9 sodium acetate , Carethate the pir of the out A current of 9.5 A flows through the electrolyte solution for 40 minutes.How many grams and milliliters and what substances will be released at the anode and cathode during the electrolysis of the Zn(NO_(3))_(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 and oxidation reactions that occur at the respective electrodes.<br /><br />At the cathode, the reduction reaction will occur, where the zinc ions ($Zn^{2+}$) will gain electrons and be reduced to metallic zinc ($Zn$). The reaction is as follows:<br /><br />$Zn^{2+} + 2e^- \rightarrow Zn$<br /><br />At the anode, the oxidation reaction will occur, where the hydroxide ions ($OH^-$) from the water will lose electrons and be oxidized to oxygen gas ($O_2$). The reaction is as follows:<br /><br />$4OH^- \rightarrow O_2 + 2H_2O + 4e^-$<br /><br />Now, let's calculate the amount of zinc and oxygen produced based on the given current and time.<br /><br />Given:<br />Current (I) = 9.5 A<br />Time (t) = 40 minutes = 40 × 60 seconds = 2400 seconds<br /><br />The total charge (Q) passed through the solution can be calculated using the formula:<br /><br />$Q = I \times t$<br /><br />Substituting the given values:<br /><br />$Q = 9.5 A \times 2400 s = 22800 C$<br /><br />Now, we can calculate the number of moles of electrons (n) using the formula:<br /><br />$n = \frac{Q}{F}$<br /><br />where F is the Faraday constant, approximately equal to 96485 C/mol.<br /><br />$n = \frac{22800 C}{96485 C/mol} \approx 0.236 mol$<br /><br />Since the reaction at the cathode involves the reduction of zinc ions, we can calculate the mass of zinc produced using the molar mass of zinc (65.38 g/mol):<br /><br />Mass of zinc = n × Molar mass of zinc<br /><br />Mass of zinc = 0.236 mol × 65.38 g/mol ≈ 15.5 g<br /><br />Therefore, approximately 15.5 grams of metallic zinc will be deposited at the cathode.<br /><br />At the anode, the oxidation reaction involves the production of oxygen gas. We can calculate the volume of oxygen gas produced using the ideal gas law:<br /><br />$PV = nRT$<br /><br />where P is the pressure (assuming standard atmospheric pressure), V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.<br /><br />Assuming standard temperature (298 K), we can rearrange the equation to solve for the volume:<br /><br />$V = \frac{nRT}{P}$<br /><br />Substituting the values:<br /><br />$V = \frac{0.236 mol \times 0.0821 L·atm/mol·K \times 298 K}{1 atm} \approx 5.7 L$<br /><br />Therefore, approximately 5.7 liters of oxygen gas will be released at the anode.<br /><br />In summary, during the electrolysis of the $Zn(NO_{3})_{2}$ solution, approximately 15.5 grams of metallic zinc will be deposited at the cathode, and approximately 5.7 liters of oxygen gas will be released at the anode.