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Gibb's Energy and Nernst Equation

What ratio of...

Question

What ratio of $P b^{2 +}$ to $S n^{2 +}$ concentration is needed to reverse the following cell reaction?
$S n (s) + P b^{2} (a q .) \to S n^{2 +} (a q .) + P b (s)$
$E_{S n^{2 +} / S n}^{\circ} = - 0.136 v o l t$ and $E_{P b^{2 +} / P b}^{\circ} = - 0.126 v o l t$ .

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Solution

In the given cell reaction,
$E_{c e l l}$ = $E_{c e l l}^{0}$ $-$ $\frac{0.059}{2}$ $l o g$ $\frac{S n^{2 +}}{P b^{2 +}}$

$E_{c e l l}$ = $E_{c e l l}^{0}$ $+$ $\frac{0.059}{2}$ $l o g$ $\frac{P b^{2 +}}{S n^{2 +}}$

$E_{c e l l}$ = 0.010 + 0.0295 $l o g$ $\frac{P b^{2 +}}{S n^{2 +}}$
when $E_{c e l l}$ is negative then cell reaction reverses,
if $\frac{P b^{2 +}}{S n^{2 +}}$ is in between 0.1 to 0.9 then, cell reaction reverse.

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Similar questions

S n (s) | S n^{2 +} (a q ., 1 M) | | P b^{2 +} (a q ., 1 M) | P b (s)

\frac{[S n^{2 +}]}{[P b^{2 +}]}

E_{\frac{S n^{2 +}}{S n}}^{\circ} = - 0.14 V; E_{\frac{P b^{2 +}}{P b}}^{\circ} = - 0.13 V, \frac{2.303 R T}{F} = 0.06 V)

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

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

\frac{[P b^{2 +}]}{[S n^{2 +}]}

E_{c e l l} = 0

E_{P b / P b^{2 +}}^{\circ} = 0.13 v o l t

E_{S n^{2 +} / S n}^{\circ} = - 0.14 v o l t

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}

E^{0}

S n | S n^{2 +} (1 M) | | P b^{2 +} (10^{- 3} M) | P b, E^{0} (S n^{2 +} | S n) = - 0.14 V E^{0} (P b^{2 +} | P b) = - 0.13

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