MyQuestionIcon

1

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

Gibb's Energy and Nernst Equation

Calculate the...

Question

Calculate the standard cell potentials of galvanic cell in which the following reactions take place :

(i) $2 C r (s) + 3 C d^{2 +} (a q) \to 2 C r^{3 +} (a q) + 3 C d$

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

Calculate the $Δ_{r} G^{\circ}$ and equilibrium constant for the reactions

Open in App

Solution

Given $E_{c e l l}^{\circ} \frac{C r^{3 +}}{C r} = - 0.74 V_{i} E_{c e l l}^{\circ} (\frac{C d^{2 +}}{C d}) = - 0.40 V$

(a) Calculation of standard cell potential $(E^{\circ})$ $E_{c e l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{a n o d e}^{\circ}$
$= (- 0.40) - (- 0.74) = + 0.34 V$
$E_{c e l l}^{\circ} = + 0.34 V$

(b) Calculation of $Δ_{r} G^{\circ}$
$Δ_{r} G^{\circ} = - n F E_{c e l l}^{\circ} = - (6 m o l) \times (96500 C m o l^{- 1}) \times (0.34 V) = - 196860 C V = - 196860 J = - 196.86 k J Δ_{r} G^{\circ} = - 196.86 k J$

(c) Calculation of equilibrium constant $(K_{c})$
$Δ_{r} G^{\circ} = - 2.303 R T l o g K_{c} l o g K_{c} = \frac{Δ_{r} G^{\circ}}{2.303 R T} = (-) \frac{(- 196860 J)}{2.303 \times (8.314 J K^{- 1}) \times (298 K)} = 34.501 K_{c} = A n t i l o g (34.501) = 3.17 \times 10^{34} K_{c} = 3.17 \times 10^{34}$

(ii)

$G i v e n E_{c e l l}^{\circ} (\frac{A g^{+}}{A g}) = 0.80 V E_{c e l l}^{\circ} (\frac{F e^{3 +}}{F e^{2 +}}) = 0.77 V$

(a) Calculation of standard cell potential $(E^{\circ})$

$E_{c e l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{a n o d e}^{\circ}$

$= (0.80) - (0.77) V = 0.03 V$

$E_{c e l l}^{\circ} = + 0.03 V$

(b) Calculation of $Δ_{r} G^{\circ}$

$Δ_{r} G^{\circ} = n F E_{c e l l}^{\circ}$

= - $(1 m o l) \times (\frac{96500 C}{m o l}) \times (0.03 V)$

$Δ_{r} G^{\circ} = - 2.895 k J$

(c) Calculation of equilibrium constant $(K_{C})$

$Δ_{r} G^{\circ} = 2.303 R T l o g K_{c}$

log $K_{c} = \frac{Δ_{r} G^{\circ}}{2.303 R T}$

$= (-) \frac{(- 2895 J)}{2.303 \times (8.314 J K^{- 1}) \times (298 K)} = 0.5074$

$K_{C} =$ Antilog (0.5074) = 3.22

$K_{c} = 3.22$

Suggest Corrections

thumbs-up

47

Similar questions

Calculate the standard cell potentials of galvanic cells in which the following reactions take place:

(i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd

(ii) Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)

Calculate the Δ_rG^θ and equilibrium constant of the reactions.

Q. Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
(i)

2 C r (s) + 3 C d^{2 +} (a q) \to 2 C r^{3 +} (a q) + 3 C d

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

Δ_{r} G^{⊖}

and equilibrium constant of the reactions

Q. What will be standard cell potential of galvanic cell with the following reaction?

2 C r_{(s)} + 3 C d_{(a q)}^{2 +} \to 2 C r_{(a q)}^{3 +} + 3 C d_{(s)}

E_{C r^{3 +} / C r}^{o} = - 0.74

E_{C d^{2 +} / C d}^{o} = - 0.40

Q. What is the relationship between the standard cell potentials

E^{o}

, for the following two galvanic cell reactions?
(I)

2 A g^{+} (a q) + S n^{2 +} (a q) \to S n^{4 +} (a q) + 2 A g (s)

A g (s) + \frac{1}{2} S n^{4 +} (a q) \to \frac{1}{2} S n^{2 +} (a q) + A g^{+} (a q)

Q. Using the standard electrode potentials given in Table

3.1

, predict if the reaction between the following is feasible:
(i)

F e^{3 +} (a q)

I^{-} (a q)

A g^{+} (a q)

C u (s)

F e^{3 +} (a q)

B r^{-} (a q)

A g (s)

F e^{3 +} (a q)

B r_{2} (a q)

F e^{2 +} (a q)

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Gibb's Energy and Nernst Equation

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon