In the electrochemical cell :
Zn|ZnSO4(0.01 M)||CuSO4(1.0 M)|Cu,the emf of this Daniell cell is E1.When the concentration of ZnSO4 is changed to 1.0 M and that of CuSo4 changed to 0.01 M,the emf changes to E2.From the following,which one is the relationship between E1 and E2?(Given,RT/F=0.059)
a) E1 < E2
b) E1 > E2
c) E2 = 0 =\ E1
d) E1 = E2
Answers
Answered by
1
Answer:
nb
Explanation:
ANSWER
E
cell
=E
cell
0
−
n
0.059
log
[Cu
2+
]
[Zn
2+
]
E
1
=1.1−
2
0.059
log
1.0
0.01
E
1
=1.1−
2
0.059
×(−2)
E
1
=1.1−(−0.0592)
E
1
=1.16V
Again,
E
2
=1.1−
2
0.059
log
0.01
1.0
E
1
=1.1−
2
0.059
×(2)
E
1
=1.1−(0.059)
E
2
=1.041 V
Hence, E
1
>E
2
Answered by
17
Answer:
Answer is option(b)
Explanation:
E cell = E° cell - 0.059 ÷ n × log × [Zn²+] ÷ [Cu²+]
E1 = E° - 0.059 ÷ 2 × log × 0.01 ÷ 1
E1 = E° - 0.059 ÷ 2 × (-2) = E° + 0.059
E2 = E° - 0.059 ÷ 2 × log × 1 ÷ 0.01 = E° - 0.059
Hence, E1 > E2
Hope it will help you.Thanks
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