Question

1.

a. How is electrode potential related to electrochemical equilibrium?

b. When the standard hydrogen electrode is connected to a half-cell containing magnesium metal immersed in 1.0 M MgSO_4,, which electrode is the anode? Why?

c. When the standard hydrogen electrode is connected to a half-cell containing an inert electrode in contact with liquid bromine and with 1.0M KBr, which electrode is the anode? Why?

Answer 1

Ans 1a: E^ocell = (RT/nF)lnK

Let electrode potential be Ecell and standard electrode potential be E^ocell.

Then Gibbs free energy, ∆G = -nFEcell and standard Gibbs free energy, ∆G^o =-nFE^ocell

Where, n = no of electrons exchanged between anode and cathode half-cells.

F = Faraday’s constant = 96500 C/mol

∆G = ∆G^o + RTlnQ

where, ∆G^o = Standard Gibbs free energy, R = Universal gas constant, T = temperature in K

            Q = reaction quotient.

At equilibrium, ∆G = 0 and Q = K (K = equilibrium constant)

So, ∆G^o = – RTlnK & ∆G^o =-nFE^ocell

E^ocell = (RT/nF)lnK

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Data collected	Standard reduction potential = Eo at 298 K
2H+ + 2e --> H2(g) (SHE)	0.0 V
Mg2+ + 2e---> Mg	-2.36 V
Br2(l) + 2e ---> 2Br-	1.09 V
SHE = standard hydrogen electrode

In galvanic cell, anode is where oxidation occurs and cathode is where reduction occurs.

For a cell to be galvanic, E^ocell > 0

E^ocell = E^ocathode - E^oanode > 0

Ans 1b: Anode electrode is Magnesium dipped in 1.0 M MgSO₄.

E^ocell = E^ocathode - E^oanode

E^ocell = E^o_SHE - E^o_Mg2+/Mg = (0.0) – (-2.36) = 2.36 V >0

Anode electrode is Magnesium dipped in 1.0 M MgSO₄.

Ans 1c: Anode electrode is standard hydrogen electrode

E^ocell = E^ocathode - E^oanode

E^ocell = E^o_Br2/2Br- - E^o_SHE = (1.09) – (0.0) = 1.09 V >0

Anode electrode is standard hydrogen electrode