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Question

2 electrode cell containing silver nitrate solution and copper sulphate solution in series a current of 2.5 ampere pass through till0.78 gram of silver was deposited how llong will the current flow and how long will be copper deposited

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Solution

1 mole of silver contains 47 g

Therefore, 0.78 g of silver contains 0.016 moles of silver: 0.016 moles of silver electrons

Therefore quantity of charge Q = 0.016 * 96500

= 1544 C

Q = I * t

t = Q / I

= 1544 / 2

= 772 seconds


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