1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
34kurt
3 years ago
5

Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --

Cu (s) 0.34V lal if you were to design a voltaic cell, what would be the anode and cathode? (2 Marks) (b) What is the potential of a cell when [Fe3+] = 0.0001 M, [Cu2+] = 0.25 M, and [Fe2+] = 0.20 M? (3 Marks)
Chemistry
1 answer:
frosja888 [35]3 years ago
5 0

Answer :

(a) The anode and cathode will be E^o_{(Cu^{2+}/Cu)}

and E^o_{(Fe^{3+}/Fe^{2+})} respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  Fe^{3+}+1e^-\rightarrow Fe^{2+}

Thus, the anode and cathode will be E^o_{(Cu^{2+}/Cu)}

and E^o_{(Fe^{3+}/Fe^{2+})} respectively.

(b) Now we have to calculate the potential of a cell.

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  Fe^{3+}+1e^-\rightarrow Fe^{2+}

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}

The overall cell reaction will be,

2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}

E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V

E^o_{[Cu^{2+}/Cu]}=0.34V

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

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

E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}

E^o=1.54V-(0.34V)=1.20V

Now we have to calculate the cell potential.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = emf of the cell = ?

Now put all the given values in the above equation, we get:

E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}

E_{cell}=1.022V

Therefore, the emf of cell potential is 1.022 V

You might be interested in
Place the particles in order from smallest to largest.
yulyashka [42]

Answer:

smallest to largest:

Electron, Neutron, Atom, Molecule, Nucleus

Explanation:

sorry if it's not right

7 0
3 years ago
When having an equal number of protons and electrons, what does that do to the overall charge of the element? Why do you think t
gladu [14]
It means it’s neutrally charged! Protons are positive while electrons are negative, so they balance each other out. Neutrons are neutral so they don’t weigh in.
4 0
3 years ago
Find degenerate orbitals 3dxy, 2px, 4dz2, 3dyz, 3pz,3dxz, 4sm​
IceJOKER [234]
Ok boomer..................
7 0
3 years ago
When a solid undergoes _____ it takes up less space due to cooling
almond37 [142]

Answer:

solificATION

Explanation:

4 0
2 years ago
Read 2 more answers
A sample of 17.0 M concentrated H2SO4 stock solution with a volume of 25.0 cm3 was diluted to final concentration of 5.0 M H2SO4
Art [367]

Answer:

60 cm³ of water

Explanation:

We'll begin by calculating the volume of the diluted solution. This can be obtained as follow:

Concentration of stock solution (C₁) = 17 M

Volume of stock solution (V₁) = 25 cm³

Concentration of diluted solution (C₂) = 5 M

Volume of diluted solution (V₂) =?

C₁V₁ = C₂V₂

17 × 25 = 5 × V₂

425 = 5 × V₂

Divide both side by 5

V₂ = 425 / 5

V₂ = 85 cm³

Thus, the volume of the diluted solution is 85 cm³

Finally, we shall determine the volume of water needed to dilute the solution. This can be obtained as follow:

Volume of stock solution (V₁) = 25 cm³

Volume of diluted solution (V₂) = 85 cm³

Volume of water =?

Volume of water = V₂ – V₁

Volume of water = 85 – 25

Volume of water = 60 cm³

Therefore, 60 cm³ of water is needed to dilute the solution.

6 0
2 years ago
Other questions:
  • What is the change in enthalpy for the single replacement reaction between solid zinc and Cs2SO4 in solution to produce cesium a
    7·1 answer
  • A cruise ship with an outdoor pool is traveling south when it comes to a sudden stop to avoid a reef directly ahead. In which di
    7·1 answer
  • Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water. Suppose 2.1 grams of ethane i
    14·1 answer
  • Given the following list of atomic and ionic species, find the appropriate match for questions 2-4. Please Explain.
    6·1 answer
  • Will mark Brainliest, thanks, and 5 stars!
    6·1 answer
  • Particles vibrate in a rigid structure and<br> do not move relative to their neighbors.
    15·2 answers
  • Which of the following best represents the reaction between sulfuric acid and calcium hydroxide? f H2SO4 + 2Ca(OH)2 → 2CaSO4 + 3
    11·1 answer
  • The respiratory system transfers what to the blood that is used by the cells of the body in order to produce energy?
    7·1 answer
  • What is a balanced nuclear equation for the alpha decay of Strontium-90?
    7·1 answer
  • Equal masses of carbon dioxide (CO2) and oxygen (O2) are mixed. What is the mole
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!