Explanation Reaction at cathode: M n O 4 − ( a q ) + 8 H + ( a q ) + 5 e − → M n 2 + ( a q ) + 4 H 2 O ( l ) E 0 = ( + ) 1.51 V Reaction at anode: F e 3 + ( a q ) + e − → F e 2 + ( a q ) E 0 = ( + ) 0.77 V Net reaction: Cathode reaction − (Anode reaction × 5) M n O 4 − ( a q ) + 8 H + ( a q ) + 5 F e 2 + ( a q ) ⇌ M n 2 + ( a q ) + 5 F e 3 + ( a q ) + 4 H 2 O ( l ) 1.50M H 2 S O 4 is equivalent to ( 1.5 × 2 ) = 3 M H + ions. Before addition any permanganate: The conical flask contains only ferrous ion. The potential of the system will be given by that for a ferrous-ferric couple as per the Nernst equation: E = E 0 + 0.059 1 log [ F e 2 + ] [ F e 3 + ] After addition of few drops K M n O 4 solution the reaction sets in as: M n O 4 − ( a q ) + 8 H + ( a q ) + 5 F e 2 + ( a q ) ⇌ M n 2 + ( a q ) + 5 F e 3 + ( a q ) + 4 H 2 O ( l ) Now it makes a confusion on these two system F e 3 + ( a q ) + e − → F e 2 + ( a q ) M n O 4 − ( a q ) + 8 H + ( a q ) + 5 e − → M n 2 + ( a q ) + 4 H 2 O ( l ) Which one will determine the net potential of the solution. Actually as soon as the permanganate solution has been added from the burette, it has oxidised some of the ferrous ion present and the system rapidly reached a position of equilibrium. Now, equilibrium between F e 2 + , F e 3 + , M n 2 + and M n O 4 − means that two half-cell potential for the two redox systems present are identical

7th Edition

ISBN: 9780321940872

Author: John E. McMurry, Robert C. Fay, Jill Kirsten Robinson

Publisher: PEARSON

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Electrolysis

Electrolysis is defined as the process of breaking down ionic compounds into their constituent elements by passing a direct electric current through the compound in a fluid state. The cathode reduces cations, while the anode oxidizes anions. An electroly…

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Chapter 18, Problem 18.167MP

Interpretation Introduction

To determine:

The cell potential after addition of
a) 5.0mL, b) 10.0mL, c) 19.0mL and d) 21.0mL of KMnO4 solution.

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Chapter 18, Problem 18.166MP

Chapter 19, Problem 19.1P

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Ch. 18 - Prob. 18.1P Ch. 18 - Prob. 18.2A Ch. 18 - Prob. 18.3P Ch. 18 - APPLY 18.4 Balance the following net ionic...Ch. 18 - Prob. 18.5P Ch. 18 - Prob. 18.6A Ch. 18 - PRACTICE 18.7 Write a balanced equation for the...Ch. 18 - Prob. 18.8A Ch. 18 - Prob. 18.9P Ch. 18 - Prob. 18.10A

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