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.

To determine the substances released at the anode and cathode during the electrolysis of the $Zn(NO_{3})_{2}$ solution, we need to consider the standard reduction potentials of the ions present in the solution.<br /><br />The standard reduction potential for $Zn^{2+}$ is $-0.76$ V, while the standard reduction potential for $H^{+}$ is $0.00$ V. Since the standard reduction potential for $Zn^{2+}$ is lower than that for $H^{+}$, $Zn^{2+}$ will be reduced at the cathode.<br /><br />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 $NO_{3}^{-}$ ions will be oxidized to $N_{2}$ gas.<br /><br />Now, let's calculate the amount of $Zn$ and $N_{2}$ produced.<br /><br />First, we need to calculate the total charge passed through the electrolyte solution. The total charge can be calculated using the formula:<br /><br />$Q = I \times t$<br /><br />where $Q$ is the total charge, $I$ is the current, and $t$ is the time in seconds.<br /><br />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:<br /><br />$Q = 9.5 \, A \times 2400 \, s = 22800 \, C$<br /><br />Next, we need to calculate the amount of $Zn$ produced. The amount of $Zn$ can be calculated using the formula:<br /><br />$Zn = \frac{Q}{n \times F}$<br /><br />where $Zn$ is the amount of $Zn$ in moles, $n$ is the number of electrons transferred, and $F$ is the Faraday constant ($96485 \, C/mol$).<br /><br />Since the reduction reaction at the cathode is $Zn^{2+} + 2e^{-} \rightarrow Zn$, the number of electrons transferred is 2. Therefore, the amount of $Zn$ produced is:<br /><br />$Zn = \frac{22800 \, C}{2 \times 96485 \, C/mol} = 0.118 \, mol$<br /><br />To convert the amount of $Zn$ to grams, we can use the molar mass of $Zn$, which is $65.38 \, g/mol$:<br /><br />$Zn = 0.118 \, mol \times 65.38 \, g/mol = 7.75 \, g$<br /><br />Therefore, 7.75 grams of $Zn$ will be deposited at the cathode.<br /><br />Next, let's calculate the amount of $N_{2}$ produced. The amount of $N_{2}$ can be calculated using the formula:<br /><br />$N_{2} = \frac{Q}{n \times F}$<br /><br />where $N_{2}$ is the amount of $N_{2}$ in moles, $n$ is the number of electrons transferred, and $F$ is the Faraday constant ($96485 \, C/mol$).<br /><br />Since the oxidation reaction at the anode is $2NO_{3}^{-} \rightarrow N_{2} + 6H_{2}O$, the number of electrons transferred is 10. Therefore, the amount of $N_{2}$ produced is:<br /><br />$N_{2} = \frac{22800 \, C}{10 \times 96485 \, C/mol} = 0.000236 \, mol$<br /><br />To convert the amount of $N_{2}$ to grams, we can use the molar mass of $N_{2}$, which is $28.02 \, g/mol$:<br /><br />$N_{2} = 0.000236 \, mol \times 28.02 \, g/mol = 0.00664 \, g$<br /><br />Therefore, 0.00664 grams of $N_{2}$ will be released at the anode.<br /><br />In summary, during the electrolysis of the $Zn(NO_{3})_{2}$ solution, 7.75 grams of $Zn$ will be deposited at the cathode, and 0.00664 grams of $N_{2}$ will be released at the anode.