In the electrolysis of a ${C u S O}_{4}$ solution, how many grams of $C u$ are plated out on the cathode in the time that it takes to liberate $5.6 l i t r e$ of $O_{2} (g)$ , measured at $1 a t m$ and $273 K$ , at the anode?

31.75

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14.2

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4.32

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13.65

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Solution

The correct option is A $31.75$
$31.75$

given, volume of O2 $= 5.6 L i t r e$

so, number of mole of O2 $= \frac{5.6}{22.4} = 0.25 m o l$

equivalents of O2 = n-factor * number of mole of O2 $= 4 \times 0.25 = 1$

equivalents of Cu = equivalent of O2

or, mass of Cu/equivalent mass = 1

or, mass of Cu/(63.5/2) = 1

or, mass of Cu $= \frac{63.5}{2} = 31.75 g r a m$

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Similar questions

Q. Assertion: Starting from aq. copper [II] sulphate solution, equations for the reactions at the cathode and anode during electrolysis of aq.

C u S O_{4}

using active copper electrodes is;
at anode:

C u^{+ 2} + 2 e^{-} \to C u

at cathode:

C u \to C u^{+ 2} + 2 e^{-}

Reason: Oxidation always take place at anode and reduction at cathode.

Q. In electrolysis of dilute solution of

C u S O_{4}

, products formed at cathode and at anode are

respectively :

Which of the following statements are correct?

a) The electrolysis of aqueous NaCl produces hydrogen gas at the cathode and chlorine gas at the anode .

b) The electrolysis of a $C u S O_{4}$ solution using Pt electrodes causes the liberation of $O_{2}$ at the anode and the deposition of copper at the cathode .

c) Oxygen and hydrogen are produced at the anode and cathode during the electrolysis of dilute aqueous solution of $H_{2} S O_{4}$ .

d) All electrolytic reactions are redox reactions .

Q. During electrolysis of an aqueous solution of

C u S O_{4}

using copper electrodes, if 2.5 g of Cu is deposited at cathode, then at anode:

Q. In the electrolysis of an aqueous solution of

N a O H, 2.8

litres of oxygen was liberated at the anode at NTP. How much hydrogen in litres, was liberated at the cathode?

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In the electrolysis of a CuSO4 solution, how many grams of Cu are plated out on the cathode in the time that it takes to liberate 5.6litre of O2(g), measured at 1atm and 273K, at the anode?

The correct option is A 31.7531.75given, volume of O2 =5.6Litreso, number of mole of O2 =5.622.4=0.25molequivalents of O2 = n-factor * number of mole of O2 =4×0.25=1equivalents of Cu = equivalent of O2 or, mass of Cu/equivalent mass = 1 or, mass of Cu/(63.5/2) = 1 or, mass of Cu =63.52=31.75gram

In the electrolysis of a ${C u S O}_{4}$ solution, how many grams of $C u$ are plated out on the cathode in the time that it takes to liberate $5.6 l i t r e$ of $O_{2} (g)$ , measured at $1 a t m$ and $273 K$ , at the anode?