A Hg/Hg^2+ (1.0 M) half-cell was connected to a A/A^2+ (1.0 M) half-cell that ha

Question

A Hg/Hg^2+ (1.0 M) half-cell was connected to a A/A^2+ (1.0 M) half-cell that has E degree_red equal to 0.167 volts The cell potential was measured as 0.684 Volts (V). The conventional notation for the resulting galvanic cell is: A(s)/A^2+ (1.0 M)//Hg^2+ (1.0 M)/Hg(l) In this galvanic cell, A(s)/A^2+ (1.0 M) is the anode (oxidation half-cell) and Hg^2+ (1.0 M)/Hg(l) is the cathode (reduction half-cell). Calculate the standard reduction potential for the Hg/Hg^2+ half-cell. Another B(s)/B^2+ (1.0 M) anode half-cell (E degree_Red = -0.637 Volts) was connected to the Hg/Hg^2+ (1.0 M) half-cell described in part 1 above Calculate the potential of the resulting galvanic cell. A B(s)/B^2+ (0.200 M) non-standard anode half-cell (E degree_Red = -0.637 Volts) was connected to a standard hydrogen half-cell. Calculate the oxidation potential (E_ox) of the non-standard B(s)/B^2+ (0.200 M) half-cell. Calculate the potential of the galvanic cell described in part 3 above.

Explanation / Answer

A(S)/ A^2+ II Hg^2+/Hg

E(cell) = E(anode) + E(cathode)

or, 0.684 = -0.167 + E(cathode)

or, E(cathode) = 0.684 + 0.167 = 0.851 V

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