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Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Electrochemistry

In an electrolysis experiment electric current was passed for 5hr through two cells connected in series.The first cell contains gold salt and second contains $CuSO_4$ solution.9.83g of gold was deposited in first cell.The magnitude of current in amperes is :-

$\begin{array}{1 1}(a)\;0.7amp&(b)\;0.8amp\\(c)\;0.6amp&(d)\;0.5amp\end{array}$

1 Answer

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$W=Z\times i\times t=\large\frac{E}{96500} $$i\times t$
$4.7=\large\frac{\Large\frac{63.5}{2}}{96500}\times$$ 1\times 5\times 60\times 60$
$\therefore i=\large\frac{4.75\times 96500\times 2}{5\times 60}$
$\Rightarrow 0.8amp$
Hence (b) is the correct answer.
answered Feb 17, 2014 by sreemathi.v

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