Question

Calculate the emf of the following cell at 298 K:
Al(s)/Al3+ (0.15M)//Cu2+(0.025M) /Cu(s)
(Given Eo(Al3+/Al) = -1.66 V, Eo(Cu2+/Cu) = 0.34V, log 0.15 = -0.8239, log 0.025 = -1.6020)

Answer 1

Answer:

Explanation:

The anwer will be 1.9689V

Answer 2

Answer:  EMF is $1.9689V$

Explanation:

$E^o_{cell} = E^o_{cathode}-E^o _{anode} = 0.34-(-1.66) = 2.00 V$

$E_{cell} = E^o_{cell}-\frac{0.059}{n} log\frac{[ A l ^3^+ ] ^2}{[ C u ^2 ^+ ]^ 3}$

here

$n = 6\\E_{cell} =2-\frac{0.059}{6} log\frac { [ 0.15 ]^2 }{[ 0.025 ]^3}$

$= 2-\frac{0.059}{6} (2log0.15-3log0.025)\\ = 2-\frac{0.059} {6}(-1.6478 +4.8062) \\=2- 0.0311 = 1.9689V$

EMF of the cell:

• The maximum potential difference between two electrodes of a cell   is known as the electromotive force (EMF) of a cell.
• The difference in voltage between the half-reactions of oxidation and reduction can also be used to characterize it.
• An electrochemical cell's EMF is primarily used to assess whether or not it is galvanic.

What is an electrochemical cell?

• An electrochemical cell is a machine that generates electricity through a chemical reaction.
• In a nutshell, it's a device that converts chemical energy into electrical energy.
• Without a chemical reaction involving the exchange of electrons, an electrochemical cell cannot work.
• Redox reactions are the name for these processes.
• Voltage is a defining characteristic of a cell.
• No matter how big the cell is, a certain type of cell produces the same voltage.
• If the cell is operated under ideal conditions, the chemical makeup of the cell is the only factor that depends on the voltage.

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