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Types of Redox Reactions

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Question

Calculate the $△ G^{\circ}$ and equilibrium constant of the reaction
$F e_{(a q)}^{+ 2} + A g_{(a q)}^{+} ⟶ F e_{(a q)}^{+ 3} + A g_{(s)}$
$E_{A g^{+} | A g}^{\circ} = 0.80$ $E_{F e^{+ 3} | F e^{+ 2}}^{\circ} = 0.77 V$

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Solution

$E_{c e l l}^{\circ} = + 0.80 V - 0.77 V = 0.03 V$
$△ G^{\circ} = - n F E_{c e l l}^{\circ}$
or $△ G^{\circ} = - 1 \times 96500 \times 0.03$
or $△ G^{\circ} = - 2895 J$ $m o l^{- 1}$
$△ G^{\circ} = - 2.303$ $R T$ $log K$
or $- 2895 = - 2.303 \times 8.314 \times 298 log K$
or $K = 3.22$

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Similar questions

Q. 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)}

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

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

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. The equilibrium constant at

25^{\circ} C

for the reaction

F e^{2 +} + A g^{+} ⇌ F e^{3 +} + A g

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. a) Draw labelled diagram of Standard Hydrogen Electrode ( SHE ).
Write its half cell reaction of

E^{o}

value.
b) Calculate

△ r G^{⊝}

for the following reaction :

F e_{(a q)}^{+ 2} + A g_{(a q)}^{+} \to {F e}_{(a q)}^{+ 3} + A g_{(s)}

E_{C e l l}^{0} = + 0.03 V, F = 96500 C

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