You visited us 1 times! Enjoying our articles? Unlock Full Access!

Gibb's Free Energy

Calculate the...

Question

Calculate the reduction potential for the given half cells at $25^{o} C$ .
$P t (s) | F e^{2 +} (a q, 0.1 M), F e^{3 +} (a q, 0.01 M); E_{F e^{3 +} / F e^{2 +}}^{0} = + 0.77 V$

E_{F e^{3 +} / F e^{2 +}} = 0.77 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

E_{F e^{3 +} / F e^{2 +}} = - 0.83 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

E_{F e^{3 +} / F e^{2 +}} = 0.711 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

E_{F e^{3 +} / F e^{2 +}} = 0.83 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $E_{F e^{3 +} / F e^{2 +}} = 0.711 V$
The given half cell electrode is a redox electrode.
Here, electrode potential results due to the presence of ions of the same substance in different oxidation states.
The given redox electrode reaction is,
$F e^{3 +} (a q) + e^{-} ⇌ F e^{2 +} (a q)$ (reduction)

Nernst equation for
$a A + b B ⇋ c C + d D$

$E = E^{0} - \frac{2.303 R T}{n F} l o g \frac{[C]^{c} [D]^{d}}{[A]^{a} [B] b} . . . . . (E q n . 1)$

Substituting,

$R = 8.314 J K^{- 1} m o l^{- 1}$

$F = 96500 C / m o l$

$T = 25 + 273 = 298 K$ , we get,

$E = E^{0} - \frac{0.059}{n} l o g \frac{[C]^{c} [D]^{d}}{[A]^{a} [B] b} . . . . (E q n . 2)$
where,
n is the number of electrons inovolved in the reaction.

For given half cell reaction,
$E_{F e^{3 +} / F e^{2 +}} = E_{F e^{3 +} / F e^{2 +}}^{0} - \frac{0.059}{n} l o g \frac{[F e^{2 +}]}{[F e^{3^{+}}]}$

$E_{F e^{3 +} / F e^{2 +}} = 0.77 - \frac{0.059}{1} l o g \frac{0.1}{0.01}$

$E_{F e^{3 +} / F e^{2 +}} = 0.77 - \frac{0.059}{1} l o g 10$

$E_{F e^{3 +} / F e^{2 +}} = 0.711 V$

Suggest Corrections

Similar questions

25^{o}

F e^{3 +} | F e^{2 +}

F e^{2 +}

10

F e^{3 +}

E_{F e^{3 +} | F e^{2 +}}^{0} = 0.77 V

25^{o} C .

F e^{3 +} (a q, 0.1 M), F e^{2 +} (a q, 1 M) | P t (s); E_{F e^{3 +} / F e^{2 +}}^{0} = 0.77 V

A g (s) | A g C l (s) ∣ C l^{-} (a q, 0.001 M); E_{A g C l / A g}^{0} = 0.22 V

E^{0}

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

E^{0}

F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = 0.77 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V

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)

Z n (s) | Z n^{2 +} (a q) | | M^{x +} (a q) | M (s)

$M^{x +} (a q / M (s))$	$A u^{3 +} (a q) / A u (s)$	$A g^{+} / A g (s)$	$F e^{3 +} (a q) / F e^{2 +} (a q)$	$F e^{2 +} (a q) / F e (s)$
$E_{M^{x +} / M^{/ (V)}}^{o}$	$1.40$	$0.80$	$0.77$	$- 0.44$

E_{Z n^{2 +} / Z n}^{o} = - 0.76 V

E_{c e l l}^{o}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program