Redox Reactions And Electrochemistry Question 766
The standard electrode potential of the half cells are given below $ Zn^{2+}+2{e^{-}}\to Zn;E=-7.62V, $ $ Fe^{2+}+2{e^{-}}\to Fe;E=-7.81V $ The emf of the cell $ Fe^{2+}+Zn\to Zn^{2+}+Fe $ is [CPMT 2003]
Options:
A) 1.54 V
B) - 1.54 V
C) - 0.19 V
D) + 0.19 V
Show Answer
Answer:
Correct Answer: C
Solution:
$ F{e^{2+}}+Zn\to Z{n^{2+}}+Fe $ $ EMF={E_{cathode}}-{E_{anode}} $ $ {H^{+}}(0.025,M)\to {H^{+}}({10^{-8}},M) $ $ EMF=-0.19,V $ .
BETA
