Redox Reactions And Electrochemistry Question 36
Question: A galvanic cell with electrode potential of $ ‘A’=+2.23\ V $ and $ ‘B’=-1.43\ V $ . The value of $ E{{{}^\circ }_{cell}} $ is [Pb.CET 2003]
Options:
A) 3.66 V
B) 0.80 V
C) - 0.80 V
D) - 3.66 V
Show Answer
Answer:
Correct Answer: A
Solution:
$ E_{A}=2.23,V>E_{B}=1.43\ V $ So A will act as cathode in galvanic cell. Hence $ E_{cell}^{0}=E_{Cathode}-E_{Anode} $ $ =E_{A}-E_{B} $ $ =(2.23)-(-1.43) $ $ =3.66\ V $ .
BETA
