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ww.saplinglearning.com/ibiscms/mod/ibis/view.php?id-4001131 rning ahoma State University, Stillwater-Chem 1515-Fall17-LAVINE Activities and Due Dates Chapter 18 HW Set 2 12/8/2017 11:55PM 047.1/100 Gradeb PrintCalculatorPeriodic Table Question 8 of 10 General Chemistryth Edition University Science Books presented by Sapling Leaming Map For the following electrochemical cell Cu(s)Cu(aq. 0.0155 M)lAg (aq, 3.50 M)Agis) write the net cell equation. Phases are optional. Do not include the concentrations Culs) + 2Ag.(aq) Cu: + 2 Ag(s) Calculate the following values at 250 using standardnotentials as needed. Number Number 1.643v Gr.- 11317.009 kJ/mol Number Number kJ/mol There is a hint availabiel View the hint bottom divider on the divider bar again to hide the hint. Close Hint Prevous ®Give Up & View SoLikn Check Answer Next Ex earch

Explanation / Answer

1)

Lets find Eo 1st

from data table:

Eo(Cu2+/Cu(s)) = 0.337 V

Eo(Ag+/Ag(s)) = 0.7996 V

the electrode with the greater Eo value will be reduced and it will be cathode

here:

cathode is (Ag+/Ag(s))

anode is (Cu2+/Cu(s))

The chemical reaction taking place is

2 Ag+(aq) + Cu(s) --> Ag(s) + 2 Cu2+(aq)

2)

Eocell = Eocathode - Eoanode

= (0.7996) - (0.337)

= 0.463 V

3)

number of electrons being transferred, n = 2

F = 96500.0 C

we have below equation to be used:

deltaGo = -n*F*Eo

= -2*96500.0*0.463

= -89359 J/mol

= -89.4 KJ/mol

4)

Number of electron being transferred in balanced reaction is 2

So, n = 2

we have below equation to be used:

E = Eo - (2.303*RT/nF) log {[Cr2+]^1/[Fe2+]^1}

Here:

2.303*R*T/F

= 2.303*8.314*298.0/96500

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Cr2+]^1/[Fe2+]^1}

E = 0.4626 - (0.0591/2) log (0.884^1/0.016^1)

E = 0.4626-(5.151*10^-2)

E = 0.411 V

5)

number of electrons being transferred, n = 2

F = 96500.0 C

we have below equation to be used:

deltaG = -n*F*E

= -2*96500.0*0.411

= -79323 J/mol

= -79.3 KJ/mol

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