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EMF

The standard ...

Question

The standard reduction potential for $F e^{2 +} | F e$ and $S n^{2 +} | S n$ electrodes are $- 0.44 V$ and $- 0.14 V$ respectively. For the cell reaction, $F e^{2 +} + S n \to F e + S n^{2 +}$ , the standard $E . M . F .$ is:

+ 0.30 V

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- 0.58 V

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+ 0.58 V

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- 0.30 V

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Solution

The correct option is D $- 0.30 V$
$F e^{2 +} + 2 e^{-} \to F e . . . . . . . . . . (I)$ $E_{F e^{2 +} | F e} = - 0.44 V$

$S n^{2 +} + 2 e^{-} \to S n . . . . . . . . (I I)$ $E_{S n^{2 +} | S n} = - 0.14 V$

Net reaction-
$F e^{2 +} + S n \to F e + S n^{2 +}$

$Δ G_{n e t}^{0} = Δ G_{I} - Δ G_{I} I$

which gives :
$- n F E_{n e t}^{0} = - n_{I} F E_{I}^{0} - (- n_{I I} F E_{I I}^{0})$
$n_{I} = 2$
$n_{I I} = 2$
For the resultant equation, $n = 2$

$E_{n e t}^{0} = \frac{2 E_{I}^{0} + (- 2 E_{I I}^{0})}{2} = - 0.44 + 0.14 = - 0.30 V$
Option D is correct.

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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 +}

S n^{2 +} | S n

Z n^{2 +} | Z n

E^{\circ}

Z n | Z n^{2 +} | | S n^{2 +} | S n

F e^{3 +} + 3 e^{-} \to F e

E^{\circ} = - 0.036 V

F e^{2 +} + 2 e^{-} \to F e

E^{\circ} = - 0.44 V

F e^{3 +} + e^{-} \to F e^{2 +}

E^{0}

E^{0_{F e 3 + / F e 2 +}} = + 0.77 V

E_{S n^{2 +} / S n}^{0} = - 0.14 V

S n (s)

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

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)

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