A battery with 1.0 L half-cells is constructed as represented in the following l
ID: 635835 • Letter: A
Question
A battery with 1.0 L half-cells is constructed as represented in the following line notation:
Ni(s)?Ni2+(0.600 M)??Cu+(0.400 M)?Cu(s). What is the potential of the battery after it delivers a current of 2.00 A for 3.68 hours?
Use this table of standard electrode potentials:
Enter your answer to three significant figures in units of volts.
Standard Electrode Potentials E°half-cell Cu+(aq) + e? ? Cu(s) +0.52 V 2H+(aq) + 2e? ? H2(g) 0.00 V Ni2+(aq) + 2e? ? Ni(s) ?0.23 V Zn2+(aq) + 2e? ? Zn(s) ?0.76 VExplanation / Answer
Ni(s)/Ni2+(0.600 M)//Cu+(0.400 M)/Cu(s)
cell reaction : Ni(s)+Cu^2+(aq) -----> Ni^2+(aq) + Cu(s)
faradays first law
equivalent weight of Cu (E)= Mwt /no of e- exchanged
= 63.5/2 = 31.75 g/equiv
W = Zit
Z= E/F
w = (E/F)it
w = weight of HgO deposited = ? grams
F = 96500 c
w = (31.75/96500)*2*3.68*60*60
w = 8.72 g
No of mol of (Cu) = w/M = 8.72/63.5 = 0.137 mol
concentration of cu disapperared = 0.137/1 = 0.137 M
final concentration,
[Ni+2] = 0.6+0.137 = 0.737 M
[Cu+2] = 0.4-0.137 = 0.263 M
E0cell = E0cathode - E0anode
=
Ecell = E0cell - (0.0591/n)log([Ni+2]/[Cu+2]
= 0.34-0.25
= 0.09 v
Ecell = 0.09-(0.0591/2)log(0.737/0.263)
= 0.0768 v
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