You are given the following reduction reactions and E- values-Fe3-aq-e- Fe2-aq E- -0-771 VFe2-aq-2e- Fes E- -0-447 VCalculate E- for the half-cell reaction Fe3-aq-3e- Fes-

We calculate the desired value of E^∘ by converting the given nbsp;E^∘ values to Δ G^∘ values, and combining these reduction reactions to obtain the desired e ...

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Standard XII

Chemistry

Ellingham Diagram

You are given...

Question

You are given the following reduction reactions and $E^{\circ}$ values:
$F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q) E^{\circ} = + 0.771 V$
$F e^{2 +} (a q) + 2 e^{-} \to F e (s) E^{\circ} = - 0.447 V$
Calculate $E^{\circ}$ for the half-cell reaction $F e^{3 +} (a q) + 3 e^{-} \to F e (s)$ .

+ 0.32 V

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+ 1.22 V

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- 0.04 V

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+ 0.04 V

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Solution

The correct option is D $+ 0.04 V$
We calculate the desired value of $E^{\circ}$ by converting the given $E^{\circ}$ values to $Δ G^{\circ}$ values, and combining these reduction reactions to obtain the desired equation.
$F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q)$
$Δ G^{\circ} = - n F E^{\circ} = - 1 \times 96485 {C mol}^{- 1} \times 0.771 V = - 74.39 {k J mol}^{- 1}$
$F e^{2 +} (a q) + 2 e^{-} \to F e (s)$
$Δ G^{\circ} = - n F E^{\circ} = - 2 \times 96485 {C mol}^{- 1} \times (- 0.447 V) = 86.26 {k J mol}^{- 1}$
We next add the two equations as well as their $Δ G^{\circ}$ to obtain
$F e^{3 +} (a q) + 3 e^{-} \to F e (s)$
$Δ G^{\circ} = - 74.39 k J m o l^{- 1} + 86.26 {k J mol}^{- 1} = 11.87 {k J mol}^{- 1}$
$E_{F e^{3 +} / F e}^{\circ} = - \frac{Δ G^{\circ}}{n F} = \frac{- 11.87 \times 10^{3} J m o l^{- 1}}{3 \times 96485 {C mol}^{- 1}} = - 0.041 V$

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Similar questions

Q. What will be the

E^{0}

value for the given half cell reaction?

F e^{3 +} (a q) + 3 e^{-} \to F e (s)

Given:

E^{0}

value for the half cell reaction,

F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = 0.77 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V

Q. The standard reduction potentials at 298K for the following half cell reaction are given below:

Z n^{2 +} (a q) + 2 e^{-} ⇌ Z n (s); E^{o} = - 0.762 V

C r^{3 +} (a q) + 3 e^{-} ⇌ C r (s); E^{o} = + 0.762 V

2 H^{+} (a q) + 2 e^{-} ⇌ H_{2} (g); E^{o} = + 0.00 V

F e^{3 +} (a q) + e^{-} ⇌ F e^{2 +} (a q); E^{o} = + 0.770 V

Based on the above reactions, select the strongest reducing agent?

Q. The standard reduction potential at 298 K for the following half all reactions are given below:

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s), E^{0} = - 0.762 V C r^{3 +} (a q) + 3 e^{-} \to C r (s), E^{0} = - 0.740 V 2 H^{+} (a q) + 2 e^{-} \to H_{2} (g), E^{0} = 0.0 V F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q), E^{0} = 0.77 V

Which is the strongest reducing agent?

Q. The standard reduction potentials at 298 K for the following half reactions are given,

Z n_{(a q)}^{2 +} + 2 e^{-} ⇌ Z n_{(s)}

E^{\circ}

- 0.762 V

C r_{(a q)}^{3 +} + 3 e^{-} ⇌ C r_{(s)}

E^{\circ}

- 0.740 V

2 H_{(a q)}^{+} + 2 e^{-} ⇌ H_{2 (g)}

E^{\circ}

0.000 V

F e_{(a q)}^{3 +} + e^{-} ⇌ F e_{(s)}^{2 +}

E^{\circ}

0.770 V

Which is the strongest reducing agent?

Q. For

M^{2 +} / M and M^{3 +} / M^{2 +}

systems,

E^{0}

values for some metals are as follows:

C r^{2 +} (a q) + 2 e^{-} \to C r (s); E^{0} = - 0.91 V C r^{3 +} (a q) + e^{-} \to C r^{2 +} (a q); E^{0} = - 0.41 V

M n^{2 +} (a q) + 2 e^{-} \to M n (s); E^{0} = - 1.2 V M n^{3 +} (a q) + e^{-} \to M n^{2 +} (a q); E^{0} = + 1.51 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = + 0.77 V

Which of the following statement(s) is/are correct regarding the given standard reduction potentials?

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You are given the following reduction reactions and E∘ values: Fe3+(aq)+e−→Fe2+(aq) E∘=+0.771 V Fe2+(aq)+2e−→Fe(s) E∘=−0.447 V Calculate E∘ for the half-cell reaction Fe3+(aq)+3e−→Fe(s).

You are given the following reduction reactions and $E^{\circ}$ values:
$F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q) E^{\circ} = + 0.771 V$
$F e^{2 +} (a q) + 2 e^{-} \to F e (s) E^{\circ} = - 0.447 V$
Calculate $E^{\circ}$ for the half-cell reaction $F e^{3 +} (a q) + 3 e^{-} \to F e (s)$ .