In the cell represented by Pbs-Pb2-1M-Ag-1M-Ags- the reducing agent is -

The correct option is A Pb Pb is the reducing agent. It itself is oxidised to Pb^2+ . Pb → Pb^2+ + 2e^ Note: Oxidation occurs at the elect ...

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Electrode Potential

In the cell r...

Question

In the cell represented by $P b_{(s)} | P b_{(1 M)}^{2 +} | | A g_{(1 M)}^{+} | A g_{(s)}$ , the reducing agent is :

P b

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P b^{2 +}

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A g

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A g^{+}

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Solution

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P b^{2 +} + 2 e \to P b

(E^{o} = - 0.13 v)

A g^{+} + e \to A g

(E^{o} = + 0.80 v)

A g^{+}

P b

{P b}^{2 +}

{E^{o}}_{A g} = 0.799 V

{E^{o}}_{P b} = - 0.126 V

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) $P b$

d) None of these

(E^{\circ})

A g / A g^{+} (1 M) ∥ C u^{2 +} (1 M) | C u; E^{\circ} = - 0.46 v o l t

Z n / Z n^{2 +} (1 M) ∥ C u^{2 +} (1 M) | C u; E^{\circ} = + 1.10 v o l t

Z n | Z n^{2 +} (1 M) ∥ A g^{+} (1 M) | A g

C u | C u^{2 +} (1 M) | | A g^{+} (1 M) | A g, E^{0} = 0.46 V

Z n | Z n^{2 +} (1 M) | | C u^{2 +} (1 M) | C u, E^{0} = 1.10 V

Z n | Z n^{2 +} (1 M) | | A g^{+} (1 M) | A g

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