Question

I am asked to create a table of reducing potentials by setting ANY of the half-cell potential to 0V. Show calculations.

My half-cells and their Eocell is :

Ag+ + e- = Ag(s) 0.80V

Cu2+ + 2e- = Cu(s) 0.34V

Pb2+ + 2e- = Pb(s) -0.13V

Sn2+ + 2e- = Sn(s) -0.14V

Fe2+ + 2e- = Fe(s) -0.44V

Zn2+ + 2e- = Zn(s) -0.76V

Mg2+ + 2e- = Mg(s) -2.37V

Answer 1

Universally, hydrogen has been recognized as having reduction and oxidation potentials of zero. Therefore, when the standard reduction and oxidation potential of chemical species are measured, it is actually the difference in the potential from hydrogen.

so the half cell potential is actually the differenct from the zero cell potential.

suppose we assume the reduction cell potential of the below reaction to be zero

Sn²⁺ + 2e^- Sn(s) actual potential = -0.14 V, we assume it to be zero than for any other reaction let's say for

Pb²⁺ + 2e^- Pb (s) actual potential = -0.13 V but

potential difference from Sn = (-0.13 V) - (-0.14 V) = 0.01 V

similarly for

Zn²⁺ + 2e^- Zn(s)

potential difference from Sn= (-0.76 V)- (-0.14 V) = -0.62 V

similarly we can calculate potential difference for all of the reactions and arranage them in order

Note:- i have assume the potential of Sn²⁺ + 2e^- Sn(s) to be zero, you can assume for any other reaction and calculate the difference as mentioned above

Ag⁺ + e^- Ag(s) 0.94 V

Cu²⁺ + 2e^-    Cu(s) 0.48 V

Pb²⁺ + 2e^-    Pb (s) 0.01 V

Sn²⁺ + 2e^- Sn(s) 0 V

Fe²⁺ + 2e^- Fe(s) -0.30 V

Zn²⁺ + 2e^-    Zn(s) -0.62 V

Mg²⁺ + 2e^- Mg(s) - 2.59 V