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# The formal potential of $F{{e}^{3+}}/F{{e}^{2+}}$ in a sulphuric acid and phosphoric acid mixture (${{E}^{o}}$ = +0.61V) is much lower than the standard potential (${{E}^{o}}$ = +0.77V). This is due to,i) Formation of the species ${{\left[ \text{FeHP}{{\text{O}}_{\text{4}}} \right]}^{\text{+}}}$ii) Lowering of potential upon complexationiii) Formation of the species ${{\left[ \text{FeS}{{\text{O}}_{\text{4}}} \right]}^{\text{+}}}$iv) High acidity of the medium.Determine the truth of the statements and mark the correct option.A. i) and ii) onlyB. i), ii) and iv) onlyC. iii) onlyD. All i), ii), iii), and iv) Verified
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Hint: Think about the factors on which reduction potential depends. Also consider how $Fe$ ions may react with sulphuric acid and phosphoric acid and if it will form any complex compounds.

Here, it is observed that the formal potentials of $F{{e}^{3+}}/F{{e}^{2+}}$ in a sulphuric acid and phosphoric acid mixture is much lower than the standard potentials. This happens because, ${{H}_{2}}S{{O}_{4}}$and ${{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}$ are both strong acids. So, the presence of ${{H}_{2}}S{{O}_{4}}$ and ${{\text{H}}_{\text{3}}}\text{P}{{\text{O}}_{\text{4}}}$ makes the medium more acidic than the others. Since the reduction potential depends on the charges and transfer of electrons the presence of more protons due to acidity will definitely affect the reduction potential. The ${{\left[ \text{FeHP}{{\text{O}}_{\text{4}}} \right]}^{\text{+}}}$ complex will form by Fe ions with these acids and hence the concentration of $F{{e}^{3+}}$ ion reduces. This results in the decrease of the formal reduction potential.
Although the ion $S{{O}_{4}}^{2-}$ exists due to the presence of sulphuric acid, it does not react with the iron ions in the presence of phosphoric acid. Thus, it will not affect the reduction potential.