Answer to Question #223233 in General Chemistry for Mei

Question

b) Given in the following electrochemical cell at 25o C:

(−) Cu çCu2+ ( 0.10 M) ççAg+ (0.10 M) çAg (+) (*)

How the voltage of cell will change (increase or decrease) when the following reagents be added? Shortly explaination. Assume the volume of anode/cathode solution does not change by addition of reagent.

-A small amount of NaCl onto solution at cathode of the cell (*).

-Addition EDTA solution onto solution at anode of the cell (*).

c) Solution A was prepared by mixing 10,0 mL NH3 solution with concentation of 0,500 M and 10,0 mL AgNO3 solution with concentration of 0,020 M.

i) Calculate pH value of solution A and equilibrium concentration of Ag+ .

ii) Replace the solution in cathode of the above cell (cell (*)) by solution A, in this case, the voltage of cell will change. Determine the new value of cell voltage.

Eo (Ag+ /Ag) = 0.799 ; Eo (Cu2+/Cu) = 0.337 ; bAg(NH3) + = 103,32; bAg(NH3)2 += 07,23; pKaNH4 +=9,24. Ignore hydroxo formation of Ag+ và Cu2+ ions

Answer 1

In an electrochemical cell, increasing the concentration of reactants will increase the voltage difference, as you have indicated. A higher concentration of reactant allows more reactions in the forward direction so it reacts faster, and the result is observed as a higher voltage.

If you have adjusted the cell volume to keep the total amount of reactants the same, the rate of electron flow through the cell will be higher at first. Then the cell will run down faster, but the total amount of electron flow will be the same.

Electrical energy is the product of the voltage times the total amount of current (total number of electrons flowing).