What current would be required to deposit `1.00 m^(2)` of chromium plate having a thickness of 0.052mm in 4.5h from a solution of `H_(2)CrO_(4)` ? The current efficiency is 74% and density of chromium is `7.19 g cm^(-3)`. [Atomic mass of Cr= 52u]

Options

A. 347A
B. 185A
C. 125A
D. 0.25A

Answer

A

Explanation

We know that mass of Cr deposited = Volume `xx` Density
`=(1xx10^(4) cm^(2) xx 0.052 xx 10^(-1) cm) xx 7.19 g cm^(3)= 52 xx 7.19g`
Oxidation state of Cr in `H_(2)CrO_(4)= +6 rArr Cr^(6+) + 6e^(-) rarr Cr`
Therefore, 1 mol or 52g of Cr is reduced by 6 mol electrons or 6F charge. Hence, `52xx 7.19g` would be deposited by `(6F)/(52) xx 52 xx 7.19C`
`rArr Q= ixx t rArr i= ((6F//52)xx 52 xx 7.19)/(4.5 xx 60 xx 60)= 256.9A ~~ 257A`
Current efficiency is 74% so that actual curretn required is `(257)/(0.74)= 347.3 ~~ 347A`