For the following cell reaction,$Ag | Ag^{+} | AgCl | Cl^{-} | Cl_2, Pt$
$\Delta G_f^{\circ}(AgCl) = -109 \ kJ/mol$
$\Delta G_f^{\circ}(Cl^{-}) = -129 \ kJ/mol$
$\Delta G_f^{\circ}(Ag^{+}) = 78 \ kJ/mol$
$E^{\circ}$ of the cell is

For four metals $A, B, C$ and $D$,the standard reduction potentials are $-3.05 \ V, -1.66 \ V, -0.40 \ V$ and $0.80 \ V$ respectively. Which metal exhibits the highest chemical reactivity?

Write the equation relating thermodynamics and electrochemistry.

Which expression represents the cell potential $(E_{cell})$?

$A$ $1.0 \ M$ solution with respect to each of the metal halides $AX_3, BX_2, CX_3$ and $DX_2$ is electrolysed using platinum electrodes. If
$E^o_{A^{3+}/A} = 1.50 \ V, \quad E^o_{B^{2+}/B} = 0.3 \ V,$
$E^o_{C^{3+}/C} = -0.74 \ V, \quad E^o_{D^{2+}/D} = -2.37 \ V.$
The correct sequence in which the various metals are deposited at the cathode is

Standard electrode potentials are given:
$A^{+}/A = -2.93 \ V$ $B^{+}/B = 0.80 \ V$
$C^{2+}/C = -2.37 \ V$ $D^{3+}/D = -0.74 \ V$
Increasing order of reducing power of these metals: