Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given: Molar mass of Cu = 63 g , 1F = 96487 C) [2024]

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Solution

Q.
Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is

(Given: Molar mass of Cu = 63 g ${mol}^{- 1}$ , 1F = 96487 C) [2024]

1 3.15 g

2 0.315 g

3 31.5 g

4 0.0315 g

Ans.
(2)

Given,

Current $(I) = 9.6487$ $A, Time (t) = 100$ $seconds$

1F = 96487 C

Reaction : ${Cu}_{(aq)}^{2 +} + 2 e^{-} \to {Cu}_{(s)}$

Here, 2 electrons are exchanged, so valency factor $(n) = 2$ .

Applying Faraday’s first law, $W = Z I t$ $(∵ Z = \frac{Equivalent weight}{F} = \frac{63}{2 \times 96487})$

So, $W = \frac{63 \times 9.6487 \times 100}{2 \times 96487}$ $= 0.315$ $g$

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