Question

$A{l}_2{O}_3$ is reduced by electrolysis at low potentials and high currents. If $4.0\times {10}^4$ amperes of current is passed through molten $A{l}_2{O}_3$ for $6$ hours, what mass of aluminium is produced?

(Assume $100\%$ current efficiency and atomic mass of $Al$ $=27$ g $mo{l}^{-1}$)

• A. $1.3\times {10}^4$ g
• B. $9.0\times {10}^3$ g
• C. $8.05\times {10}^4$ g
• D. $2.4\times {10}^5$ g

Answer 1

Answer: (C)

$8.05\times {10}^4$ g

$n=3$ $A{l}^{3+}+3e^{-}\to Al$

$Molarmass=27gmo{l}^{-1}$

$t=6hours=6\times 60\times 60sec$

$i=4\times {10}^{4}A$

$m=\frac{AtomicMass}{n\times F}\times i\times t$

$m=\frac{27}{3\times 96500}\times 4\times {10}^4\times 6\times 60\times 60$

$=\frac{23328}{3\times 965}\times {10}^4$

$=8.05\times {10}^{4}g$

Mass of $Al$ produced is $8.05\times {10}^{4}g$