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3 years ago

An electrolytic cell is based upon a nonspontaneous reaction; the Eocell for the reaction is _____.

1 answer:

6 0

In an electrochemical cell in which the oxidation reaction is nonspontaneous the oxidation will not occur spontaneously at the anode and the reduction will not be spontaneous at the cathode. And according to the law for the calculation of the voltage potential of the electrochemical cell (Ecell):

Ecell = Eox. - Ere. where (Eox. is the potential of the oxidation at the anode and Ere. is the potential of the reduction at the cathode). The standard potential for an electrolytic cell is negative, because of the Ere. which is greater than Eox.

The answer is : less than zero.

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Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re

borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

A + B ⇄ C

where;

numbers of moles = 0.386 mol C (g)

Volume = 7.29 L

Molar concentration of C = $\frac{0.386}{7.29}$

= 0.053 M

A + B ⇄ C

Initial 0 0 0.530

Change +x +x - x

Equilibrium x x (0.0530 - x)

$K = \frac{[C]}{[A][B]}$

where

K is given as ; 78.2 atm-1.

So, we have:

$78.2=\frac{[0.0530-x]}{[x][x]}$

$78.2= \frac{(0.0530-x)}{(x^2)}$

$78.2x^2= 0.0530-x$

$78.2x^2+x-0.0530=0$

Using quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where; a = 78.2 ; b = 1 ; c= - 0.0530

= $\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $\frac{-b-\sqrt{b^2-4ac} }{2a}$

= $\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$ or $\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}$

= 0.0204 or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

= 0.0530 - 0.0204

= 0.0326

Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

$V = \frac{nRT}{P}$

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K (fixed constant temperature )

V (volume) = ???

$V=\frac{(0.0734*0.0821*273.15)}{(1.00)}$

V = 1.64604

V ≅ 1.65 L

3 0

3 years ago

Use the following half-reactions to write three spontaneous reactions, calculate E°cell for each reaction, and rank the oxidizin

Ainat [17]

Answer:

See explaination

Explanation:

we know that

half cell with higher reduction potential is cathode

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

anode :

Cr(s) ---> Cr+3 + 3e-

overall reaction is

3 N20 + 6H+ + 2 Cr ---> 3N2 + 3H20 + 2Cr+3

now

Eo cell = Eo cathode - Eo anode

EO cell = 1.77 + 0.74

Eo cell = 2.51 V

now

in this case

oxidizing agents are N20 and Cr+3

reducing agents are Cr and N2

higher the reduction potential , stronger the oxidizing agent

lower the reduction potential , stronger the reducing agent

oxidzing agents

N20 > Cr+3

reducing agents

Cr > N2

cathode :

Au+ + e- --> Au

anode :

Cr ---> Cr+3 + 3e-

overall reaction

3Au+ + Cr ---> 3Au + Cr+3

Eo cell = 1.69 + 0.74

Eo cell = 2.43

now

oxidizing agents :

Au+ > Cr+3

reducing agents :

Cr > Au

cathode :

N20 + 2H+ + 2e- ---> N2 + H20

andoe :

Au ---> Au+ + e-

overall

2 Au + N20 + 2H+ --> 2 Au+ + N2 + H20

Eo cell = 1.77 - 1.69

Eo cell = 0.08

oxidizing agents

N20 > Au+

reducing agents

Au > N2

8 0

3 years ago

The pH of a 0.150 molar solution of a weak acid is 4.10. What is the pKa of the acid?

Kamila [148]

Answer: 4.21×10⁻⁸

Explanation:

1) Assume a general equation for the ionization of the weak acid:

Let HA be the weak acid, then the ionization equation is:

HA ⇄ H⁺ + A⁻

2) Then, the expression for the ionization constant is:

Ka = [H⁺][A⁻] / [HA]

There, [H⁺] = [A⁻], and [HA] = 0.150 M (data given)

3) So, you need to determine [H⁺] which you do from the pH.

By definition, pH = - log [H⁺]

And from the data given pH = 4.1

⇒ 4.10 = - log [H⁺] ⇒ [H⁺] = antilog (- 4.10) = 7.94×10⁻⁵

4) Now you have all the values to calculate the expression for Ka:

ka = 7.94×10⁻⁵ × 7.94×10⁻⁵ / 0.150 = 4.21×10⁻⁸

3 0

3 years ago

Read 2 more answers

what are the relative strengths of intermolecular forces compared to the forces between ions in a salt or the forces between met

Sophie [7]

There are two kinds of forces, or attractions, that operate in a molecule—intramolecularand intermolecular. Let's try to understand this difference through the following example.

Figure of towels sewn and Velcroed representing bonds between hydrogen and chlorine atoms

We have six towels—three are purple in color, labeled hydrogen and three are pink in color, labeled chlorine. We are given a sewing needle and black thread to sew one hydrogen towel to one chlorine towel. After sewing, we now have three pairs of towels: hydrogen sewed to chlorine. The next step is to attach these three pairs of towels to each other. For this we use Velcro as shown above.

So, the result of this exercise is that we have six towels attached to each other through thread and Velcro. Now if I ask you to pull this assembly from both ends, what do you think will happen? The Velcro junctions will fall apart while the sewed junctions will stay as is. The attachment created by Velcro is much weaker than the attachment created by the thread that we used to sew the pairs of towels together. A slight force applied to either end of the towels can easily bring apart the Velcro junctions without tearing apart the sewed junctions.

Exactly the same situation exists in molecules. Just imagine the towels to be real atoms, such as hydrogen and chlorine. These two atoms are bound to each other through a polar covalent bond—analogous to the thread. Each hydrogen chloride molecule in turn is bonded to the neighboring hydrogen chloride molecule through a dipole-dipole attraction—analogous to Velcro. We’ll talk about dipole-dipole interactions in detail a bit later. The polar covalent bond is much stronger in strength than the dipole-dipole interaction. The former is termed an intramolecular attraction while the latter is termed an intermolecular attraction.

7 0

3 years ago

A quantity of 85.0 mL of 0.900 M HCl is mixed with 85.0 mL of 0.900 M KOH in a constant-pressure calorimeter that has a heat cap

san4es73 [151]

Answer:

There is a 1:1 neutralization reaction, so, number of moles:

0.085 l * 0.900 mol/ L = 0.0765 mol HCL

The heat produced: 0.0765 mol * -56.2 kJ/mol = -4.229 kJ (this is the heat of neutralization)

Change in temperature: The mass to use in the equation Q=cmT.

4229 J / (4.186 J/gC*170 g) = 6.042 C

Add to the initial temperature:

18.24 + 6.042 = 24.29 C°

3 0

3 years ago