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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
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B
0.3125 V
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C
1.114 V
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D
1.114 V
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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

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