Explanation Given information: Anode half-reaction: Ti ( s ) + 2 H 2 O ( l ) → TiO 2 ( s ) + 4 H + ( a q ) + 4 e − Cathode half-reaction: 6 H + ( a q ) + 6 e → 3 H 2 ( a q ) E 0 for the anode half reaction: + 1.066 V In oxidation, loss of electron/s takes place and in reduction gain of electron/s takes place. During a redox reaction , electrons are transfer from one substance to another. Half reaction method represents the electron transfer method. In a redox reaction there are two parts, one is reduction reaction and other one is oxidation reaction. There are few steps to be followed to get the balanced equation of a reaction. The element which has an increasing in the oxidation number is considered as the oxidizing element and the element which has a decrease in the oxidation number is considered as the reduction element. Anode half-reaction: Ti ( s ) + 2 H 2 O ( l ) → TiO 2 ( s ) + 4 H + ( a q ) + 4 e − Cathode half-reaction: 6 H + ( a q ) + 6 e → 3 H 2 ( a q ) In order to add both reactions together, all H + and e should be balanced in both reactions. Then only all the electrons and other species that occur in both sides of the equation will cancel off. In order to make the electron count in same in both half-reactions they are multiplied by integers. In a balanced redox reaction, the number of released electrons in the oxidation half-reaction must be the same as the number of consumed electrons in the reduction half-cell reaction. Hence,cathode reaction should be multiplied with 2 / 3 to balance the electrons in both anode and cathode. Cathode half reaction add to the anode half-reaction to obtain overall cell reaction

8th Edition

ISBN: 9780135216972

Author: Robinson

Publisher: PEARSON CO

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Chapter 19, Problem 19.145SP

Interpretation Introduction

Interpretation:

Overall cell reaction and the cell potential for anodizing titanium should be identified.

Concept introduction:

Anodizing is the process that oxidized a metal anode to yield a porous metal oxide coating that can incorporate dye molecules to give brilliant colors.

Overall cell potential: E0cell=E0ox+E0red

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Chapter 19, Problem 19.144SP

Chapter 19, Problem 19.146SP

Chapter 19 Solutions

EBK CHEMISTRY

Ch. 19 - Balance the following net ionic equation by the...Ch. 19 - Balance the following net ionic equation by the...Ch. 19 - Prob. 19.3P Ch. 19 - Balance the following net ionic equation by the...Ch. 19 - Prob. 19.5P Ch. 19 - Prob. 19.6A Ch. 19 - PRACTICE 18.7 Write a balanced equation for the...Ch. 19 - Consider the following galvanic cell with...Ch. 19 - The standard cell potential at 25oC is 1.20 V for...Ch. 19 - The standard free-energy change is 59.8kJ for the...

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Solution Summary: The author explains the overall cell reaction and the cell potential for anodizing titanium.

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