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Nernst Equation

The two half-...

Question

The two half-cell reactions of an electrochemical cell is given as
${A g}^{+} + e^{-} ⟶ A g; E_{{A g}^{+} | A g}^{o} = - 0.3995 V$
${F e}^{2 +} ⟶ {F e}^{3 +} + e^{-}; E_{{F e}^{3 +} | {F e}^{2 +}}^{o} = - 0.7120 V$
The value of cell EMF will be:

A

- 0.3125 V

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B

0.3125 V

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C

1.114 V

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D

- 1.114 V

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Solution

The correct option is D $0.3125 V$
Species with more negative $E$ (standard reduction potential) generally acts as reducing agent while with less negative $E$ acts as the oxidising agent. Thus, the overall reaction is:

${A g}^{+} + {F e}^{2 +} ⟶ {F e}^{3 +} + A g$

So,
$E_{c e l l}^{o} = E_{R H} - E_{O H}$
$= (- 0.3995 - (- 0.7120)) V$
$= (- 0.3995 + 0.7120) V$
$= + 0.3125 V$

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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. Consider an electrochemical cell in which the following reaction occurs and predict which changes will decreases the cell voltage:

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

(I) decreases the

[A g^{+}]

(II) increases in

[F e^{3 +}]

(III) increase the amount of

A g

Q. For the given cell reaction

F e^{2 +} + C e^{4 +} ⇌ F e^{3 +} + C e^{3 +}

The equilibrium constant is:

E_{C e^{4 +} | C e^{3 +}}^{o} = 1.44 V; E_{F e^{3 +} | F e^{2 +}}^{o} = 0.68 V

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.

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

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