Determine the amount of silver liberated at the cathode if a current of 0.500 A flows through an AgNO_3 electrolyte for 1 hour. The atomic weight of silver is 107.9 g mol^ -1 .

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Quantrex Academy · Accepted Answer

Correct answer: A. 2.01 g

Using Faraday's first law of electrolysis, the mass of the substance deposited at the electrode is given by:

m = ZIt

where

Z

is the electrochemical equivalent,

I

is the current, and

t

is the time.

Z = \dfrac{M}{nF}

For the deposition of silver from

\text{AgNO}_3

, the reaction is

\text{Ag}^+ + e^- \rightarrow \text{Ag}

. Thus,

n = 1

.

Given:

M = 107.9 \text{ g mol}^{-1}

I = 0.500 \text{ A}

t = 1 \text{ hour} = 3600 \text{ s}

F = 96500 \text{ C mol}^{-1}

Substituting the values:

m = \dfrac{107.9 \times 0.500 \times 3600}{1 \times 96500}

m = \dfrac{107.9 \times 1800}{96500}

m = \dfrac{1942.2}{96500} \approx 2.01 \text{ g}

Determine the amount of silver liberated at the cathode if a...

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