Consider the following $E^{0}$ values: $E^{0_{F e 3 + / F e 2 +}} = + 0.77 V$
$E_{S n^{2 +} / S n}^{0} = - 0.14 V$
Under standard conditions the potential 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:

1.68

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1.40

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0.91

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0.63

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Solution

The correct option is A $0.91$ V
For the redox system:
$S n (s) + 2 F e^{3 +} (a q) \to 2 F e^{2 +} (a q) + S n^{2 +} (a q)$
Reaction at cathode:
$2 F e^{3 +} (a q) + 2 e^{-} \to 2 F e^{2 +} (a q)$
Reaction at anode:
$S n^{2 +} (a q) + 2 e^{-} \to S n (s)$
$E_{F e^{3 +} / F e^{2 +}} = + 0.77 V$
$E_{S n^{2 +} / S n}^{0} = - 0.14 V$
Under standard conditions the potential for the reaction
$E_{c e l l} = E_{c a t h o d e} - E_{a n o d e}$
$E_{c a t h o d e} = + 0.77 V$
$E_{a n o d e} = - 0.14 V$
$E_{c e l l} = + 0.77 - (- 0.14)$
$= + 0.91 V$
C is correct

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Similar questions

E^{0}

E_{F e^{3 +} / F e^{2 +}}^{0}

E_{S n^{2 +} / S n}^{0}

S n (s) + 2 F e^{3 +} (a q) \to 2 F e^{2 +} (a q) + S n^{2 +} (a q)

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 :

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 +}

E^{o}

E_{{F e}^{3 +} / {F e}^{2 +}} = + 0.76 V

E_{{F e}^{2 +} / {F e}} = - 0.44 V

E_{{S n}^{2 +} / {S n}} = - 0.14 V

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

E_{F e^{3 +} / F e}^{0} = 0.036 V, E_{F e^{2 +} / F e}^{0} = - 0.439 V

F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q)

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