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Use the Standard Reaction Potentials as a sorce of half-reaction equations and E^o values for the following reaction, and use them to answer part a) and b).
Mg^2+ + 2 e- -->Mg E^o = -2.37V
Al^3+ + 3 e - --> Al E^o = - 1.66V

Solid Mg reacts with a solution containing aluminum ions , forming a solution that contains magnesium ions and solid Al. Write the balanced net equation:

Then the sum half reactions:

a) calculate E^o for the reaction as written
b) State and breifly justify the favored direction of the reaction based on Gibbs Free Energy. Faraday Constant F= 96485 J/Volt•mol e-

Respuesta :

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Answer:

a) E°=3*2.37 -2*1.66 = 3.79 V

b) the favored direction is 3Mg +2Al^(3+) ----> 3Mg^(2+) + 2Al

Explanation:

3Mg + 2Al^(3+) ---> 3Mg^(2+) + 2Al

a)

Mg                ---> Mg^(2+) + 2e-  , E°= +2.37V     /*3

Al^(3+) + 3 e - --> Al                     , E° = - 1.66V  /*2

3Mg +2Al^(3+) ----> 3Mg^(2+) + 2Al ,     E°=3*2.37 -2*1.66 = 3.79 Volt

b)  ΔG°= − nFE° = - 6 mol e-*96485 J/Volt*mol e- * 3.79Volt  < 0,

so the favored direction is 3Mg +2Al^(3+) ----> 3Mg^(2+) + 2Al

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