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zysi [14]
3 years ago
12

Iron(II) can be oxidized by an acidic K2Cr2O7 solution according to the net ionic equation: Cr2O72− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe

3+ + 7H2O If it takes 35.5 mL of 0.0250 M K2Cr2O7 to titrate 25.0 mL of a solution containing Fe2+, what is the molar concentration of Fe2+ in the original solution?
Chemistry
1 answer:
cestrela7 [59]3 years ago
4 0

Answer:

The molar concentration of Fe²⁺ in the original solution is 1.33 molar

Explanation:

Moles of K₂Cr₂O₇ = Molarity x Volume (lit)

                             = 0.025 x 35.5 x 10⁻³

                             = 0.0008875

Cr₂O₇²⁻ + 6 Fe²⁺ + 14 H⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O

From equation

              1 mole K₂Cr₂O₇ used for the oxidation of 6 moles Fe²

0.0008875 mole K₂Cr₂O₇ used for the oxidation of =\frac{6 X 0.0008875}{1} = 0.005325 mole of  Fe²

      Molarity = \frac{No. of moles of solute}{Volume     of solution(lit)}

Molar concentration of  Fe² = \frac{0.005325 X 1000}{25} = 1.33 molar

So molar concentration of Fe²⁺ in the original solution = 1.33 molar

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From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : H_2\rightarrow 2H^{+}+2e^-    

Reaction at cathode (reduction) : Ag^{+}+1e^-\rightarrow Ag    

The balanced cell reaction will be,  

H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag

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E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}

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Therefore, the standard cell potential will be +0.799 V

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