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Консультация № 153454
08.12.2008, 14:50
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Увожаемые эксперты помогите пожалуста решить задачки 
Составте электронные уравнения процессов, протекающих на утольных электродах при электролизе водного раствора MgCl2. вычеслите силу тока, если изавестно, что при электролизе MgCl2 в течение 30 минут на катоде выделилось 8,4 л водорода, измеренного при нормальных условиях. за рание мпасибо!!!!
Составте электронные уравнения процессов, протекающих на утольных электродах при электролизе водного раствора MgCl2. вычеслите силу тока, если изавестно, что при электролизе MgCl2 в течение 30 минут на катоде выделилось 8,4 л водорода, измеренного при нормальных условиях. за рание мпасибо!!!!
Обсуждение
Неизвестный
08.12.2008, 15:54
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это ответ
Здравствуйте, Шарафулина Ирина Илдаровна!
Перечислим все процессы, которые могут идти на аноде:
2H2O - 4e = 4H+ + O2, E = + 1.42 В(потенциалы здесь и далее я привожу так, как они даются в справочнике - то есть, для реакции восстановления; то есть, в данном случае, потенциал соответствует обратной реакции)
2Cl- - 2e = Cl2, E = + 1.36 В
Катодные процессы:
Mg2+ + 2e = Mg, E = -2.363 В
2H2O + 2e = H2 + 2OH-, E = -0.828 В
Теперь надо выбрать предпочтительные процессы. Принцип тут какой? Предпочтительный процесс идет легче. В этом нам помогут потенциалы. Анодный процесс с меньшим потенциалом протекает легче, следовательно, на аноде идет окисление хлора. Катодный процесс идет легче при большем потенциале, то есть, на катоде идет восстановление воды.
(А): 2Cl- - 2e = Cl2
(К): 2H2O + 2e = H2 + 2OH-
Суммарный процесс:
2Cl- + 2H2O + Mg2+ = Cl2 + H2 + Mg(OH)2[$8595$]
MgCl2 + 2H2O = Mg(OH)2[$8595$] + H2 + Cl2
Количество эквивалентов вещества, выделяющегося при электролизе, выражается следующей формулой:
n = I*t/F, где I - сила тока в амперах, t - время в секундах, F = 96485 Кл/моль*экв.
Количество эквивалентов водорода можно легко найти:
n = 8.4 л/ (22.4 л/моль) = 0.375 моль, а количество эквивалентов вдвое больше - 0.375 * 2 = 0.75 моль*экв. Отсюда I = n*F/t = 0.75 * 96485 / (30 * 60) = 40.2 А. Жутковатая какая-то величина :)
С уважением, Товарищ Бородин
Перечислим все процессы, которые могут идти на аноде:
2H2O - 4e = 4H+ + O2, E = + 1.42 В(потенциалы здесь и далее я привожу так, как они даются в справочнике - то есть, для реакции восстановления; то есть, в данном случае, потенциал соответствует обратной реакции)
2Cl- - 2e = Cl2, E = + 1.36 В
Катодные процессы:
Mg2+ + 2e = Mg, E = -2.363 В
2H2O + 2e = H2 + 2OH-, E = -0.828 В
Теперь надо выбрать предпочтительные процессы. Принцип тут какой? Предпочтительный процесс идет легче. В этом нам помогут потенциалы. Анодный процесс с меньшим потенциалом протекает легче, следовательно, на аноде идет окисление хлора. Катодный процесс идет легче при большем потенциале, то есть, на катоде идет восстановление воды.
(А): 2Cl- - 2e = Cl2
(К): 2H2O + 2e = H2 + 2OH-
Суммарный процесс:
2Cl- + 2H2O + Mg2+ = Cl2 + H2 + Mg(OH)2[$8595$]
MgCl2 + 2H2O = Mg(OH)2[$8595$] + H2 + Cl2
Количество эквивалентов вещества, выделяющегося при электролизе, выражается следующей формулой:
n = I*t/F, где I - сила тока в амперах, t - время в секундах, F = 96485 Кл/моль*экв.
Количество эквивалентов водорода можно легко найти:
n = 8.4 л/ (22.4 л/моль) = 0.375 моль, а количество эквивалентов вдвое больше - 0.375 * 2 = 0.75 моль*экв. Отсюда I = n*F/t = 0.75 * 96485 / (30 * 60) = 40.2 А. Жутковатая какая-то величина :)
С уважением, Товарищ Бородин
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