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Electrode Potential

In simple ele...

Question

In simple electrochemical cell, which is in its standard state, the
half cell reactions with their appropriate reduction potential are?
$P b^{2 +} + 2 e \to P b$ $(E^{o} = - 0.13 v)$
$A g^{+} + e \to A g$ $(E^{o} = + 0.80 v)$

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Solution

$P b^{+ 2} + 2 e^{-} \to P b$ $(E^{0} = 0.13 v)$
$A g^{+} + e^{-} \to A g$ $(E^{0} = + 0.80 v)$
as Reduction potential of $A g^{+}$ is large that $P b^{+ 2}$
So, $A g^{+}$ will be at cathode.
$E^{0} = (E^{0} r e a)_{c} + (E_{o x a}^{0})_{A}$
$E_{c e l l}^{0} = + 0.080 v + 0.13 v = 0.93 v$
So, $E^{o} c e l l = 0.93 V$

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Similar questions

Given: $F e_{(s)} \to F e^{2 +} + 2 e^{-}; E^{o} = + 0.44 V$

$P b_{(s)} \to P b^{2 +} + 2 e^{-}; E^{o} = + 0.13 V$

$A g^{+} e^{-} \to A g; E^{o} = + 0.8 V$

$C u^{2 +} 2 e^{-} \to C u; E^{o} = + 0.34 V$

Which of the following metals will oxidise iron?

a) $A g$

b) $C u$

c) $P b$

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Z n^{+ 2} + 2 e^{-} ⇌ Z n; E^{o} = - 0.76 V

A g^{+} + e^{-} ⇌ A g; E^{o} = + 0.80 V

P b + S n^{2 +} \to P b^{2 +} + S n

E^{o}

P b \to P b^{2 +}

E^{o}

S n^{2 +} + 2 e^{-} \to S n

E^{o}

P b + S n^{2 +} \to P b^{2 +} + S n

P b \to P b^{2 +}, E^{o} = 0.13 V

S n^{2 +} + 2 e^{-} \to S n; E^{o} = - 0.14 V

E^{o}

Z n \to Z n^{2 +} + 2 e^{-}; E^{o} = - 0.76 V

F e^{2 +} + 2 e^{-} \to F e; E^{o} = + 0.41 V

F e^{2 +} + Z n \to Z n^{2 +} + F e

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