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Electrochemistry question

2021 · 26 Feb · Shift 1 · Q20
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  5. /2021 · 26 Feb · Shift 1 · Q20

Electrochemistry question

2021 · 26 Feb · Shift 1 · Q20

JEE MainChemistryElectrochemistryNumerical+4 / −1
Consider the following reaction MnO4−+8H++5e−→Mn+2+4H2O,Eo=1.51VMnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51VMnO4−​+8H++5e−→Mn+2+4H2​O,Eo=1.51V. The quantity of electricity required in Faraday to reduce five moles of MnO4−MnO_4^ -MnO4−​ is ‾\underline{\hspace{2cm}}​. (Integer answer)
Numerical answer
View written solutionFree

Correct answer: 25

  1. Write the given reduction half-reaction

MnO4−+8H++5e−→Mn2++4H2OMnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2OMnO4−​+8H++5e−→Mn2++4H2​O

From the equation, 1 mole of MnO4−MnO_4^-MnO4−​ requires 5 moles of electrons for reduction.

  1. For 5 moles of MnO4−MnO_4^-MnO4−​

If 1 mole of MnO4−MnO_4^-MnO4−​ needs 5 moles of electrons, then 5 moles of MnO4−MnO_4^-MnO4−​ will need:

5×5=255 \times 5 = 255×5=25

moles of electrons.

  1. Convert moles of electrons to Faraday

By definition,

1 Faraday=1 mole of electrons1 \text{ Faraday} = 1 \text{ mole of electrons}1 Faraday=1 mole of electrons

Therefore, required electricity is:

25 Faraday25 \text{ Faraday}25 Faraday

  1. Final integer answer

25\boxed{25}25​

  1. Comparison with stored correct answer

Stored correct answer = 252525

Our derived answer also = 252525, so they agree.

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