Question

While performing Lab 11, Electrochemistry, students connected 1 M solutions of the following metals and used electrodes to monitor the cell potential. Using the table of theoretical values and what you know of the metals to fill in the blanks below.

When writing equations: Use spaces only before and after a plus sign. For example, the reaction: Fe2++2e−→FeFe2++2e-→Fe  would be entered as: Fe^2+ + 2e^- →→ Fe

Galvanic Cell Consisting of:	Zn2++2e−	→→	Zn	-0.76
	Mg2++2e−	→→	Mg	-2.37

E0cell Measured	1.38
Write the equation occuring at the anode.	→→
Write the equation occurring at the cathode.	→→
Shorthand Notation for the Cell:	l  lI  l
Oxidizing Agent:	Mg `Mg2+ Zn Zn2+
Reducing Agent:	Mg Zn `Mg2+ Zn2+
E0cell from Table Values (Theoretical Numbers above).
E0cell Percent Error

Answer 1

The one with the highest reduction potential will be cathode and which with the lowest reduction potential will be anode.

Anode reaction : Zn----> Zn²⁺ + 2e

Cathode reaction: Mg²⁺ +2e -------> Mg

Notation of cell: Anode half cell Salt bridge Cathode half cell

                        :Zn/Zn²⁺//Mg²⁺/Mg

Oxidising Agent: oxidizing agent got reduced at the cathode -- Mg²⁺

Reducing agent: reducing agent got oxidized at the anode – Zn

E⁰_cell = E⁰_cathode - E⁰_anode

E⁰_cell from table = -2.372 – (-0.7618) = -1.6102 V

E⁰_cell calculated = -2.37 –(-0.76) = -1.61 V

E⁰_cell Percent Error ={[ -1.61 –(-1.6102)]/ -1.6102}×100 = 0.01%