Consider figure from the above question and answer the questions (i) to (vi) given below. (i) Redraw the diagram to show the direction of electron flow. (ii) Is silver plate the anode or cathode? (iii) what will happen it salt bridge is removed? (iv) When will the cell stop functining? ( v) How will concentration of Zn^(2+) ions and Ag^(+) ions be affected when the cell function? (vi) How will the concentration of Zn^(2+) ions and Ag^(+) ions be affected after the cell becomes dead?

Consider figure from the above question and answer the questions (i) t

1.	Consider figure from the above question and answer the questions (i) to (vi) given below. (i) Redraw the diagram to show the direction of electron flow. (ii) Is silver plate the anode or cathode? (iii) what will happen it salt bridge is removed? (iv) When will the cell stop functining? ( v) How will concentration of Zn^(2+) ions and Ag^(+) ions be affected when the cell function? (vi) How will the concentration of Zn^(2+) ions and Ag^(+) ions be affected after the cell becomes dead?
Answer» Solution : (i) Electrons MOVE from Zn to Ag as `E^(@)` is more negative for Zn so Zn undergoes OXIDATION and ` Ag^(+)` undergoes reduction. (II) Ag is the cathode as it is the site of reduction where `Ag^(+)` takes electrons from medium and deposit at cathode. (iii) Cell will stop functioning because cell POTENTIAL drops to zero. AT E=0 reaction reaches equilibrium. (iv) When `E_("cell")`=0 because at this condition reaction reaches to equilibrium. (v) CONCENTRATION of `Zn^(2+)` ions will increase and concentartion of `Ag^(+)` ions will decreases because Zn is converted into `Zn^(2+)` and `Ag^(+)` is converted into Ag. (vi) when `E_("cell")=0` equilibrium is reached and concentration of `Zn^(2+)` ions and `Ag^(+)` will not change.