   Chapter 18, Problem 90E

Chapter
Section
Textbook Problem

# For the following half-reaction, = −2.07 V: A1F 6 3 − ( a q ) + 3e − → Al ( s ) + 6F − ( a q ) Using data from Table 17-1, calculate the equilibrium constant at 25°C for the reaction A1 3+ ( a q ) + 6F − ( a q ) ⇌ A1F 6 3 − ( a q )           K = ?

Interpretation Introduction

Interpretation:

A half-reaction and its reduction potential is given. By using these values, the equilibrium constant K for the given equilibrium reaction is to be calculated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

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)

This relation is further used to determine the relation between ΔG° and K, ΔG° and E°cell. The value of equilibrium constant is used to predict the extent of the reaction.

To determine: The value of equilibrium constant K for the given equilibrium reaction.

Explanation

Given

The half-cell reaction is,

AlF63(aq)+3eAl(s)+6F(aq)E°=2.07V

The equilibrium reaction is,

Al3+(aq)+6F(aq)AlF63(aq)

The reaction taking place at cathode is,

Al3+(aq)+3eAl(s)E°red=1.66V

The reaction taking place at anode is,

Al(s)+6F(aq)AlF63(aq)+3eE°ox=2.07V

Add both the reduction and oxidation half-reaction,

Al(s)+6F(aq)AlF63(aq)+3eAl3++3eAl(s)

The final equation is,

Al3+(aq)+6F(aq)AlF63(aq)

The overall cell potential is calculated as,

E°cell=E°ox+E°red=2

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