Determine the potential of a platinum electrode (vs. SHE) when placed in the fol

Question

Determine the potential of a platinum electrode (vs. SHE) when placed in the following solutions at 25°C. A) A mixture of 0.00231 M TlCl3 and 0.0455 M TlCl. B) A mixture of 0.0132 M potassium permanganate (KMnO4) and 0.0475 M manganese chloride (MnCl2) at pH 3.0

Map o Sapling Learning macmillan loarnig Determine the potential of a platinum electrode (vs. SHE) when placed in the following solutions at 25°C. A) A mixture of 0.00231 M TICl3 and 0.0455 M TICI aq) +2e- E"= 1.25 V Number EPt = 111.16 B) A mixture of 0.0132 M potassium permanganate (KMnO4) and 0.0475 M manganese chloride (MnCl2) at pH 3.0 Number E,, = | | .8402 Pt ^?Previous ? Give Up & View Solution O Check Answer Next Exit Hint

Explanation / Answer

(A) Ti3+(aq) + 2e- ----> Tl+(aq)

According to Nernst Equation ,

E = Eo - (0.059 / n) log ([Products] / [reactants] )

   = Eo - (0.059 / n) log ([Tl+] / [Tl3+] )

Where

E = electrode potential of the cell = ?

Eo = standard electrode potential = 1.25 V

n = number of electrons involved in the reaction = 2

[Tl3+] = 0.00231 M

[Tl+] = 0.0455 M

Plug the values we get

E = 1.25 - (0.059 / 2 ) x log ( 0.0455/0.00231 )

   = 1.21 V

Therefore the potential of the cell is 1.21 V

(B) MnO4-(aq) + 5e- +8H+(aq) ----> Mn2+(aq) + 4H2O(l)

According to Nernst Equation ,

E = Eo - (0.059 / n) log ([Products] / [reactants] )

   = Eo - (0.059 / n) log ([Mn2+] / ([MnO4-][H+] )

Where

E = electrode potential of the cell = ?

Eo = standard electrode potential = 1.51 V

n = number of electrons involved in the reaction = 5

[Mn2+] = 0.0475 M

[MnO4-] = 0.0132 M

[H+] = 10-pH = 10-3.0 = 10-3 M

Plug the values we get

E = 1.51 - (0.059 / 5 ) x log ( 0.0475/(0.0132x(10-3))

   = 1.47 V

Therefore the potential of the cell is 1.21 V

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