1. Evaluate the reduction potential (in mV, at 25°C) of a nickel sheet in an aqueous solution of nickel (II) sulfate obtained by dissolving 53.2 g of salt (95% purity) in 5.00 liters of water. Evaluate the electromotive force (Ecell) of the voltaic cell obtained by coupling the above electrode with a Zn electrode (Zn sheet in a 0.250 M solution of zinc sulfate).
1. Evaluate the reduction potential (in mV, at 25°C) of a nickel sheet in an aqueous solution of nickel (II) sulfate obtained by dissolving 53.2 g of salt (95% purity) in 5.00 liters of water. Evaluate the electromotive force (Ecell) of the voltaic cell obtained by coupling the above electrode with a Zn electrode (Zn sheet in a 0.250 M solution of zinc sulfate).
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter17: Electrochemistry
Section: Chapter Questions
Problem 77AP
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I uploaded the question and answer
The part I don’t understand is why is there written mass of NiSo4/molar mass/ volume then they substituted the mass of the salt along side the other stuff
And the second thing I don’t understand is what do I need to substitute in [ni2+] and [Zn2+] and Ered to get the voltage
Transcribed Image Text:1. Evaluate the reduction potential (in mV, at 25°C) of a nickel
sheet in an aqueous solution of nickel (II) sulfate obtained by
dissolving 53.2 g of salt (95% purity) in 5.00 liters of water.
Evaluate the electromotive force (Ecell) of the voltaic cell obtained
by coupling the above electrode with a Zn electrode (Zn sheet in a
0.250 M solution of zinc sulfate).
![Anode
(Oxidation)
Zn
Exercise 1
Voltmeter
V
salt bridge
KCI
ZnSO4(aq)
Ni
NiSO4(aq)
Exercise 2
+
Cathode
(Reduction)
[NiSO4] =
Ered =
(mass of NiSO4) (53.2 g. 154.779)
*
Molar mass
Volume
Ered
Ecell = Ecathod
0.059
2
0.059
5L
Ni2+ + 2e Ni (s)
1
-log [Ni²+]
= 0.065M
= -0.29V = -2.9 * 10²mV
Zn²+ + 2e → Zn(s)
1
Ered = Ered
-log
2 [Zn²+]
Ered,Ni> Ered,Zn Zn is the anode and Ni is the cathode
= -0.78V = -7.8 * 10²mV
Eanode = -0.29V- (-(0.78V) = 0.49V 500m
sible to calculate the H](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6099d21a-e15a-47f8-adbb-0c871c33581f%2F5be04a85-a7c1-4f09-84ba-a213b3fe2f34%2F38kht9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Anode
(Oxidation)
Zn
Exercise 1
Voltmeter
V
salt bridge
KCI
ZnSO4(aq)
Ni
NiSO4(aq)
Exercise 2
+
Cathode
(Reduction)
[NiSO4] =
Ered =
(mass of NiSO4) (53.2 g. 154.779)
*
Molar mass
Volume
Ered
Ecell = Ecathod
0.059
2
0.059
5L
Ni2+ + 2e Ni (s)
1
-log [Ni²+]
= 0.065M
= -0.29V = -2.9 * 10²mV
Zn²+ + 2e → Zn(s)
1
Ered = Ered
-log
2 [Zn²+]
Ered,Ni> Ered,Zn Zn is the anode and Ni is the cathode
= -0.78V = -7.8 * 10²mV
Eanode = -0.29V- (-(0.78V) = 0.49V 500m
sible to calculate the H
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Step 1: Explaining the voltaic cell in this situation.
VIEWStep 2: Q1- subpart of calculating half-cell reduction potential of Ni
VIEWStep 3: Calculation of molar conc of NiSO4
VIEWStep 4: Calculating Ehalf-cell (reduction) of Ni+2/Ni(s) half-cell in mV
VIEWStep 5: Calculating Ehalf-cell(reduction) for the Zn+2/Zn(s) half-cell
VIEWStep 6: Calculating Ecell (reduction) of the full voltaic cell
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