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Mkey [24]
3 years ago
14

Given that E°red = -0.40 V for Cd2+/Cd at 25°C, find E° and E for the concentration cell expressed using shorthand notation belo

w.
Cd(s) ∣ Cd2+(1.0 × 10-5 M) ∣∣ Cd2+(0.100 M) ∣ Cd(s)

E° = 0.00 V and E = +0.12 V


E° = 0.00 V and E = +0.24 V


E° = -0.40 V and E = -0.16 V


E° = -0.40 V and E = -0.28 V
Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
3 0

From the information in the question, the  E° and E for the cell is 0.00 V and 0.12 V.

Using the Nernst equation;

Ecell =  E°cell - 0.0592/n log Q

We know that  E°cell = 0.00 V since the anode and cathode are both made up of cadmium.

Substituting the given values into the Nernst equation;

Ecell = 0.00 V - 0.0592/2 log (1.0 × 10-5 M/0.100 M)

Ecell =  0.00 V - 0.0296 log(1 × 10^-4)

Ecell = 0.12 V

Learn more: brainly.com/question/8646601

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An ionic bond can be formed when one or more electrons are 1) equally shared by two atoms 2) unequally shared by two atoms 3) tr
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Answer:

4) transferred from the valence shell of one atom to the valence shell of another atom

Explanation:

Electrons are located outside of the nucleus which contains the protons and the neutrons.

For bonds to form, valence electrons located in the outermost shell electrons are involved. These are the valence electrons. These outer shell electrons can be shared or transferred between two combining atoms to form stable atoms.

In ionic bonds, the electrons are transferred from one specie to another. The atom that loses the electrons becomes positively charged and the receiving atom becomes negatively charged. This is the crux of ionic bonds.

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Use Boyle's Law to explain what would happen to the volume of a helium- filled balloon if it was carried underwater by a diver t
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The deeper the diver takes the helium balloon, the more it reduces in size. This is due to the pressure of the water column above pressing on the balloon. According to Boyle’s law (P= k*1/V.), as the volume of the balloon decreases, the pressure of the helium inside increases.

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3 years ago
When calculating specific heat, what does the variable 'Q'
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The answer is AIM NOT POSTIVE
4 0
2 years ago
Read 2 more answers
Electrons are:
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B. Are fixed within electron orbitals

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4 0
3 years ago
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

4 0
3 years ago
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