One litre of 1 M CuSO4 solution is electrolysed. After passing 2 F of electricity, the molarity of solution will be

Question

One litre of 1 M CuSO4 solution is electrolysed. After passing 2 F of electricity, the molarity of solution will be (A) M/3 (B) M/2 (C) M/4 (D) 0

Tardigrade Admin · Accepted Answer

Cu2+ +2e- arrow Cu 2 F will decompose 1 mol of Cu2+ and will deposit 1 mol of copper 1L of 1M CuSO4 solution contains 1 mol of CuSO4. Therefore, molarity after electrolysis = 0

Q. One litre of $1 M C u S O_{4}$ solution is electrolysed. After passing $2 F$ of electricity, the molarity of solution will be

M/3

M/2

M/4

0

$C u^{2 +} + 2 e^{-} \to C u$
$2 F$ will decompose $1$ mol of $C u^{2 +}$ and will deposit
$1$ mol of copper
$1 L$ of $1 M C u S O_{4}$ solution contains $1$ mol of $C u S O_{4}$ . Therefore, molarity after electrolysis $= 0$

Q. One litre of 1MCuSO4​ solution is electrolysed. After passing 2F of electricity, the molarity of solution will be

M/3

M/2

M/4

0

Cu2++2e−→Cu 2F will decompose 1 mol of Cu2+ and will deposit 1 mol of copper 1L of 1MCuSO4​ solution contains 1 mol of CuSO4​. Therefore, molarity after electrolysis =0

Q. One litre of $1 M C u S O_{4}$ solution is electrolysed. After passing $2 F$ of electricity, the molarity of solution will be

$C u^{2 +} + 2 e^{-} \to C u$
$2 F$ will decompose $1$ mol of $C u^{2 +}$ and will deposit
$1$ mol of copper
$1 L$ of $1 M C u S O_{4}$ solution contains $1$ mol of $C u S O_{4}$ . Therefore, molarity after electrolysis $= 0$