CameraIcon

SearchIcon

MyQuestionIcon

222

You visited us 222 times! Enjoying our articles? Unlock Full Access!

Electrode Potential

At equimolar ...

Question

At equimolar concentrations of $F e^{2 +}$ and $F e^{3 +}$ , what must [ $A g^{+}$ ] be so that the voltage of the galvanic cell made from the ( $A g^{+} | A g)$ and ( $F e^{3 +} | F e^{2 +})$ electrodes equals zero ?
$F e^{2 +} + A g^{+} ⇌ F e^{3 +} + A g$
$E_{A g^{o} | A g}^{o} = 0.7991; E_{F e^{3 +} | F e^{2 -}}^{o} = 0.771$

A

0.34

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

B

0.44

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

C

0.47

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

D

0.61

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

Open in App

Solution

The correct option is B $0.34$
${F e}^{2 +} + {A g}^{+} ⇌ {F e}^{3 +} + A g$
Given $[{F e}^{2 +}] = [{F e}^{3 +}], E_{c e l l} = 0$
$E_{c e l l} = {E^{0}}_{c e l l} - \frac{0.059}{n} l o g \frac{[{F e}^{3 +}]}{[{F e}^{2 +}] [{A g}^{+}]}$
$↓$
Neunsl equation.
${E^{0}}_{c e l l} = E_{{A g}^{+} / A g} - E_{{F e}^{3 +} / {F e}^{2 +}}$
$= 0.7991 - 0.771 = 0.0281$
$0 = 0.0281 - \frac{0.059}{1} l o g [\frac{1}{({A g}^{+})}]$
$\frac{0.0281}{0.059} = l o g [\frac{1}{({A g}^{+})}]$
$0.4754 = l o g (\frac{1}{{A g}^{+}})$
$2.988 = 1 / [{A g}^{+}]$
$[{A g}^{+}] = 0.34 M$

Suggest Corrections

thumbs-up

0

Similar questions

Q. At equimolar concentrations of

F e^{2 +}

F e^{3 +}

[A g^{+}]

be so that the voltage of the galvanic cell made from

A g^{+} / A g

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

electrodes equals zero? The reaction is

F e^{2 +} + A g^{+} ⇌ F e^{3 +} + A g

. Determine the equilibrium constant at

25^{\circ} C

for the reaction. (Given:

E_{A g^{+} / A g}^{\circ} = 0.799 v o l t

E_{F e^{3 +} / F e^{2 +}}^{\circ} = 0.771 v o l t

Q. At equimolar concentration of

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

[A g^{+}]

be so that the voltage of the galvanic cell made from

(A g^{+} / A g and F e^{3 +} / F e^{2 +})

electrodes equals zero? The reaction is

F e^{2 +} + A g^{+} ⇌ F e^{3 +} + A g

E_{A g^{+} / A g}^{0} = 0.799

E_{F e^{+} / F e^{2 +}}^{0} = 0.771 volt., 10^{- 0.47} = 2.94)

Give your answer upto two decimal only.

Q. A galvanic cell is constructed with

A g^{+} / A g

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

electrodes. The concentrations of

A g^{+}

at which the e.m.f. of the cell is zero at equimolar concentration of

F e^{2 +}

F e^{3 +}

. Write answer as nearest integer after multiplying it with

10

E_{A g^{+} / A g}^{o} = 0.80 V

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

E_{F e^{3 +} / F e^{2 +}}^{o} = 0.771 V

E_{A g^{+} / A g}^{o} = 0.799 V

The minimum conc. of Ag

^{+}

that would remain unreduced by a standard

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

electrode at equilibrium is:

E^{\circ}

values to calculate

△ G^{\circ}

for the reaction

F e^{2 +} + A g^{+} \to F e^{3 +} + A g

E_{A g^{+} / A g}^{\circ} = 0.80 v o l t

E_{P t / F e^{3 +}, F e^{2 +}}^{\circ} = 0.77 v o l t

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