A current of 2.68 A is passed for one hour through an aqueous solution of $C u S O_{4}$ using copper electrodes. Calculate the change in mass of cathode and that of the anode. (At. mass of copper = 63.5).

Increase in the mass of cathode = 3.174g

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Decrease in the mass of anode = 3.174g

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No change in the mass of electrodes

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The ratio between the change in the mass of cathode to anode is 1:2

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Solution

The correct options are
A Increase in the mass of cathode = 3.174g
B Decrease in the mass of anode = 3.174g
At Cathode : $C u^{2} + 2 e^{-} ⟶ C u (s)$

At Anode : $C u (s) ⟶ C u^{2 +} + 2 e^{-}$

Increase in mass of cathode = decrease in mass of Anode = $\frac{2.68 \times 3600}{96500} \times \frac{63.5}{2} = 3.174 g$

Options A and B are correct.

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A current of 2.68 A is passed for one hour through an aqueous solution of CuSO4 using copper electrodes. Calculate the change in mass of cathode and that of the anode. (At. mass of copper = 63.5).

The correct options are A Increase in the mass of cathode = 3.174g B Decrease in the mass of anode = 3.174gAt Cathode : Cu2+2e−⟶Cu(s)At Anode : Cu(s)⟶Cu2++2e−Increase in mass of cathode = decrease in mass of Anode = 2.68×360096500×63.52=3.174gOptions A and B are correct.

A current of 2.68 A is passed for one hour through an aqueous solution of $C u S O_{4}$ using copper electrodes. Calculate the change in mass of cathode and that of the anode. (At. mass of copper = 63.5).