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EMF

Calculate E...

Question

Calculate $E^{0}$ and $E$ for the cell $S n | {S n}^{2 +} (1 M) ∥ {P b}^{2 +} (10^{- 3} M) | P b, E^{0} ({S n}^{2 +} / S n) = - 0.14 V, E^{0} ({P b}^{2 +} / P b) = - 0.13 V$ .
Is cell representation is correct?

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Solution

The cell representation is :-
$S n ∣ ∣ {S n}^{2 +} (1 M) ∣ ∣ ∣ ∣ {P b}^{2 +} (10^{- 3} m) ∣ ∣ P b$
${E^{0}}_{c e l l} = E_{R} - E_{L}$
$= E_{{P b}^{2 +} / P b} - E_{{S n}^{2 +} / S n}$
$= 0.13 - (- 0.14) = 0.01 V$
Nearest equation :
$E_{c e l l} = E^{0} - \frac{0.059}{n} l o g ⎡ ⎢ ⎢ ⎣ \frac{({S n}^{2 +})}{({P b}^{2 +})} ⎤ ⎥ ⎥ ⎦$
$E_{c e l l} = 0.01 - \frac{0.059}{2} l o g [\frac{1}{10^{- 3}}]$
$E_{c e l l} = 0.01 - 0.0885$
$E_{c e l l} = - 0.0785 V$
eqn : $S n + {P b}^{2 +} ⟶ {S n}^{2 +} + P b$
$n = {2 e}^{-}$

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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

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

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

\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

[1] P b^{2 +} > P b^{4}, S n^{2 +} > S n^{4 +}

[2] P b^{2 +} < P b^{4}, S n^{2 +} < S n^{4 +}

[3] P b^{2 +} > P b^{4}, S n^{2 +} < S n^{4 +}

[4] P b^{2 +} < P b^{4}, S n^{2 +} > S n^{4 +}

[5] S n^{2 +} < P b^{2 +}, S n^{4 +} > P b^{4 +}

[6] S n^{2 +} < P b^{2 +}, S n^{4 +} < P b^{4 +}

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