Question

NEED HELP

. Calculate the number of moles of MnO4_ reduced to Mn2+ by the addition of 7.5 mole electrons in MnO4_ .

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Answer 1

Answer:

If we write the balanced (half) reduction reaction,

8H+ + MnO4- + 5e- → Mn2+ + 4H2O

Here, 1 mole of MnO4– requires 5 moles of electrons to give 1 mole of Mn2+.

So, charge required for 1 molecule = 5 × charge of e = 5 × 1.6 × 10^(-19)C

= 8 × 10^(-19) C

So, charge required for 1 mole = 8 × 10^(-19) C × 6.023 × 10²³ = 48.184 × 10⁴ C

I think I am right