   # An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. If the copper electrode is placed in a solution of 0.10 M NaOH that is saturated with Cu(OH) 2 , what is the cell potential at 25°C? [For Cu(OH) 2 , K sp = 1.6 × 10 −19 .] ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 17, Problem 81E
Textbook Problem
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## An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. If the copper electrode is placed in a solution of 0.10 M NaOH that is saturated with Cu(OH)2, what is the cell potential at 25°C? [For Cu(OH)2, Ksp = 1.6 × 10−19.]

Interpretation Introduction

Interpretation:

An electrochemical consisting of standard hydrogen electrode and copper metal electrode placed in a solution of 0.10M NaOH that is saturated with Cu(OH)2 is given. The value of cell potential is to be calculated.

Concept introduction:

Solubility product is applied only for those ionic compounds that are sparingly soluble. The product of solubility of ions is called solubility product and solubility is present in moles per liter.

The value of Ecell is calculated using Nernst formula,

E=E°(RTnF)ln(Q)

At room temperature the above equation is specifies as,

E=E°(0.0591n)log(Q)

To determine: The value of cell potential at 25°C for the given reaction.

### Explanation of Solution

Given

The concentration of NaOH is 0.10M.

The value of Ksp is 1.6×1019.

The dissociation of Cu(OH)2 in NaOH occurs as,

Cu(OH)2Cu+2OH

The concentration of Cu2+ ions is calculated using the formula,

Ksp=[Cu2+][OH]21.6×1019=[Cu2+][OH]2[Cu2+]=1.6×1019[OH]2

Substitute the value of OH- in the above equation.

[Cu2+]=1.6×1019[0.1M]2=1.6×1017

The reaction taking place on cathode,

Cu2++2eCuE°red=0.34V

The reaction taking place at anode,

H22H++2eE°ox=0.00V

Add both the reduction half and oxidation half-reaction

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