Explanation Given information: Pt|V(OH) 4 + (2 .67 × 10 − 4 M ) ; VO 2 + (3 .42 × 10 − 2 M), H + ( 4.81 × 10 − 3 M)|| Cu 2 + (2 .15 × 10 − 2 ) | Cu Internal resistance: 4.71 ohm. The expression for the potential of the electrolytic cell is: E = E C − E A − V ...... (I) Here, the electrode potential for cathode is E C , electrode potential for anode is E A , and the potential difference is V . The expression for the potential difference is: V = I R Here, the current flowing through the cell is I and the resistance of the cell is R . Substitute I R for V in Equation (I) E = E C − E A − I R ...... (II) The half cell reaction for cathode is: Cu 2 + + 2 e − ⇌ Cu ( s ) The value of standard electrode potential for the reaction is 0.337 V E C 0 = 0.337 V Here, the standard electrode potential is E C 0 . The expression for the electrode potential for cathode is: E C = E C 0 − 0.0592 n log ( 1 [ Cu 2 + ] ) ...... (III) Here, the electrons involved in the reaction is n , the concentrations of Cu 2 + is [ Cu 2 + ] . Substitute 0.337 V for E C 0 in Equation (III). E C = ( 0.337 V ) − 0.0592 n log ( 1 [ Cu 2 + ] ) ...... (IV) The half cell reaction for anode is: V ( OH ) 4 + + 2 H + ⇌ VO 2 + + 3 H 2 O + e − The value of standard electrode potential for the reaction is 1 V E A 0 = 1 V Here, the standard electrode potential is E A 0 . The expression for the electrode potential for anode is: E A = E A 0 − 0.0592 n log ( [ VO 2 + ] [ H + ] 2 [ V ( OH ) 4 + ] ) ...... (V) Here, the electrons involved in the reaction is n , the concentrations of VO 2 + is [ VO 2 + ] , the concentrations of H + is [ H + ] , and concentrations of V ( OH ) 4 + is [ V ( OH ) 4 + ]

7th Edition

ISBN: 9781305577213

Author: Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher: Cengage Learning

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Potentiometric Titrations

Potentiometric titration is one of the important titration methods and is mainly carried out to characterize acid solutions. It follows a method similar to that of the normal titration technique, but the main difference is that there is no indicator used…

Question

Chapter 22, Problem 22.16QAP

Interpretation Introduction

Interpretation:

The initial potential of the given cell has to be found.

Concept introduction:

In an electrolytic cell, for the generation of an electric current, an enough amount of potential difference is required. This potential difference overcomes the resistance offered by the ions and electrons present in the cell. According to the Ohm’s law, the potential difference can be expressed as the product of current flowing in the circuit and the resistance of the cell.

Expert Solution & Answer

Chapter 22 Solutions

Principles of Instrumental Analysis

Ch. 22 - Calculate the electrode potentials of the...Ch. 22 - Prob. 22.2QAP Ch. 22 - For each of the following half-cells, compare...Ch. 22 - For each of the following half-cells, compare...Ch. 22 - Prob. 22.5QAP Ch. 22 - Calculate the electrode potentials for the...Ch. 22 - Calculate the theoretical potential of each of the...Ch. 22 - Calculate the theoretical potential of each of the...Ch. 22 - Prob. 22.9QAP Ch. 22 - Prob. 22.10QAP

Ch. 22 - Calculate the standard potential for the...Ch. 22 - Calculate the standard potential for the...Ch. 22 - From the standard potentials Tl++eTl(s)E0=0.336V...Ch. 22 - From the standard potentials...Ch. 22 - Prob. 22.15QAP Ch. 22 - Prob. 22.16QAP Ch. 22 - Prob. 22.17QAP Ch. 22 - Prob. 22.18QAP

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