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In an analytical determination of arsenic, a solution containing arsenious acid, H 3 AsO 3 , potassium iodide, and a small amount of starch is electrolyzed. The electrolysis produces free iodine from iodide ion, and the iodine immediately oxidizes the arsenious acid to hydrogen arsenate ion, HAsO 4 2− . I 2 ( a q ) + H 3 AsO 3 ( a q ) + H 2 O ( l ) → 2 I − ( a q ) + HAsO 4 2 − ( a q ) + 4 H + ( a q ) When the oxidation of arsenic is complete, the free iodine combines with the starch to give a deep blue color. If, during a particular run, it takes 65.4 s for a current of 10.5 mA to give an endpoint (indicated by the blue color), how many grams of arsenic are present in the solution?

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General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343

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Section
BuyFindarrow_forward

General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343
Chapter 19, Problem 19.115QP
Textbook Problem
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In an analytical determination of arsenic, a solution containing arsenious acid, H3AsO3, potassium iodide, and a small amount of starch is electrolyzed. The electrolysis produces free iodine from iodide ion, and the iodine immediately oxidizes the arsenious acid to hydrogen arsenate ion, HAsO42−.

I 2 ( a q ) + H 3 AsO 3 ( a q ) + H 2 O ( l ) 2 I ( a q ) + HAsO 4 2 ( a q ) + 4 H + ( a q )

When the oxidation of arsenic is complete, the free iodine combines with the starch to give a deep blue color. If, during a particular run, it takes 65.4 s for a current of 10.5 mA to give an endpoint (indicated by the blue color), how many grams of arsenic are present in the solution?

Interpretation Introduction

Interpretation:

The grams mass of arsenic present in given sample solution should be calculated.

Concept introduction:

Faraday's first law:

The amount of substance, which is liberated from electrolysis, is directly proportional to the amount of current passed through an electrode.

M α QMass α chargeM α It g

I= current in amperes t = time in seconds M= produced mass in grams.

Explanation of Solution

To calculate the grams mass of arsenic present in given sample solution

Given:

Time is 65.4 s

Amount of current passed is 10.5 mA

Reaction of arsenic acid and potassium iodide is,

I2(aq)+H2AsO3(aq)+H2O(l)2I-(aq)+HAsO42-(aq)+4H+(aq)

From the above equation, the mole of Iodine formed is equal to mole of arsenic acid in the reaction.

One mole of Arsenic acid contains one mole of Arsenic.  Therefore the mole of Iodine formed is equal to mole of Arsenic present in given sample solution.

=65.4 s(10.5×10-3A)×1C1A×s1mole-9.6485×104C×1molI22mole-=3.558×10-6mol

Amount of current passed, time for current passed, moles of iodine and electron are plugged in above equation to give a mole of Iodine formed

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Chapter 19 Solutions

General Chemistry - Standalone book (MindTap Course List)
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