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  • Consider the figure given below and answer the questions (i) to (vi) that follow. (i) Redraw the diagram to show the direction of electron flow. (ii) Is silver plate anode or cathode?

Consider the figure given below and answer the questions (i) to (vi) that follow.

(i) Redraw the diagram to show the direction of electron flow.
(ii) Is silver plate anode or cathode?
(iii) What will happen if salt bridge is removed?
(iv) When will the cell stop functioning?
(v) How will concentration of Zn^{+2} ions and Ag^{+} ions be affected when the cell functions ?
(vi) How will the concentration of Zn^{+2} ions and Ag^{+} ions be affected after the cell becomes ‘dead’?

 

Answers (1)

(i) The cell reaction can be summarised as:

Zn(s)\left | Zn^{+2} \right |\left | Ag^{+} \right |Ag

Electrons move from Zn to Ag.

(ii) Due to a higher standard reduction potential, Silver will act as Cathode and in the external circuit, electrons will flow from zinc anode to silver cathode.

(iii) Removal of salt bridge will lead to a sudden drop in potential to zero.

(iv) If the potential reaches zero (or the cell is discharged), all reactions will cease and the cell will stop functioning.

(v) Nernst equation for the cell is: 0.059,

E=E^{0}-\frac{0.059}{2}log\frac{[Zn^{2+}]}{[Ag^{+}]^{2}}

With increase in concentration of [Zn^{+2}], cell potential will decrease, and with an increase in concentration of [Ag^{+}], cell potential will increase.

(vi) At equilibrium (discharged state, potential drop to zero), the concentration of [Zn^{+2}] and  [Ag^{+}] will not change.

 

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