Question

Two voltaic cells are to be joined so that one will run the other as an electrolytic cell. In the first cell, one half-cell has Au foil in 1.00 M Au(NO₃)₃, and the other half-cell has a Cr bar in 1.00 M Cr(NO₃)₃. In the second cell, one half-cell has a Co bar in 1.00 M Co(NO₃)₂, and the other half-cell has a Zn bar in 1.00 M Zn(NO₃)₂.(a) Calculate E°cell for each cell.

Answer 1

The $E^o_{cell}$ Au/Cr cell and Co/Zn are 2.24 V and 0.48 V respectively.

What is standard potential?

A measurement of the potential for equilibrium is known as standard electrode potential.

The potential of the electrode is defined as the difference in potential between the electrode and the electrolyte.

The electrode potential is referred to as the standard electrode potential when unity represents the concentrations of all the species involved in a semi-cell.

It is given by

$E^o_{cell} = E^o_{cathode} - E^o_{anode}$

The reactions of each half-cell are as follows:

$Au^{3+}(aq) + 3e^- \rightleftharpoons Au(s)$ ; E° = 1.50 V

$Cr^{3+}(aq) + 3e^- \leftrightharpoons Cr(s)$ ; E° = –0.74 V

$Co^{2+}(aq) + 2e^- \leftrightharpoons Co(s)$ ; E° = –0.28 V

$Zn^{2+}(aq) + 2e^- \leftrightharpoons Zn(s)$ ; E° = –0.76 V

For Au/Cr, Au acts as the cathode and Cr acts as the anode,

$E^o_{cell} = E^o_{cathode} - E^o_{anode}$ = 1.50 -(-0.74 ) = 2.24 V

For Co/Zn, Zn acts as the anode and Co acts as the cathode

$E^o_{cell} = E^o_{cathode} - E^o_{anode}$ = -0.28 - (-0.76) = 0.48 V

When the two cells are added in series, the voltages will add

$E^o_{series} = E^o_{Au/Cr}+ E^o_{Co/Zn}$

              = 2.24 V + 0.48 V = 2.72 V

Hence, The $E^o_{cell}$ Au/Cr cell and Co/Zn are 2.24 V and 0.48 V respectively.

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