   Chapter 17, Problem 6P

Chapter
Section
Textbook Problem

If 3.25 × 10−3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.78 h, what is the current in the cell during that period? Assume the gold ions carry one elementary unit of positive charge.

To determine
The current developed in electrolytic cell by deposition of gold.

Explanation

Given Info: The mass of gold deposited in negative electrode is 3.25×103kg for a period of 2.78 h, The mass of gold deposited in negative electrode is 3.25×103kg for a period of 2.78 h. Each Gold ion carry only one positive charge.

Explanation:

Formula to calculate the number of gold atoms is,

n=mNAM

• n is the number of gold atoms ,
• NA is the Avogadro number,
• M is the molecular mass of gold,

Substitute 3.25×103kg for m , 6.022×1023atomsmol-1 for NA 197gmol-1 in the above equation to find n .

n=(3.25×103kg)(6.022×1023atomsmol1)(197gmol1)(103kg1g)=9.93×1021

Thus, 3.25×103kg Gold contain 9

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