Question

# The two half-cell reactions of an electrochemical cell is given as$${ Ag }^{ + }+{ e }^{ - }\longrightarrow Ag;\ { E }_{ { { Ag }^{ + } }|{ Ag } }^{o }=-0.3995\ V$$$${ Fe }^{ 2+ }\longrightarrow { Fe }^{ 3+ }+{ e }^{ - };\ { E }_{ { { Fe }^{ 3+ } }|{ { Fe }^{ 2+ } } }^{o}=-0.7120\ V$$The value of cell EMF will be:

A
0.3125 V
B
0.3125 V
C
1.114 V
D
1.114 V

Solution

## The correct option is D $$0.3125\ V$$Species with more negative $${ E }^{ }$$ (standard reduction potential) generally acts as reducing agent while with less negative $${ E }^{ }$$ acts as the oxidising agent. Thus, the overall reaction is:$${ Ag }^{ + }+{ Fe }^{ 2+ }\longrightarrow { Fe }^{ 3+ }+Ag$$So,$${ E }^{ o}_{cell}={ E }_{ RH }^{ }-{ E }_{ OH }^{ }$$          $$=(-0.3995-\left( -0.7120 \right)) V$$          $$=(-0.3995+0.7120)\ V$$          $$=+0.3125\ V$$Chemistry

Suggest Corrections

0

Similar questions
View More

People also searched for
View More