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 solution, we need to consider the standard reduction and oxidation reactions that occur at the respective electrodes.At the cathode, the reduction reaction will occur, where the zinc ions ( ) will gain electrons and be reduced to metallic zinc ( ). The reaction is as follows: At the anode, the oxidation reaction will occur, where the hydroxide ions ( ) from the water will lose electrons and be oxidized to oxygen gas ( ). The reaction is as follows: Now, let's calculate the amount of zinc and oxygen produced based on the given current and time.Given:Current (I) = 9.5 ATime (t) = 40 minutes = 40 × 60 seconds = 2400 secondsThe total charge (Q) passed through the solution can be calculated using the formula: Substituting the given values: Now, we can calculate the number of moles of electrons (n) using the formula: where F is the Faraday constant, approximately equal to 96485 C/mol. 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):Mass of zinc = n × Molar mass of zincMass of zinc = 0.236 mol × 65.38 g/mol ≈ 15.5 gTherefore, approximately 15.5 grams of metallic zinc will be deposited at the cathode.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: 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.Assuming standard temperature (298 K), we can rearrange the equation to solve for the volume: Substituting the values: Therefore, approximately 5.7 liters of oxygen gas will be released at the anode.In summary, during the electrolysis of the 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.