Page **1** of **1**

### Electron-transfer reaction

Posted: **Fri Jul 23, 2021 9:13 am**

by **Dhamnekar Winod**

1)What current will be required to produce 1.0 g of Na in 10.0 min. by electrolysis of molten NaCl?

2) What quantity of electricity will provide 0.624 mol of electrons?

3)If the current is 3.5 A, how many Coulombs per second are passing through the cell?

4) How many minutes will be required for a current 3.5 A , to produce 6.02 × 10⁴ C?

### Re: Electron-transfer reaction

Posted: **Sat Jul 24, 2021 2:50 am**

by **ChenBeier**

The basic is the law of faraday

m = I * t * c

c = electrochemical equivalent =M/ (z *F)

m= mass, I = current, t= time, M = molar mass, z = amount of transferred electrons, F = Faraday constant 96485 As,

### Re: Electron-transfer reaction

Posted: **Tue Jul 27, 2021 9:59 pm**

by **Dhamnekar Winod**

Answers to 1),2),3) and 4)

1)\(\frac{1 g}{22.99 g/mol Na} \times \frac {1 mol e^-}{1mol Na} \times \frac{96485.3399 C}{1 mol e^-} \frac{1}{10 min \times 60 s/min}= 7.0 C = 7.0 A\) **current will be required to produce 1 g of Na in 10.0 minutes by the electrolysis of molten NaCL**

2)\(0.624 mol e^- \times \frac{96485.3399 C}{1 mol e^- }= 6.02 \times 10^4 C \) **elecectricity will provide 0.624 mol e⁻.**

3) **If the current is 3.5 A, 3.5 Coulombs per second are passing through the cell. **

4)\( \frac{6.02 \times 10^4 C}{3.5 C/s} \times \frac {1 min}{60 s} = 2.90 \times 10^2 min\) will be required for a current of 3.5 A to produce \(6.02 \times 10^4 C \)** electricity**

### Re: Electron-transfer reaction

Posted: **Tue Jul 27, 2021 11:38 pm**

by **ChenBeier**

Looks ok.