Calculate ΔG° for the following reaction:

Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)

Given:

E°Fe³⁺/Fe²⁺ = +0.77 V
E°Ag⁺/Ag = +0.80 V

1. +$2.89$ $\mathrm{kJ}$

2. $-2.89$ $\mathrm{kJ}$

3. $-2.89$ $\mathrm{J}$

4. -186.83 J

Given:
(i) ${\mathrm{Cu}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$    E^o = 0.337 V 
(ii) ${\mathrm{Cu}}^{2+}+{\mathrm{e}}^{-}\to {\mathrm{Cu}}^{+}$  E^o = 0.153 V 
Electrode potential, E^o for the reaction, 
${\mathrm{Cu}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Cu}$, will be: 

1. 0.52 V

2. 0.90 V

3. 0.30 V

4. 0.38 V

Given the following cell reaction:
 \(\mathrm{2Fe^{3+}(aq) \ + \ 2I^{-}(aq)\rightarrow 2Fe^{2+}(aq) \ + \ I_{2}(aq)}\)

[Given: \(F  = 96500\) \(C\) \(mol^{- 1}\)]

1. \(23 . 16\) \(kJ\) \(mol^{- 1}\)

2. \(- 46 . 32\) \(kJ\) \(mol^{- 1}\)

3. \(- 23 . 16\) \(kJ\) \(mol^{- 1}\)

4. \(46 . 32\) \(kJ\) \(mol^{- 1}\)

For a cell involving one electron \(E_{cell}^{\ominus} = 0 . 59  V\) at 298 K. The equilibrium constant for the cell reaction is :
\(\mathrm{[Given~ that~ \frac {2.303 ~RT}{F} = 0.059 ~V~ at~ T = 298 K]}\)