Question

How many moles of electron are required for the reduction of
(i) 3 moles of Zn²⁺ to Zn
(ii) 1 mole of Cr³⁺ to Cr?

How many faradays of electricity will be required in each case?

Answer 1

A)For 1mole of Zn2+ ion,2electrons~2Faradays of electricity is required.So,by pure quick analysis or long form of unitary method,we arrive at 6Faradays of electricity.

B)Solving it by redox(acidic or basic medium),we will arrive at 3Faradays.

Hope it helps.

Answer 2

(i) $Zn^{2+}+2e^-\rightarrow Zn$ : 2 moles : 2 Faraday

(ii) $Cr^{3+}+3e^-\rightarrow Cr$ : 3moles : 3 Faraday

Explanation:

1 electron carry charge= $1.6\times 10^{-19}C$

1 mole of electrons carry charge=[tex]\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C=1 Faraday [/tex]

(i) $Zn^{2+}+2e^-\rightarrow Zn$

1 mol of $Zn^{2+}$ requires 2 moles of electrons

1 mole of electrons carry charge = 1 Faraday

Thus 2 moles of electrons carry charge=\frac{1F}{1}\times 2=2Faraday[/tex]

(ii) $Cr^{3+}+3e^-\rightarrow Cr$

1 mol of $Cr^{3+}$ requires  = 3 moles of electrons

1 mole of electrons carry charge = 1 Faraday

Thus 3 moles of electrons carry charge=\frac{1F}{1}\times 3=3Faraday[/tex]

Learn More about electronic charge

https://brainly.in/question/2728460

https://brainly.in/question/2981640