Chemistry - Electrochemistry Question with Solution | TestHub

(a) $M^{n+}+n{e}^{-}⟶M(s)$

$nF$gives ⇒ M gm metal

1 F gives ⇒ $M/n$ metal =$E$

(b) 1 F = charge of one mole electron

$1F=6.023\times 10^{23}\times 1.6\times 10^{-19}C$

=$96500$ C .

(c) $C{u}^{2+}+2e^{-}⟶Cu(s)$

2 F gives = $63.5g$.

Therefore, 2 F deposits $63.5g$

1 F gives = $\frac{63.5g}{2}=31.75g$

(d) 2F gives = 1 mole Cu.

1 F gives = $\frac{1}{2}moleofCu$