How much charge is required for the following reductions: (i) 1 mol of Al 3 + to Al? (ii) 1 mol of Cu 2 + to Cu? (iii) 1 mol of MnO – 4 to Mn 2+ ?

Question

Philoid · Accepted Answer

(i) The electrode reaction is given as,

Al^{3 +} _(aq) + 3e ^- → Al_(s)

∴ The quantity of charge required for the reduction of 1 mol of Al³⁺ = 3F

= 3×96487 C

= 289461 C

(ii) The electrode reaction is given as,

Cu^{2 +} _(aq) + 2e ^- → Cu_(s)

∴ The quantity of charge required for the reduction of 1 mol of Cu²⁺ = 2F

= 2×96487 C

= 192974 C

(iii) The electrode reaction is given as,

MnO₄→ Mn^{2 +}

i.e., Mn^{7 +} + 5e ^- → Mn^{2 +}

∴ The quantity of charge required for the reduction of 1 mol of Mn⁷⁺ = 5F

= 5×96487 C

= 482435 C