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

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.