A solution containing trivalent ions M^{3+} of an unknown metal M is electrolyzed using in… — JAMB UTME

A solution containing trivalent ions $M^{3+}$ of an unknown metal $M$ is electrolyzed using inert electrodes. If a steady current of $2.50\text{ A}$ passed for $5790\text{ s}$ deposits $1.35\text{ g}$ of metal $M$ at the cathode, what is the relative atomic mass of metal $M$ ? [ $1\text{ Faraday} = 96500\text{ C/mol}$ ]

Cevap: 27 g/mol

Cevap

The relative atomic mass of metal

M

27\text{ g/mol}

Using Faraday's first law (

Q = I \times t

), the total charge passed is

2.50 \times 5790 = 14475\text{ C}

. Dividing by Faraday's constant (

96500\text{ C/mol}

) gives

0.15\text{ mol}

of electrons. Since the metal ion is trivalent (

M^{3+}

), the half-reaction is

M^{3+} + 3e^- \rightarrow M

, meaning

0.15\text{ mol}

of electrons deposits

0.05\text{ mol}

of metal. Dividing the mass (

1.35\text{ g}

) by the amount in moles (

0.05\text{ mol}

) yields a relative atomic mass of

27\text{ g/mol}

Adım Adım Çözüm

Calculate the total electric charge passed (

Q

Q = 2.50\text{ A} \times 5790\text{ s} = 14475\text{ C}

Electric charge is the product of current in amperes and time in seconds.

Calculate the moles of electrons transferred.

n_e = \frac{14475\text{ C}}{96500\text{ C/mol}} = 0.15\text{ mol } e^-

One mole of electrons carries a charge of 1 Faraday (

96500\text{ C}

Determine the amount (in moles) of metal

M

deposited.

From

M^{3+} + 3e^- \rightarrow M

3\text{ moles of } e^-

deposit

1\text{ mole of } M

. Thus,

n_M = \frac{0.15\text{ mol}}{3} = 0.05\text{ mol}

The reduction of a trivalent ion requires 3 moles of electrons per mole of metal deposited.

Calculate the relative atomic mass (

M_r

M_r = \frac{1.35\text{ g}}{0.05\text{ mol}} = 27\text{ g/mol}

Molar mass is calculated by dividing the mass of the substance by the amount in moles.

Anahtar Kavram

Faraday's Laws of Electrolysis and Quantitative Calculations

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