A) 1.24 min
B) 124 h
C) 1.24 s
D) none of these
Correct Answer: B
Solution :
\[C{{r}^{3+}}+3{{e}^{-}}\xrightarrow{{}}Cr(s)\] 3 mol or 3 x 96500 C of electricity are needed to deposit 1 mol or 52 g of Cr. 52 g of Cr require current \[=3\times 96500\,C\] 1 g of Cr will require current \[=\frac{3\times 96500}{52}C\] \[=5567.3\,C\] Now, number of coulombs = current (ampere) x t (seconds) \[\therefore \,\,\text{Time}\,\text{(s)}\,\text{required}\,\text{=}\,\frac{\text{no}\text{. of coulombs}}{\text{current }\left( \text{ampere} \right)}\] \[=\frac{5567.3\,C}{1.25\,A}\] \[=4453.8\,\,s\] or \[=1.24\,h\]
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