Calculate the time required to deposit 1.27 g of copper at cathode when a current of 2A was passed through the solution of `CuSO_(4)`.
`("Molar mass of "Cu=63.5g mol^(-1), 1F =96500 C mol^(-1)).`

Explanation

To calculate the time required to deposit 1.27 g of copper at the cathode when a current of 2 A is passed through a solution of CuSO₄, we can use the formula related to electrolysis:

\[ \text{Deposit} = Z \cdot I \cdot T \]

Where:
- \( Z \) is the electrochemical equivalent (in g/C),
- \( I \) is the current (in A),
- \( T \) is the time (in seconds).
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