A galvanic cell at a temperature of 42 degrees Celcius is powered by the following redox reaction:
3CU2+(aq) + 2Al(s) > 3Cu(s) + 2Al3+(aq)
Suppose the cell is prepared with 3.43 M Cu2+n one half-cell and 1.63 M Al3+in the other.
A) Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.

The cell voltage of the given reaction is 1.86 V

Explanation:

The given chemical equation follows:

Oxidation half reaction:       ( × 2)

Reduction half reaction:       ( × 3)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the of the reaction, we use the equation:

Putting values in above equation, we get:

To calculate the EMF of the cell, we use the Nernst equation, which is:

where,

= electrode potential of the cell = ? V

= standard electrode potential of the cell = +1.82 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol Kl

T = temperature =

F = Faraday's constant = 96500

Putting values in above equation, we get:

Hence, the cell voltage of the given reaction is 1.86 V

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.