General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 19, Problem 19.63QP
Interpretation Introduction

Interpretation:

The emf Eo value after 1 hour has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: Eocell is composed of a contribution from the anode and a contribution from the cathode is given by,

Ecello=EcathodeoEanodeo

Where both Ecathodeo and Eanodeo are the standard reduction potentials of the electrodes.

Effect of concentration on cell Emf:

The mathematical relationship between the emf of galvanic cell and the concentration of reactants and products in a redox reaction under nonstandard-state conditions is,

ΔG=ΔG0+RTlnQwhere, ΔG0isstandardGibb'sfreeenergy            Qisreactionquotient.

As known ΔG0=-nFEcell0 and ΔG=-nFEcell, above expression can be written as,

ΔG=ΔG0+RTlnQ-nFEcell=-nFEcell0+RTlnQ

Dividing by –nF, the above equation becomes,

-nFEcellnF=-nFEcell0nF+RTlnQnFEcell=Ecell0RTnFlnQNernst equation

Nernst equation: The Nernst equation is used to calculate the cell voltage under nonstandard-state conditions.

Expert Solution & Answer
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Explanation of Solution

Calculation of standard emf Eocell value:

A Daniel (Galvanic cell) cell consists of Zinc electrode in ZnSO4 solution, Zinc act as Anode electrode: and a Cu electrode in CuSO4 solution, it act as Cathode electrode. At anode oxidation occurs and at cathode reduction occurs.

The cell diagram as follows,

Zn(s)|Zn2+(1M)Cu2+(1M)|Cu(s)Ecello=?V

At Anode, possible oxidation process occurs as follows,

Zn(s)Zn2+(aq)+2eE0=+0.76V

At Cathode, possible reduction process occurs as follows,

Cu2+(1M)+2e-Cu(s)E0=+0.34V

Hence, the standard emf of the cell reaction are,

Anode(reduction):Zn(s)Zn2+(aq)+2eEanodeo=+0.76VCathode(oxidation):Cu2+(1M)+2e-Cu(s)Ecathode0=+0.34V_overall:Cu2+(1M)+Zn(s)Zn2+(aq)+Cu(s)_

Ecello=Ecathodeo+Eanodeo=(+0.34)V+(+0.76)V=+1.1V

Therefore, the standard emf (Eocell) of the given cell reaction is +1.1V

As the concentration of given ions are nonstandard concentrations, the reaction spontaneity is determined by emf of a given galvanic reaction using Nernst equation.

Calculation of non-standard emf value using Nernst equation:

The reaction quotient for the given reaction is, Q=[Zn2+][Cu][Zn][Cu2+]

The concentration of pure solids and pure liquids do not appear in the expression for Q.

Hence, the reaction quotient becomes, Q=[Zn2+][Cu2+]

As to know the concentration of Zn2+ and Cu2+ after 1 hour, the concentration of Zn2+ will increase as the reaction runs and concentration of Cu2+ will decrease.

Given: Current = 20.0 A; Time, t=1hr=1×60×60=3600sec

Convert Current into coulomb:

(Current)×(time)=no.ofCoulomb

As known, 1Ampere=1CoulombSec

no.ofCoulomb=(Current)×time=(10.0 C.sec1)(3600sec)=36000C

Convert number of Coulombs into mole of electrons:

moleofe-=no.ofCoulombFaradayconstant=36000C96500C/molee-=0.373molofe-

The concentration of Zn2+ is calculated as shown below,

  • Convert mole of electrons into number of moles of Silver:

Zn(s)Zn2+(aq)+2eE0=+0.76V

1 mole of Zn2+ 2 mole of e-

‘x’ of Zn2+= 0.373 moles of e-

Number of moles of Zn2+=(1moleofZn2+)(0.373molofe-)2molofe=0.186 molofZn2+

Hence, the concentration of Zn2+ after 1 hour is 0.186 molofZn2+1L=0.186M

Construct ICE table for the cell half-reaction to calculate the concentration of Zn2+ increases after 1 hour:

Zn(s)Zn2+(aq)+2einitial:1molchange:+0.186molequilibrium:1mol+0.186mol=(1.186mol)

The concentration of Zn2+ after 1 hour is (1.186mol)1L=1.186M

Construct ICE table for the cell half-reaction to calculate the concentration of Cu2+ decreases after 1 hour:

Cu2+(1M)+2e-Cu(s)initial:1molchange:0.186molequilibrium:1mol0.186M=(0.814mol)

The concentration of Cu2+ after 1 hour is (0.814mol)1L=0.814M

Substitute known constant values of R, T and F into Nernst equation becomes as follows,

Ecell=Ecell0-(0.0257V)nln[Zn2+][Cu2+]

The number of electrons transferred in the given redox reaction is TWO (n=2) and Ecell0=+1.1V

Ecell(+1.1V)-(0.0257V)2ln[1.186M0.814M]=(+1.1V)-(0.01285)ln(1.457)=(+1.1V)-(0.01285)(0.3764)=(+1.1V)-0.00483=+1.095V

Obtained Eo value after 1 hour is +1.095V

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Chapter 19 Solutions

General Chemistry

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