Using Nernst equation for the cell reaction- Pb - Sn2- Pb2- - SnCalculate the ratio -Pb2-Sn2- for which Ecell - 0-Given- E-Pb- Pb2- - 0-13 volt and E-Sn2- Sn - -0-14 volt-

Pb+Sn^2+⟶ Pb^2++Sn E_cell=E^∘_cell RTnF× ln× Q ⟹ E^∘cell 8.314× 2982× 96500× ln× Q ⟹ 0=0.01 (0.0128)× ln× Q ⟹ ln× Q=0.78125 ∴ Q=[Pb^ ...

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Using Nernst ...

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Using Nernst equation for the cell reaction,
$P b + S n^{2 +} \to P b^{2 +} + S n$
Calculate the ratio $\frac{[P b^{2 +}]}{[S n^{2 +}]}$ for which $E_{c e l l} = 0$ .
(Given: $E_{P b / P b^{2 +}}^{\circ} = 0.13 v o l t$ and $E_{S n^{2 +} / S n}^{\circ} = - 0.14 v o l t$ ).

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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

E^{o}

P b \to P b^{2 +}

E^{o}

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

E^{o}

P b / P b^{2 +} | | S n^{2 +} / S n

- 0.01 V

E_{c e l l} = 0

l o g [S n^{2 +} / [P b^{2 +}]

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)

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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