The standard electrode potentials(reduction) of $F e^{3 +}$ , $F e^{2 +} / P t$ and $S n^{4 +}$ , $S n^{2 +} / P t$ are $+ 0.77 V$ and $0.15 V$ respectively at $25^{o} C$ . The standard EMF of the following reaction is: $S n^{4 +} + 2 F e^{2 +} \to S n^{2 +} + 2 F e^{3 +}$

- 0.62 V

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- 0.92 V

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+ 0.31 V

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+ 0.85 V

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Solution

The correct option is A $- 0.62 V$
$E_{c e l l}^{0} = S . R . P (C a t h o d e) - S . R . P . (A n o d e)$
$E_{c e l l}^{0} = 0.15 - 0.77 = - 0.62 V$

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Similar questions

E_{F e^{3 +} / F e^{2 +}}^{\circ} = + 0.77 V; E_{S n^{2 +} / S n}^{\circ} = - 0.14 V

S n_{(s)} + 2 F e_{(a q)}^{3 +} \to 2 F e_{(a q)}^{2 +} + S n_{(a q)}^{2 +}

298 K

2 F e^{3 +} + S n^{2 +} \to 2 F e^{2 +} + S n^{4 +}

E_{S n^{4 +} / S n^{2 +}}^{\circ} = 0.15 v o l t; E_{F e^{3 +} / F e^{2 +}}^{\circ} = 0.771 v o l t

Consider the following $E^{0}$ values:

$E_{F e^{3 +} / F e^{2 +}}^{0} = + 0.77 V$ ; $E_{S n^{2 +} / S n}^{0} = - 0.14 V$

Under standard conditions the EMF for the reaction

$S n_{(s)} + 2 F e {^{3 +}}_{(a q)} \to 2 F e {^{2 +}}_{(a q)} + S n {^{2 +}}_{(a q)}$ is :

Pt(s) | Sn²⁺ (aq, 1 M), Sn⁴⁺ (aq, 1 M) || Fe²⁺ (aq, 1 M), Fe³⁺ (aq, 1 M) | Pt(s)
Anode reaction is:

P b^{4 +} + S n^{2 +} \to S n^{4 +} + P b^{2 +}

S n^{4 +} + P b^{2 +} \to P b^{4 +} + S n^{2 +}

E_{P b^{4 +} / P b^{2 +}}^{o} = 1.69 V, E_{S n^{4 +} / S n^{2 +}}^{o} = 0.15 V

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