9th Edition
Steven S. Zumdahl
ISBN: 9781133611097




9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

The table below lists the cell potentials for the 10 possible galvanic cells assembled from the metals A. B. C. D. and E. and their respective 1.00 M 2+ ions in solution. Using the data in the table, establish a standard reduction potential table similar to Table 17-1 in the text. Assign a reduction potential of 0.00 V to the half-reaction that falls in the middle of the series. You should get two different tables. Explain why, and discuss what you could do to determine which table is correct.

  A(s)in A2+(aq) B(s)in B2+(aq) C(s)in V2+(aq) D(s)in D2+(aq)
E(s)in E2+(aq) 0.28V 0.81V 0.13V 1.00V
D(s)in D2+(aq) 0.72V 0.19V 1.13V
C(s)in V2+(aq) 0.41V 0.94V
B(s)in B2+(aq) 0.53V

Interpretation Introduction

Interpretation: The standard reduction potential table is established, a reduction potential of 0.00 V to the half-cell reaction that falls in the middle of the series is to be assigned and a correct table is determined.

Concept introduction: Cell potential is defined as the measure of energy per unit charge that is the potential difference in an electrochemical cell containing two half cells available from the redox reaction to carry out the reaction The limited form of galvanic cell is concentration cell consists of two equivalent half-cells of the same composition.

To determine: The standard reduction potential table, a reduction potential of 0.00 V to the half-cell reaction that falls in the middle of the series and a correct table of reduction potential.




The concentration of A, B, C, D and E metals in their ionic solution is 1.00 M .

The potentials for the 10 possible galvanic cells is,

A(s)inB(s)inC(s)inD(s)in A2+(aq)B2+(aq)C2+(aq)D2+(aq)E(s)in E2+(aq)0.28 V0.81 V0.13 V1.00 VD(s)in D2+(aq)0.72 V0.19 V1.13 VC(s)in C2+(aq)0.41 V0.94 VB(s)in B2+(aq)0.53 V

The starting reduction potential is taken as 0.00 V . In this case, it is assumed,

B2++2eB     0.00 V .

From the data, when BB2+ and EE2+ are taken together as a cell, then Eο=0.81 V . This means, E2++2eE must have a potential of either +0.81 V or0.81 V as E is involved either in reduction–half or oxidation-half reactions. For instance, it is assumed that E has a potential of 0.81 V . If it is assumed that a potential of 0.00 V is for B and 0.81 V for E, then the following table of potentials is formed.

Table (1)

B2++2eB      0.00 V

E2++2eE    0.81 V

D2++2eD      0.19 V

C2++2eC    0.94 V

A2++2eCA   0

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