CameraIcon

SearchIcon

MyQuestionIcon

1

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

Electrode Potential

In the follow...

Question

In the following redox reaction:
$5 {F e}^{2 +} + M n O_{4}^{-} + 8 H^{+} ⇌ {M n}^{2 +} + 5 {F e}^{3 +} + 4 H_{2} O$
Given: ${F e}^{3 +} + e^{-} ⟶ {F e}^{2 +}, {E_{1}}^{o}$
$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} ⟶ {M n}^{2 +} + 4 H_{2} O, {E_{2}}^{o}$
Potential at the equivalence point is :

A

E = ({E_{1}}^{o} + {E_{2}}^{o}) - 0.08 p H

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

B

E = \frac{{E_{1}}^{o} + {5 E}_{2}^{o}}{6} - 0.08 p H

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

C

E = ({E_{1}}^{o} - {E_{2}}^{o}) - 0.08 p H

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

D

E = \frac{{E_{1}}^{o} - {5 E}_{2}^{o}}{6} + 0.08 p H

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

Open in App

Solution

The correct option is C $E = \frac{{E_{1}}^{o} + {5 E}_{2}^{o}}{6} - 0.08 p H$
At the equivalent point, all species are in unit concentration and
$Δ G^{o} (n e t) = Δ G_{1}^{o} + Δ G_{2}^{o}$
$- 6 F E_{3}^{o} = - F E_{1}^{o} - 5 F E_{2}^{o}$
$∴ E_{3}^{o} = \frac{E_{1}^{o} + 5 E_{2}^{o}}{6}$
Except $H^{+}$ , all other species $= 1 M$
So, according to Nernst equation,
$∴ E = E_{3}^{o} - \frac{0.0591}{6} log {(\frac{1}{H^{+}})}^{8}$
$= \frac{E_{1}^{o} + 5 E_{2}^{o}}{6} - \frac{0.0591}{6} \times 8 log (\frac{1}{H^{+}})$
$= \frac{E_{1}^{o} + 5 E_{2}^{o}}{6} - 0.08 p H$

Suggest Corrections

thumbs-up

0

Similar questions

O x_{1} + a e^{-} ⇌ {R e d}_{1}, {E_{1}}^{o}

O x_{2} + b e^{-} ⇌ {R e d}_{2}, {E_{2}}^{o}

The potential at the equivalence point is :

Q. Calculate the value of

E_{M n O_{4}^{-} | M n O_{2}}^{\circ}

M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O

E^{\circ} = 1.51 V

M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O

E^{\circ} = 1.23 V

Q. Assertion :If two half reaction with electrode potential

E_{1}^{o}

E_{2}^{o}

gives a third half reaction, then

Δ G_{3}^{o} = Δ G_{1}^{o} + Δ G_{2}^{o}

E_{3}^{o} = E_{1}^{o} + E_{2}^{o}

K M n O_{4}

reacts with ferrous sulphate according to the following equation:

M n O_{4}^{-} + 5 F e^{2 +} + 8 H^{+} \to M n^{2 +} + 5 F e^{3 +} + 4 H_{2} O

10 m L

0.1 M K M n O_{4}

is equivalent to

[C P M T 1999]

Q. The standard redox potentials E

^{o}

of the following systems are:

	System	E $^{o}$ (volts)
(i)	$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$	1.51
(ii)	$S n^{2 +} \to S n^{4 +} + 2 e^{-}$	0.15
(iii)	$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$	1.33
(iv)	$C e^{3 +} \to C e^{4 +} + e^{-}$	1.61

The oxidising power of the various species are related as:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Electrode Potential and emf

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Electrode Potential

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon