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A battery is constructed based on the oxidation of magnesium and the reduction of Cu2+. The initial concentrations of Mg2+ and Cu2+ are 1.0×10−4 M and 1.6 M , respectively, in 1.0-liter half-cells.

What is the initial voltage of the battery?

Respuesta :

pstnonsonjoku

Answer:

2.59 V

Explanation:

Equation of the reaction:

Mg(s) + Cu^2+(aq) ------------> Mg^2+(aq) + Cu(s)

From Nernst Equation

E= E°-0.0592/n log [Red]/ [Ox]

E°= 0.34- (-2.37)= 2.71

E= 2.71 - 0.0592/2 log [1.6]/ [1×10^-4]

E= 2.71 - 0.12

E= 2.59 V

mickymike92

The initial voltage of the battery is 2.83 V

From the reduction potential table, the standard reduction potential values, E° of Mg²⁺ and Cu²⁺ are -2.37 and +0.34 respectively.

Since magnesium is oxidized and copper is reduced, the equation of the reaction is given as:

Mg (s) + Cu²⁺ (aq) ---------> Mg²⁺ (aq) + Cu (s)

E° of cell is given as: E°(oxidized) + E°(reduced)

E° of cell = 2.37 +0.34 = 2.71 V

The initial concentrations of Mg²⁺ and Cu²⁺ are 1.0 × 10⁻⁴ M and 1.6 M

Using the Nernst Equation:

Ecell = E° - 0.0592/n * logQ

n = 2 moles of electrons

Q = [Mg²⁺]/[Cu²⁺]

Q = 1.0 × 10⁻⁴ M / 1.6 M = 6.25 * 10⁻⁵

Ecell = 2.71 - (0.0592/2) * log 6.25 * 10⁻⁵

Ecell = 2.83 V

Therefore, the initial voltage of the battery is 2.83 V

learn more at: https://brainly.com/question/14312871

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