Chemistry Forum

Voltaic Cells

a)Do not require a power source.

b)Convert chemical potential energy into electrical energy.

c)The redox reaction is spontaneous. (The cell potential is positive and the change in Gibb's free energy is negative)

d)The reaction with the lower standard reduction potential is utilized as the anode, and the reaction with the high standard reduction potential is utilized as the cathode.


Electrolytic Cells

a)Require a power source.

b)Require work by converting electrical energy to chemical potential energy.

c)The redox reaction is non-spontaneous. (The cell potential is negative and the change in Gibb's free energy is positive)

d)The reaction with the lower standard reduction potential is utilized as the cathode, and the reaction with the high standard reduction potential is utilized as the anode.

Are the above differences correct? especially d) part of the both cell features

d is not correct.
The Anode is the pol where oxidation takes place, the cathode where reduction takes place .It doesnt matter if its electrolytic or Voltaic cell. In an Voltaic cell we have the power source itself. The oxidation is at the lower oxidation potential. It is the Anode and it is plus. The reduction is the higher Oxidation potential It is cathode and minus.
In an electrolytic cell we have extern power source. So now it depends how it is connected. The Anode is the negative or reducing electrode that releases electrons to the external circuit and oxidizes during and electrochemical reaction. The Cathode is the positive or oxidizing electrode that acquires electrons from the external circuit and is reduced during the electrochemical reaction. But during the electrolysis the oxidation takes place at Anode side.
Look at a car battery. Plus and minus are not changed during charging and discharging. The only thing what changes is direction of current.

In Daniell cell, which is also called voltaic cell or galvanic cell, Redox reaction takes pace as follows:

O: Zn(s) → Zn²⁺(aq) + 2e⁻ E°/V= +0.76 Anode with (-) sign

R: Cu²⁺(aq) + 2e⁻ → Cu(s) E°/V = 0.34 Cathode with (+) sign

Add the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.

Zn(s) + 2 e⁻ + Cu²⁺ → Zn²⁺ + 2 e⁻ + Cu(s)
Simplify the equation. The same species on opposite sides of the arrow can be canceled.

Zn(s) + Cu²⁺ → Zn²⁺ + Cu(s)

That means reaction with higher reduction potential works as anode and reaction with lower reduction potential works as cathode.

https://i.stack.imgur.com/221Tt.png

Is that correct?

No its not correct because the voltage for Zinc is -0,76 V
More easier Anode is there, where the oxidation takes place.

After reading above table, we can say E°/V for the reaction Zn(s) → Zn²⁺(aq) + 2 e⁻ is +0.76. Is this tabulated information wrong?

You have that arrow <-> between.


Check
https://www.google.com/search?client=fi ... 0&bih=1006