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

Acidic Buffer Action

A half cell i...

Question

A half cell is prepared by dipping a silver electrode in a solution containing KCl and some AgCl. Half of the cell is used as cathode during cell reaction. Quantity of AgCl will:

Increase

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

Decrease

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

Same

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

Can't say

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

Open in App

Solution

The correct option is B Decrease
Quantity of AgCl will decrease in the solution due to the electrochemical reaction that takes will take place between the Ag electrode and the Ag in the solution.

Hence, the correct answer will be option B.

Suggest Corrections

Similar questions

0.1 M S C N^{-}

A g S C N i s x \times 10^{v}

[E_{A g^{+} / A g}^{\circ} = 0.8 V]

{A g}^{+} | A g

0.7991 V

25^{o} C

K_{s p} = 1.56 \times 10^{- 10}

A g C l

A g | A g C l

In a galvanic cell, the reactions taking place in the anodic half cell and the cathodic half cell will be:

C l^{-} | A g C l | A g

A g^{+} | A g

E_{A g + | A g}^{o} = E_{C l^{-} | A g C l | A g}^{o} + \frac{R T}{F} l n K_{s p} (A g C l)

C o^{2 +}

C l^{-}

C l_{2} (g)

E^{⊖} (1 / 2 C l_{2} + e^{-} \to C l^{⊖}) = 1.36 V

E^{⊖}

Join BYJU'S Learning Program