Question 12

How much charge is required for the following reductions:

(i) 1 mol of Al3+ to Al.

(ii) 1 mol of Cu2+ to Cu.

(iii) 1 mol of MnO4- to Mn2+.

Answer

(i) Al3+ + 3e-  → Al

Required charge = 3 F

= 3 × 96487 C = 289461 C

 

(ii) Cu2+ + 2e→ Cu

Required charge = 2 F

= 2 × 96487 C

= 192974 C

 

(iii) MnO4-  → Mn2+

i.e.,

Mn7+ +5e- → Mn2+

Required charge

= 5 F

= 5 × 96487 C

= 482435 C

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