Draw and calculate Ecello. Ecell, and free energy for the following batteries. W

Question

Draw and calculate Ecello. Ecell, and free energy for the following batteries. Will this battery generate electricity spontaneously? Calculate the concentration of zinc ion in the Ecell is 1.31V and concentration of magnesium is 0.950M Mg2+ Mg(s) ? Mg+2(0.500M) ?? Zn+2 (2.500M) ? Zn(s) Draw and calculate Ecello. Ecell, and free energy for the following batteries. Will this battery generate electricity spontaneously? Calculate the concentration of zinc ion in the Ecell is 1.31V and concentration of magnesium is 0.950M Mg2+ Mg(s) ? Mg+2(0.500M) ?? Zn+2 (2.500M) ? Zn(s) Draw and calculate Ecello. Ecell, and free energy for the following batteries. Will this battery generate electricity spontaneously? Calculate the concentration of zinc ion in the Ecell is 1.31V and concentration of magnesium is 0.950M Mg2+ Mg(s) ? Mg+2(0.500M) ?? Zn+2 (2.500M) ? Zn(s)

Explanation / Answer

here cathode = Zinc and anode is Mg

The cell will be

Mg(s) / Mg+2 (0.5M) // Zn+2 (2.5 M) / Zn(s)

EoMg = ?2.372

EoZn = ?0.7628

Eocell = Eocathode - Eoanode

Eocell = -0.7628 - (-2.372) = 1.609 V

?Go= ?nFEocell = -2X 96500 X 1.609 = -310.53 KJ

The negative value of delta G and positive vlaue of Ecell reveals that the reaction and cell will be spontaneous

2) Now E cell = Eocell - 0.0592 / n log [Mg+2 / Zn+2] = 1.609 - 0.0296 log [0.5 / 2.5]

Ecell = 1.588

3) Now E cell = Eocell - 0.0592 / n log [Mg+2 / Zn+2] = 1.609 - 0.0296 log [0.950 / Zn+2]

1.31 = 1.609 - 0.0296 log [0.950 / Zn+2]

1.31 -1.609 = - 0.0296 log [0.950 / Zn+2]

-0.299 / -0.0296 = log [0.950 / Zn+2]

10.10 = log [0.950 / Zn+2]

Taking antilog

1.258 X 10^10 = 0.950 / Zn+2

Zn+2 = 0.755 X 10^-10 M

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