Question

How many grams of ca metal are produced by the electrolysis of molten cabr2 using a current of 30.0 amp for 8.0 hours?

Answer 1

Answer:

$m=179.4g$ of Ca

Explanation:

First we are going to write down the balanced reaction:

$CaBr_{2}=Ca^{2+}+_{2}Br^{-1}$

The reduction for the $Ca^{2+}$ ion will be:

$Ca^{2+}+2e^{-}=Ca$

It means that 2 Faraday left for each mol of Ca.

Converting from Faraday to Coulombs:

$2Faraday*\frac{96485Coulombs}{1Faraday}= 192970Coulombs$

Then we can apply the Faraday´s law:

$m=\frac{M*I*t}{e^{-}*96485}$

where

m=mass in grams

M=molar mass

I=current

t=time in seconds

e-=number of electrons per mol

Replacing values:

$m=\frac{40.078g*30.0A*28800s}{2*96485C}$

$m=179.4g$ of Ca