(Two part question. Please answer each part or allow someone who will. Enlarge p
ID: 927186 • Letter: #
Question
(Two part question. Please answer each part or allow someone who will. Enlarge picture if script is hard to read)
Use the Nernst equation and data from Appendix D in the textbook to calculate Ecell for each of the following cells.
Part A
Mn(s)|Mn2+( 0.50 M )||Cr3+( 0.35 M ),Cr2+( 0.25 M )|Pt(s)
Express your answer using two decimal places.
SubmitMy AnswersGive Up
Part B
Mg(s)|Mg2+( 0.018 M )||[Al(OH)4]?( 0.30 M ),OH?( 0.044 M )|Al(s)
Express your answer using three decimal places.
Ecell = V Exercise 20.38 Part A Use the Nernst equation and data from Appendix D in the textbook to calculate Ecell for each of the following cells. Mn(s)Mn2 Express your answer using two decimal places (0.50 M)lICr3 (0.35 M), Cr(0.25 M)|Pt(s) Ex Boell = Submit My Answers Give Up Part B Mg(s)Mg (0.018 M)lIAI(OH) (0.30 M), OH (0.044 M)IAl(s) Express your answer using three decimal places Ecell = Submit My Answers Give UpExplanation / Answer
PART A:
Given : Mn(s)|Mn2+( 0.50 M )||Cr3+( 0.35 M ),Cr2+( 0.25 M )|Pt(s)
Concentration of Mn2+ = 0.50 M
Concentration of Cr3+ = 0.35 M
Concentration of Cr2+ = 0.25 M
To Find: Ecell of the reaction=?
Solution:
Firstly, let us write the cell reaction in the form of the chemical reaction:
Mn (s) + 2Cr3+ (aq) --> Mn2+ (aq) + 2Cr2+ (aq)
Let us divide the given cell reaction in two halves:
Mn(s)|Mn2+( 0.50 M ): This is the oxidation half cell.
Mn2+ (aq) + 2 e- --> Mn (s)
For which the standard reduction potential is Eo(Mn2+/Mn) = -1.18 V
Cr3+( 0.35 M ),Cr2+( 0.25 M ): This is the reduction half cell.
Cr3+(aq) + 1 e- --> Cr2+ (aq)
For which the standard reduction potential is Eo(Cr3+/Cr2+) = -0.41 V
Now, we write the Nernst equation for the following reaction:
Ecell = Eocell – 0.0591/n log [oxidised] / [reduced]
At 25oC.
Thus,
Ecell = Eocell – 0.0591/n log [Mn2+] / [Cr3+]
Where Eocell = ECr3+/Cr2+ - EMn2+/Mn
And n = number of electrons involved in the reaction.
Now, substituting the given values in the above equation:
Ecell = (-0.41 – (-1.18)) – 0.0591/2 * log [0.50] / [0.35]
= (0.77) - 0.0591/2 * log (1.4286)
= 0.77 – 0.0046
= 0.7654 ~ 0.76 V
PART B:
Given:
Mg(s)|Mg2+( 0.018 M )||[Al(OH)4]( 0.30 M ),OH( 0.044 M )|Al(s)
Concentration of Mg2+ = 0.018 M
Concentration of [Al(OH)4]- = 0.30 M
Concentration of OH- = 0.044 M
To Find: Ecell of the reaction=?
Solution:
Firstly, let us write the cell reaction in the form of the chemical reaction:
2Mg (s) + [Al(OH)4]- (aq) --> 2Mg(OH)2 (aq) + Al (s)
Let us divide the given cell reaction in two halves:
Mg(s)|Mg2+( 0.018 M ): This is the oxidation half cell.
Mg2+ (aq) + 2 e- --> Mg (s)
For which the standard reduction potential is Eo(Mg2+/Mg) = -2.37 V
[Al(OH)4]( 0.30 M ),OH( 0.044 M )|Al(s): This is the reduction half cell.
Al3+(aq) + 3 e- --> Al (s)
For which the standard reduction potential is Eo(Al3+/Al) = -1.66 V
Now, we write the Nernst equation for the following reaction:
Ecell = Eocell – 0.0591/n log [oxidised] / [reduced]
At 25oC.
Thus,
Ecell = Eocell – 0.0591/n log [Mg2+] / [[Al(OH)4]-]
Where Eocell = EAl3+/Al - EMg2+/Mg
And n = number of electrons involved in the reaction.
Now, substituting the given values in the above equation:
Ecell = (-1.66 – (-2.37)) – 0.0591/6 * log [0.018] / [0.30]
= (0.71) - 0.0591/6 * log (0.06)
= 0.71 – (-0.012)
= 0.722 V
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