Question

Please consider the following galvanic cell setup: the cell is composed of a silver rod dipped into a silver nitrate solution connected (via a salt bridge) to a solution of a magnesium rod dipped into a magnesium chloride solution. (a) Calculate the standard cell potential. (b) Define the anode and cathode. (c) What do you need to do if you want make the cell potential more positive? (d) Sketch the voltaic cell and label the anode and cathode, and indicate the direction of electron flow.

Answer 1

For part A is given in the below image.

Std ng cag) + 2e = ng Gs) E = -2-37V - E = +0-7991 reduction Agt + t é - - ty é Ages potential Now in above cell Magnesium metal is being Oxidized to magnesium cation E oxidaton = - E - (-2.370) = 2-37v. Half rell reacton. E =+2-37U Dudaton ng mgr the 2 Ag+ +7e 2Ag F mg + 2 Ay taqy - mgr ons + 2A9 (6) +0.7991 Scanned with CamScanner

Ereu = Ereduction + F'oudetton - 0.7991 V + 2-37V = 3.17 V e flow Electron cano de flow from magnesium electrode ne eilver electrode [cato de ng - Ag Canode) (cathode)

For part B:

Anode is the rod where the electorn gets released because of the low reduction potenitial voltage and cathod is the rod which gains the electon from the cathod because of the high reduction potential.

In our case magnesium is the anode and silver is the cathod so at the end of the reaction silver electode will gain the weight while magnesium electrode will lose the weight.

For Part C:

If we increase the concentration of the electrolyte solution, we can delay the equelibrium. So incresing in the concentration of any of the electrolyte solution we can shift the reaction from equilibrium to non-equlibrium stage so flow fo the electron will be more than less concentration of the electrolyte.

For Part D, find the below image

ano de Canode Salt bridge Scanned with CamScanner