Вопрос
9. A current of 9.5 A flows through the electrolyte solution for 40 minutes. How grams and milliliters and what substances will be at the anode and cathode during the electrolysis of the Zn(NO_(3))_(2) (NO3)2 solution.
Решения
3.9
(246 Голоса)
Ульян
элита · Репетитор 8 лет
Ответ
To determine the substances released at the anode and cathode during the electrolysis of the
solution, we need to consider the standard reduction potentials of the ions present in the solution.The standard reduction potential for
is
V, while the standard reduction potential for
is
V. Since the standard reduction potential for
is lower than that for
,
will be reduced at the cathode.At the anode, the oxidation reaction will occur. Since the anode is the site of oxidation, the substance that will be released at the anode will be the one that loses electrons. In this case, the
ions will be oxidized to
gas.Now, let's calculate the amount of
and
produced.First, we need to calculate the total charge passed through the electrolyte solution. The total charge can be calculated using the formula:
where
is the total charge,
is the current, and
is the time in seconds.Given that the current is 9.5 A and the time is 40 minutes (or 2400 seconds), we can calculate the total charge as follows:
Next, we need to calculate the amount of
produced. The amount of
can be calculated using the formula:
where
is the amount of
in moles,
is the number of electrons transferred, and
is the Faraday constant (
).Since the reduction reaction at the cathode is
, the number of electrons transferred is 2. Therefore, the amount of
produced is:
To convert the amount of
to grams, we can use the molar mass of
, which is
:
Therefore, 7.75 grams of
will be deposited at the cathode.Next, let's calculate the amount of
produced. The amount of
can be calculated using the formula:
where
is the amount of
in moles,
is the number of electrons transferred, and
is the Faraday constant (
).Since the oxidation reaction at the anode is
, the number of electrons transferred is 10. Therefore, the amount of
produced is:
To convert the amount of
to grams, we can use the molar mass of
, which is
:
Therefore, 0.00664 grams of
will be released at the anode.In summary, during the electrolysis of the
solution, 7.75 grams of
will be deposited at the cathode, and 0.00664 grams of
will be released at the anode.