For the cell reaction:
$2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightarrow 2 Fe^{2+}_{(aq)} + I_{2(aq)}$
$E^{\ominus}_{cell} = 0.24 \ V$ at $298 \ K$. The standard Gibbs energy $(\Delta_r G^{\ominus})$ of the cell reaction in $kJ \ mol^{-1}$ is:
[Faraday constant $F = 96500 \ C \ mol^{-1}$]

The standard electrode potential for the Daniell cell is $1.1 \, V$. Calculate the standard Gibbs energy for the reaction:
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$

The standard reduction potentials of $Cu^{2+}$,$Zn^{2+}$,$Sn^{2+}$ and $Ag^{+}$ are $0.34 \ V$,$-0.76 \ V$,$-0.14 \ V$ and $0.80 \ V$ respectively. Which of the following solutions can be stored in the given vessel without any reaction occurring (under standard conditions)?

The standard electrode potential for the Daniell cell is $1.1 \ V$. What will be the value of standard Gibbs energy for the reaction?
$Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
$(1 \ F = 96487 \ C \ mol^{-1})$

Which one among the following is the strongest reducing agent?
$Fe^{2+} + 2e^- \to Fe \ (E^\circ = -0.44 \ V)$
$Ni^{2+} + 2e^- \to Ni \ (E^\circ = -0.25 \ V)$
$Sn^{2+} + 2e^- \to Sn \ (E^\circ = -0.14 \ V)$
$Fe^{3+} + e^- \to Fe^{2+} \ (E^\circ = +0.77 \ V)$

$A$ $1.0 \ M$ 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