   Chapter 18, Problem 82E

Chapter
Section
Textbook Problem

# An electrochemical cell consists of a nickel metal electrode immersed in a solution with [Ni2+] = 1.0 M separated by a porous disk from an aluminum metal electrode immersed in a solution with [Al3+] = 1.0 M. Sodium hydroxide is added to the aluminum compartment, causing Al(OH)3(s) to precipitate. After precipitation of Al(OH)3 has ceased, the concentration of OH− is 1.0 × 10−4 M and the measured cell potential is 1.82 V. Calculate the Ksp value for Al(OH)3. Al(OH) 3 ( s ) ⇌ Al 3+ ( a q ) + 3OH − ( a q )         K sp = ?

Interpretation Introduction

Interpretation:

The information regarding addition of sodium hydroxide to aluminum compartment that is present along with nickel metal electrode in other compartment is given. The value of Ksp for Al(OH)3 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 Ksp for Al(OH)3.

Explanation

Given

The concentration of Ni2+ is 1.0M.

The concentration of Al3+ is 1.0M.

The concentration of OH is 1.0×104M.

The measured cell potential is 1.82V.

The reaction taking place on cathode,

Ni2++2eNiE°red=0.23V

The reaction taking place at anode,

AlAl3++3eE°ox=1.66V

Multiply reduction half-reaction with a coefficient of 3 and oxidation half-reaction with a coefficient of 2 and then add both the reduction half and oxidation half-reaction.

3Ni2++6e3Ni2Al2Al3++6e

The overall cell reaction is,

2Al+3Ni2+3Ni+2Al3+

The overall cell potential is calculated as,

E°cell=E°ox+E°red=1.66V+(0.23V)=1.43V_

The reaction involves the transfer of 6 moles of electrons and the value of Ecell=1.82V.

The cell potential is calculated using the Nernst equation,

E=E°(0.0591n)log(Q)

Where,

• E is the cell potential.
• E° is the cell potential at the standard conditions.
• Q is the activity of the species in the cell.
• n is the number of electrons involved in the reaction.

Substitute the values of Eo, n and Q in above equation as,

Ecell=E°cell(0.05916)log([Al3+]2[Ni2+]3)1

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