Mg\(^{2+}\)  + 2e\(^-\) → Mg

If  2F produces 24g Mg

then x  will produce 1g Mg

 24x = 2F

 x = \(\frac{2F}{24}\)

Similarly, Al\(^{3+}\)  +  3e\(^-\) → Al

If  3F produces 27g Al

then x  will produce 10g Al

27x = 3F

 x = \(\frac{{3F}\times{10}}{27}\)

 x =  \(\frac{30F}{27}\)

Now, if \(\frac{2F}{24}\) of Mg costs N5

then  \(\frac{30F}{27}\) of Al will cost y

By cross- multiplying to make y the subject of formula.

  \(\frac{30F}{27}\times 5\) = \(\frac{2F}{24}\times y\)

  y =  \(\frac{30F}{27}\times 5\) ÷ \(\frac{2F}{24}\)

  y = \(\frac{30F}{27}\times 5\) x \(\frac{24}{2F}\)

  y = \(\frac{{30F}\times{24}\times{5}}{{27}{2F}}\)

  y = N 66.67

Therefore, it will cost N66.67 to deposit 10g of Aluminium.