Question

What is the amount of na deposited by 5 amp current for 10 mins from fused nacl?

Answer 1

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Answer 2

Answer:

The amount of $sodium$ deposited will be equal to $0.715g$ during the electrolysis of fused $NaCl$.

Explanation:

The chemical reaction of electrolysis of NaCl:

                 $Na^{+} + e^{-}$  →  $Na(s)$

Given current flows $I=5A$ and time $t=10min=10*60=600sec$

We know, $I=\frac{q}{t}$

Charge, $q=I*t$

$q=(5A)(600sec)=3000C$

1mole of sodium deposited with 1mole of electrons or 96500C charge

1mole of sodium $=23g$  of Na

$96500Coulombs$ charge required to deposit $=23grams$ of Na

$3000Coulombs$  charge will deposit Na $=\frac{(3000)(23)}{96500} =0.715g$ of Na