During the working of a galvanic cell and with the passage of time:
1) Spontaneity of the cell reaction decreases, $E_{c e l l}$ decreases.
2) Reaction quotient Q decreases, $E_{c e l l}$ increases.
3) Reaction quotient Q increases, $E_{c e l l}$ decreases.
4) At equilibrium, $Q = K_{e q}, E_{c e l l} = 0$ .
Identify the correct statement(s).

2,3,4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1,4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1,3,4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1,2,3,4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 1,3,4
$E_{c e l l} = E_{c e l l}^{\circ} - \frac{R T}{n F} l n Q$
$∴$ as time passes,
Q increases [reaches equilibrium]
and $E_{c e l l}$ decreases
At equilibrium $Q = K_{e}$
At equilibrium, $Δ G = - R T l n K_{e q} = - n F E_{c e l l}$
$∴$ spontaneity decreases with time.

Option C is correct.

Suggest Corrections

Similar questions

2 C o_{a q}^{3 +} + Z n (s) ⇌ 2 C o^{2 +} (a q) + Z n^{2 +} (a q)

Z n (s) + C u^{2 +} (a q .) ⇌ C u (s) + Z n^{2 +} (a q .)

Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}

E_{c e l l}

Q

O A = 1.10 v o l t . E_{c e l l} = 1.1591 v o l t

Z n + C u^{2 +} (a q) ⇌ C u + Z n^{2 +} (a q)

Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}

E_{c e l l}

V . E_{c e l l}

1.1591 V

$Z n + C u {^{2 +}}_{(a q)} ⇌ C u + Z n {^{2 +}}_{(a q)}$

Reaction quotient, $Q = \frac{[Z n^{2 +}]}{[C u^{2 +}]}$

Variation of $E_{c e l l}$ with log Q is of the type as shown in the graph with $O A = 1.10 V . E_{c e l l}$ of the reaction will be 1.1591 V when:

E_{c e l l} = 0

Join BYJU'S Learning Program