Three Faradays of electricity was passed through an aqueous solution of magnesium bromide. The weight of magnesium metal deposited at the cathode in grams is:

Question

Three Faradays of electricity was passed through an aqueous solution of magnesium bromide- The weight of magnesium metal deposited the cathode in grams is-568436168

Toppr Admin · Accepted Answer

MgBr2 has Mg2- ions-Mg2- -xA0- - 2e -gt- Mg2 moles of electron deposits 1 mole of Mg-1 mole of electron - 1 Faraday3 moles of the electron will yield - 3-2 moles of Mg-1 mole of Mg- 24 gm-therefore 3-2 moles of Mg- 36 gm-xA0-Option C is correct-

Three Faradays of electricity was passed through an aqueous solution of magnesium bromide. The weight of magnesium metal deposited at the cathode in grams is:

56
84
36
168

$M g B r_{2}$ has $M g^{2 +}$ ions.

$M g^{2 +}$ + 2e => Mg

2 moles of electron deposits 1 mole of Mg.

1 mole of electron = 1 Faraday

3 moles of the electron will yield = 3/2 moles of Mg.

1 mole of Mg= 24 gm.

therefore 3/2 moles of Mg= 36 gm.

Option C is correct.

Three Faradays of electricity was passed through an aqueous solution of magnesium bromide. The weight of magnesium metal deposited at the cathode in grams is:568436168

MgBr2 has Mg2+ ions.Mg2+ + 2e => Mg2 moles of electron deposits 1 mole of Mg.1 mole of electron = 1 Faraday3 moles of the electron will yield = 3/2 moles of Mg.1 mole of Mg= 24 gm.therefore 3/2 moles of Mg= 36 gm. Option C is correct.

Three Faradays of electricity was passed through an aqueous solution of magnesium bromide. The weight of magnesium metal deposited at the cathode in grams is:

56
84
36
168

$M g B r_{2}$ has $M g^{2 +}$ ions.

$M g^{2 +}$ + 2e => Mg

2 moles of electron deposits 1 mole of Mg.

1 mole of electron = 1 Faraday

3 moles of the electron will yield = 3/2 moles of Mg.

1 mole of Mg= 24 gm.

therefore 3/2 moles of Mg= 36 gm.

Option C is correct.