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Консультация № 189518
28.05.2016, 14:46
0.00 руб.
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Уважаемые эксперты! Пожалуйста, ответьте на вопрос:
Задача 1:
Раствор электролита подвергается электролизу при силе тока 1А в течение 0.5 часа. Записать уравнения электродных процессов и уравнение электролиза раствора. Определить массу и объём газа, выделившегося на аноде. Электролит: FeSO4
Задача 2:
Металлический проводник, изготовленный из кадмия, погружён в 0,1 М раствор соли SdSO4. Рассчитать величину относительно электродного потенциала данного электрода. Пользуясь таблицей стандартных электродных потенциалов, составить схему гальванического элемента, в котором рассматриваемый электрод является анодом. Для выбранного гальванического элемента записать уравнения электродных процессов и уравнение электрохимического процесса. Рассчитать величину ЭДС приняв электродный потенциал катода равный стандартному. Чему равна стандартная ЭДС?
Задача 3:
Cd+HNO3=>Cd(NO3)2+NO2+...
Используя метод электронно-ионных уравнений, составить уравнения реакции. В электронно-ионных уравнениях указать процессы окисления и восстановления, окислитель и восстановитель. Пользуясь таблицей стандартных электродных потенциалов, объяснить, почему данная реакция возможна.
Задача 1:
Раствор электролита подвергается электролизу при силе тока 1А в течение 0.5 часа. Записать уравнения электродных процессов и уравнение электролиза раствора. Определить массу и объём газа, выделившегося на аноде. Электролит: FeSO4
Задача 2:
Металлический проводник, изготовленный из кадмия, погружён в 0,1 М раствор соли SdSO4. Рассчитать величину относительно электродного потенциала данного электрода. Пользуясь таблицей стандартных электродных потенциалов, составить схему гальванического элемента, в котором рассматриваемый электрод является анодом. Для выбранного гальванического элемента записать уравнения электродных процессов и уравнение электрохимического процесса. Рассчитать величину ЭДС приняв электродный потенциал катода равный стандартному. Чему равна стандартная ЭДС?
Задача 3:
Cd+HNO3=>Cd(NO3)2+NO2+...
Используя метод электронно-ионных уравнений, составить уравнения реакции. В электронно-ионных уравнениях указать процессы окисления и восстановления, окислитель и восстановитель. Пользуясь таблицей стандартных электродных потенциалов, объяснить, почему данная реакция возможна.
Обсуждение
01.06.2016, 10:49
общий
это ответ
Здравствуйте, krisanov97!
1.
(-): Fe2++2e-=Fe
(+): 2H2O-4e-=O2+4H+
2FeSO4+2H2O=2Fe+2H2SO4+O2
количество кислорода, выделившегося на аноде
n(O2)=It/4F=1А[$183$]1800с/(4[$183$]96500Кл/моль)=0,00466 моль
m(O2)=0,15 г
V(O2)=104 мл
2.
Cd2++2e-=Cd
[$949$]0(Cd2+/Cd)=-0,403 В
[$949$](0,1M Cd2+/Cd)=[$949$]0(Cd2+/Cd)+(RT/2F)ln(C(Cd2+))=[$949$]0(Cd2+/Cd)+(0,059/2)lg(C(Cd2+))=-0,433 В
Cd|0,1M CdSO4||1M CuSO4|Cu
Cd-2e-=Cd2+
Cu2++2e-=Cu ([$949$]0(Cu2+/Cu)=+0,342 В)
Cd+CuSO4=Cu+CdSO4
E=[$949$]0(Cu2+/Cu)-[$949$](0,1M Cd2+/Cd)=0,775 В
E0=[$949$]0(Cu2+/Cu)-[$949$]0(Cd2+/Cd)=0,745 В
3.
процесс окисления (кадмий - восстановитель):
Cd-2e-=Cd2+ ([$949$]0(Cd2+/Cd)=-0,403 В)
процесс восстановления (азотная кислота - окислитель):
NO3-+2H++e-=NO2+H2O ([$949$]0(HNO3/NO2)=+0,803 В)
Cd+2NO3-+4H+=Cd2++2NO2+2H2O
Cd+4HNO3=Cd(NO3)2+2NO2+2H2O
E0=[$949$]0(HNO3/NO2)-[$949$]0(Cd2+/Cd)=1,206 В
1.
(-): Fe2++2e-=Fe
(+): 2H2O-4e-=O2+4H+
2FeSO4+2H2O=2Fe+2H2SO4+O2
количество кислорода, выделившегося на аноде
n(O2)=It/4F=1А[$183$]1800с/(4[$183$]96500Кл/моль)=0,00466 моль
m(O2)=0,15 г
V(O2)=104 мл
2.
Cd2++2e-=Cd
[$949$]0(Cd2+/Cd)=-0,403 В
[$949$](0,1M Cd2+/Cd)=[$949$]0(Cd2+/Cd)+(RT/2F)ln(C(Cd2+))=[$949$]0(Cd2+/Cd)+(0,059/2)lg(C(Cd2+))=-0,433 В
Cd|0,1M CdSO4||1M CuSO4|Cu
Cd-2e-=Cd2+
Cu2++2e-=Cu ([$949$]0(Cu2+/Cu)=+0,342 В)
Cd+CuSO4=Cu+CdSO4
E=[$949$]0(Cu2+/Cu)-[$949$](0,1M Cd2+/Cd)=0,775 В
E0=[$949$]0(Cu2+/Cu)-[$949$]0(Cd2+/Cd)=0,745 В
3.
процесс окисления (кадмий - восстановитель):
Cd-2e-=Cd2+ ([$949$]0(Cd2+/Cd)=-0,403 В)
процесс восстановления (азотная кислота - окислитель):
NO3-+2H++e-=NO2+H2O ([$949$]0(HNO3/NO2)=+0,803 В)
Cd+2NO3-+4H+=Cd2++2NO2+2H2O
Cd+4HNO3=Cd(NO3)2+2NO2+2H2O
E0=[$949$]0(HNO3/NO2)-[$949$]0(Cd2+/Cd)=1,206 В
Форма ответа
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