Question

Suppose the current flowing from a battery is used to electroplate an object with silver. Calculate the mass of silver that would be deposited by a battery that delivers 1.65 A·hr of charge.

Answer 1

Answer:

m = 0.00659 kg = 6.59 g

Explanation:

From Faraday's Law of Electrolysis, we know that:

m = ZQ

where,

m = mass of silver deposited = ?

Q = charge supplied = (1.65 A-hr)(3600 s/1 hr) = 5940 C

Z = electrochemical equivalent of silver = 1.18 x 10⁻⁶ kg/C

Therefore,

m = (1.11 x 10⁻⁶ kg/C)(5940 C)

m = 0.00659 kg = 6.59 g

Answer 2

The mass of silver that would be deposited by a battery is 6.65 grams

The precipitation of Ag requires the removal of one electron. The reduction process for silver electrode at the cathode is as follows:

$\mathbf{Ag^+ + e^- \to Ag(s)}$

• The current flowing in the battery = 1.65 A = 1.65 C/s

• The time at which the current is flowing = 1 hr = 3600sec

∴

• The charge Q = Current (I) × time (t)
• Charge Q = 1.65 C/s × 3600 s
• Charge (Q) = 5940 C

In one mole of an electron, the charge carried = 96500 C

Recall that:

The atomic mass of silver (Ag) = 108 g

∴

The mass of silver that would be deposited in a 5940 C can be computed as:

$\mathbf{=5940\ C \times \dfrac{108 \ g }{96500 \ C}}$

= 6.65 grams

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