JEE Challenger
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ChemMedAdv 2022 • Paper 1

Calculate Standard Reduction Potential of Permanganate to Manganese

The reduction potential (E0E^0, in V) of MnO4−(aq)/Mn(s)\text{MnO}_4^-(\text{aq})/\text{Mn}(\text{s}) is ______.

[Given: E(MnO4−(aq)/MnO2(s))0=1.68 VE^0_{(\text{MnO}_4^-(\text{aq})/\text{MnO}_2(\text{s}))} = 1.68\text{ V}; E(MnO2(s)/Mn2+(aq))0=1.21 VE^0_{(\text{MnO}_2(\text{s})/\text{Mn}^{2+}(\text{aq}))} = 1.21\text{ V}; E(Mn2+(aq)/Mn(s))0=−1.03 VE^0_{(\text{Mn}^{2+}(\text{aq})/\text{Mn}(\text{s}))} = -1.03\text{ V}]

Official Numerical Answer0.77

Topics & Concepts

ElectrochemistryElectrochemical CellNernst Equation

Step-by-Step Solution

To determine the standard reduction potential (E0E^0) for the reduction of MnO4−(aq)\text{MnO}_4^-(\text{aq}) to Mn(s)\text{Mn}(\text{s}), we can analyze the individual reduction steps and their associated standard Gibbs free energy changes (ΔG0\Delta G^0).

The reduction process occurs in three step-wise half-reactions:

  1. Step 1: MnO4−(aq)\text{MnO}_4^-(\text{aq}) to MnO2(s)\text{MnO}_2(\text{s})

    • Oxidation state change of Mn: +7→+4+7 \rightarrow +4
    • Number of electrons transferred (n1n_1): 33
    • Standard reduction potential (E10E_1^0): 1.68 V1.68\text{ V}
    • Standard Gibbs free energy change (ΔG10\Delta G_1^0): ΔG10=−n1FE10=−3×F×1.68=−5.04F\Delta G_1^0 = -n_1 F E_1^0 = -3 \times F \times 1.68 = -5.04 F

  2. Step 2: MnO2(s)\text{MnO}_2(\text{s}) to Mn2+(aq)\text{Mn}^{2+}(\text{aq})

    • Oxidation state change of Mn: +4→+2+4 \rightarrow +2
    • Number of electrons transferred (n2n_2): 22
    • Standard reduction potential (E20E_2^0): 1.21 V1.21\text{ V}
    • Standard Gibbs free energy change (ΔG20\Delta G_2^0): ΔG20=−n2FE20=−2×F×1.21=−2.42F\Delta G_2^0 = -n_2 F E_2^0 = -2 \times F \times 1.21 = -2.42 F

  3. Step 3: Mn2+(aq)\text{Mn}^{2+}(\text{aq}) to Mn(s)\text{Mn}(\text{s})

    • Oxidation state change of Mn: +2→0+2 \rightarrow 0
    • Number of electrons transferred (n3n_3): 22
    • Standard reduction potential (E30E_3^0): −1.03 V-1.03\text{ V}
    • Standard Gibbs free energy change (ΔG30\Delta G_3^0): ΔG30=−n3FE30=−2×F×(−1.03)=+2.06F\Delta G_3^0 = -n_3 F E_3^0 = -2 \times F \times (-1.03) = +2.06 F

Overall Reaction: MnO4−(aq)→Mn(s)\text{MnO}_4^-(\text{aq}) \rightarrow \text{Mn}(\text{s})

  • Total oxidation state change of Mn: +7→0+7 \rightarrow 0
  • Total number of electrons transferred (nn): n1+n2+n3=3+2+2=7n_1 + n_2 + n_3 = 3 + 2 + 2 = 7

Since Gibbs free energy is an extensive property, the overall standard Gibbs free energy change is the sum of the individual steps: ΔG0=ΔG10+ΔG20+ΔG30\Delta G^0 = \Delta G_1^0 + \Delta G_2^0 + \Delta G_3^0

−nFE0=−n1FE10−n2FE20−n3FE30-n F E^0 = -n_1 F E_1^0 - n_2 F E_2^0 - n_3 F E_3^0

Dividing both sides by −F-F, we get: nE0=n1E10+n2E20+n3E30n E^0 = n_1 E_1^0 + n_2 E_2^0 + n_3 E_3^0

Substitute the given values into the equation: 7×E0=(3×1.68)+(2×1.21)+(2×−1.03)7 \times E^0 = (3 \times 1.68) + (2 \times 1.21) + (2 \times -1.03)

7×E0=5.04+2.42−2.067 \times E^0 = 5.04 + 2.42 - 2.06

7×E0=5.407 \times E^0 = 5.40

E0=5.407≈0.7714 VE^0 = \frac{5.40}{7} \approx 0.7714\text{ V}

Rounding off to two decimal places, we get: E0=0.77 VE^0 = 0.77\text{ V}

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