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transcribed image text: (27pts) part ii. galvanization mass of iron nail before: 0.620 g mass of galvanized iron nail: 0.610 9 (1 pts) mass of zinc produced (9) mass of zinc electrode before: 5.720 9 mass of zinc electrode after: 5.710 g (1 pts) mass of zinc consumed (9) length of time ran: 5.00 min 1. calculate the number of moles of electrons that passed through the cell during the electrolysis based off the amount of zinc produced on the iron nail. (4pts) moles of electrons (mol of e) (1pts) for full credit, upload all of your work for question 1 here. browse your files to upload or drag and drop max attachments: 5 max size: 20.00mb each 2. calculate the current (amps) that passed through the cell during the electrolysis. (8pts) current (amps)

Sagot :

The mass of Zn produced on the iron nail  is similar/ equal to the mass of Zn lost from the Zn electrode. Hence no loss of mass.

Elaborate the above statement with calculation.

Mass of Zn produced in iron nail= 0.630g -0. 610g = 0.01g

Mass of Zn consumed = 5.720 to 5.710 g = 0.01g

Given length of time run= 5 min = 5× 60 = 300 second

Moles of Zn passed = 0.001g = 0.00015 mol

1 mol contain = 6.022×10^23 atoms

Total electron contain in 0.00015 mol of Zn atoms

= 0.00015 × 30 ×6.022× 10^23

= 0.00461 × 6.022 × 10^23

Therefore, number of moles of electron transferees = 0.00461×6.022×10^23 / 6.022×10 ^23

= 0.00461 mol

From the Faraday first law of electrolysis

I = mF / Et

= 0.01 × 96500/32.7×200

= 0.098 mps

The mass of Zn produced on the iron nail  is similar/ equal to the mass of Zn lost from the Zn electrode. Hence no loss of mass.

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