 11.64 M

b) To prepare a solution by dilution of a concentrated one, we can use the equation:

C1V1 = C2V2

Where C is the concentration, V is the volume, 1 is the concentrated solution, and 2 the final solution. So:

11.64*V1 = 2.00*0.250

V1 = 0.0429 L ≅ 43 mL

c) The neutralization will happen by the equation:

HCl + NaHCO₃ → NaCl + CO₂ + H₂O

So, 1 mol of NaHCO₃ is needed to react with 1 mol of HCl. At 1.75 L, the number of moles of the acid is:

nHCl = 1.75*11.64 = 20.37 mol

The molar mass of NaHCO₃ is 84 g/mol so the mass needed is the molar mass multiplied by the number of moles:

m = 84*20.37 = 1,711.08 g

m = 1.7 kg

6 0

2 years ago

A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit

Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

$\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}$

$P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}$

$P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times 35.0} = 0.776969038566$

∴ The final pressure P₂ ≈ 0.78 atm.

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2 years ago

g When 2.50 g of methane (CH4) burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion (in kJ) per mole

Anna [14]

Answer:

-800 kJ/mol

Explanation:

To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).

First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:

Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol

moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄

Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:

ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol

Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).

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2 years ago