According to Kohlrausch’s law, the molar conductivity of NH₄OH

$$\Lambda _{m\left( {N{H_4}OH} \right)}^0$$ = $$\Lambda _{m\left( {N{H_4}Cl} \right)}^0 + \Lambda _{m\left( {NaOH} \right)}^0 - \Lambda _{m\left( {NaCl} \right)}^0$$

Since the reduction potential of Fe³⁺/Fe²⁺ is greater than that of I₂ /I^– , Fe³⁺ will be reduced and I^– will be oxidised.

2Fe³⁺ + 2I^– $$ \to $$ 2Fe²⁺ + I₂

Cu²⁺_(aq) + e^$$-$$ $$ \to $$ Cu⁺_(aq) ; E₁^o = 0.15 V

Cu⁺_(aq) + e^$$-$$ $$ \to $$ Cu_(s) ; E₂^o = 0.50 V

Cu²⁺ + 2e^– $$ \to $$ Cu ; E^o = ?

$$\Delta $$G^o = $$\Delta $$G₁° + $$\Delta $$G₂°

$$ \Rightarrow $$ – nFE° = – n₁FE₁° – n₂FE₂°

$$ \Rightarrow $$ E^o = $${{1 \times 0.15 + 1 \times 0.50} \over 2}$$ = 0.325 V

E°_{Sn⁴⁺/Sn²⁺} = 0.15 V

E°_Cr³⁺/Cr = –0.74 V

E°_cell = E°_cathode – E°_anode

= 0.15 – (– 0.74) = 0.15 + 0.74 = 0.89 V