Question

what weight of Cr deposited when three faraday of electricity is passed through solution of Cr2(So4)3​

Answer 1

Answer:

faraday law equation...

Answer 2

Answer: 52 grams

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes

t= time in seconds

$Cr_2(SO_4)_3\rightarrow 2Cr^{3+}+3SO_4^{2-}$

$Cr^{3+}+3e^-\rightarrow Cr$    

$96500\times 3=193000Coloumb$ of electricity deposits 1 mole of Chromium

(1 Faraday = 96500 Coloumb)

3 Faraday of electricity deposits 1 mole of Chromium.

Mass of chromium =${\text {no of moles}}\times {\text {Molar mass}}$

Mass of chromium =$1\times 52g/mol=52 g$

The weight of chromium deposited is 52 grams.