In the simulation provided click on the Experiment men and then click on the Run
ID: 521318 • Letter: I
Question
In the simulation provided click on the Experiment men and then click on the Run Experiment button that appears. Run each electrolysis experiment with the settings described below to complete the following paragraphs. Fill in the blanks as you proceed through each reaction or write down you results as you go to avoid having to rerun the experiments numerous times. When moving from one experiment to the next click on the blue circular Reset button to reset the entire experiment or the blue circular Rerun button to reset only the settings of the ammeter (current and time). Match the words and numbers to the appropriate blanks in the sentences below. Make certain each sentence is complete before submitting your answer. Electrolysis experiment A is run to completion with the following parameters: The positive terminal is made of A1, the negative terminal is made of Al, the solution is Al(NO_3)_3, the current is set to 30.00 A, and the time is set to 10 min. The final mass of the positive terminal is, and the final mass of the negative terminal is. Thus, the total mass of Al metal that electroplated is. Electrolysis experiment B is run to completion with the following parameters: The positive terminal is made of Al, the negative terminal is made of Al, the solution is Al(NO_3)_3, the current is set to 10.00 A, and the time is set to 10 min. The final mass of the positive terminal is, and the final mass of the negative terminal is. Thus, the total mass of Al that is electroplated is. The ratio change in mass between experiments A and B is equal to. This corresponds to an identical ratio between the two experiments' values for .Explanation / Answer
(1) At anode ( negative termianl) oxidation takes and at cathode (positive terminal) reduction takes place.
So, at negative terminal the mass of Al will decresase and at positive terminal the mass of Al will increase.
At chathode: Al3+ (aq.) + 3 e ------------> Al (aq.)
According to Faraday's first law of electrolysis,
W = M c t / Z F
W = 27 * 30.0 * 10.0 * 60 / (3 * 96500)
W = 1.68 g.
So, 1.68 g. of Al metal is electroplated.
(2)
At anode ( negative termianl) oxidation takes and at cathode (positive terminal) reduction takes place.
So, at negative terminal the mass of Al will decresase and at positive terminal the mass of Al will increase.
At chathode: Al3+ (aq.) + 3 e ------------> Al (aq.)
According to Faraday's first law of electrolysis,
W = M c t / Z F
W = 27 * 10.0 * 10.0 * 60 / (3 * 96500)
W =0.560 g.
So, 0.560 g. of Al metal is electroplated.
(3)
Ratio mass of Al electroplated in A to B = 1.68 / 0.56 = 3 / 1 = 3 : 1
It corresponds the identical ration between tow experiments values for current. (30.0 / 10.0 = 3/1)