Calculate the equilibrium constant of the reaction-Feaq-2-Agaq- Feaq-3-AgsGiven - E- Ag-Ag-0-8 VE- Fe-3-Fe-2-0-77 V2-303 RTF-0-06 V-10-3-16

E^∘ _cell=E_red E_oxd ⇒ E^∘ _cell =0.8 0.77 = 0.03V Fe_(aq)^+2+Ag_(aq)^+→ Fe_(aq)^+3+Ag_(s) The number of electrons involved here is 1. We know, E^o_cell = 0. ...

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

Byju's Answer

Standard XII

Chemistry

Electrode Potential

Calculate the...

Question

Calculate the equilibrium constant of the reaction:
$F e_{(a q)}^{+ 2} + A g_{(a q)}^{+} \to F e_{(a q)}^{+ 3} + A g_{(s)}$
Given : $E_{A g^{+} / A g}^{\circ} = 0.8 V E_{F e^{+ 3} / F e^{+ 2}}^{\circ} = 0.77 V \frac{2.303 R T}{F} = 0.06 V \sqrt{10} = 3.16$

Open in App

Solution

$E_{c e l l}^{\circ} = E_{r e d} - E_{o x d}$
$\Rightarrow E_{c e l l}^{\circ} = 0.8 - 0.77 = 0.03 V$
$F e_{(a q)}^{+ 2} + A g_{(a q)}^{+} \to F e_{(a q)}^{+ 3} + A g_{(s)}$
The number of electrons involved here is $1$ .
We know,
$E_{cell}^{o} = \frac{0.059}{n} log K_{eq}$
$\Rightarrow K_{e q} = 10^{⎛ ⎝ \frac{n E^{\circ}}{0.06} ⎞ ⎠} = 10^{⎛ ⎝ \frac{0.03}{0.06} ⎞ ⎠} = 10^{1 / 2} = 3.16$

Suggest Corrections

Similar questions

Q. Calculate the

△ G^{\circ}

and equilibrium constant of the reaction

F e_{(a q)}^{+ 2} + A g_{(a q)}^{+} ⟶ F e_{(a q)}^{+ 3} + A g_{(s)}

E_{A g^{+} | A g}^{\circ} = 0.80

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

Q. The equilibrium constant at

25^{\circ} C

for the reaction

F e^{2 +} + A g^{+} ⇌ F e^{3 +} + A g

is:
Given

E_{A g^{+} / A g}^{0} = 0.7998 V a n d E_{F e^{3 +} / F e^{2 +}}^{0} = 0.771 V

, log 2 =0.30, log 3 = 0.48,

\frac{2.303 R T}{F} = 0.06

Q. 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

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

. Determine the equilibrium constant at

25^{\circ} C

for the reaction. (Given:

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

and

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

Q. Use

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

and

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

Q. The equilibrium constant at

25^{\circ} C

for the reaction

F e^{2 +} + A g^{+} ⇌ F e^{3 +} + A g

is:
Given

E_{A g^{+} / A g}^{0} = 0.7998 V a n d E_{F e^{3 +} / F e^{2 +}}^{0} = 0.771 V

, log 2 =0.30, log 3 = 0.48,

\frac{2.303 R T}{F} = 0.06

Join BYJU'S Learning Program

Calculate the equilibrium constant of the reaction: Fe+2(aq)+Ag+(aq)→Fe+3(aq)+Ag(s) Given : E∘Ag+/Ag=0.8 VE∘Fe+3/Fe+2=0.77 V2.303RTF=0.06 V√10=3.16

E∘cell=Ered−Eoxd ⇒E∘cell=0.8−0.77=0.03V Fe+2(aq)+Ag+(aq)→Fe+3(aq)+Ag(s) The number of electrons involved here is 1. We know, Eocell=0.059nlogKeq ⇒Keq=10⎛⎝nE∘0.06⎞⎠=10⎛⎝0.030.06⎞⎠=101/2=3.16

Calculate the equilibrium constant of the reaction:
$F e_{(a q)}^{+ 2} + A g_{(a q)}^{+} \to F e_{(a q)}^{+ 3} + A g_{(s)}$
Given : $E_{A g^{+} / A g}^{\circ} = 0.8 V E_{F e^{+ 3} / F e^{+ 2}}^{\circ} = 0.77 V \frac{2.303 R T}{F} = 0.06 V \sqrt{10} = 3.16$