Charge of one mole of electrons = 96500 C
∴ 1 mole gram equivalent of substance will be deposited by one mole of electrons.

No explanation is given for this question. Let's discuss the answer together.

$$\eqalign{ & 1\,mole\,{\text{of}}\,{e^ - } = 1F = 96500C \cr & 27g\,{\text{of}}\,Al\,{\text{is deposited by}}\,3 \times 96500\,C \cr & 5120g\,{\text{of}}\,Al\,{\text{will be deposited by}} \cr & = \frac{{3 \times 96500 \times 5120}}{{27}} \cr & = 5.49 \times {10^7}C \cr} $$

For a concentration cell having different concentrations of $$ions.$$
$$E = - \frac{{0.0591}}{n}\log \frac{{{c_1}}}{{{c_2}}}$$
If all the concentrations are identical then obviously the cell voltage is zero. But as the $$pH$$  of $$0.1\,M\,HCl$$   ( strong acid ) and $$pH$$  of $$0.1\,M\,\,C{H_3}COOH$$    is ( weak acid ) not same, therefore the cell voltage is not zero.

$$\eqalign{ & {\text{Total charge}} = 2 \cr & {\text{Number of equivalent of ion}} \cr & = \frac{{{\text{Charge}}\,\,{\text{on the }}ion}}{{{\text{Total}}\,\,{\text{charge}}}} \cr} $$
$$\therefore \,\,Eq\,{\text{of}}\,$$   $$ = \frac{2}{2} = 1$$
$$Eq\,{\text{of}}\,N{a^ + } = \frac{1}{2},Eq\,{\text{of}}\,\,{K^ + } = \frac{1}{2}$$
$$\therefore {\lambda ^ \circ }_{eq}$$ 
$$ = {\lambda ^ \circ }_{eq}$$  $$ + \frac{1}{2}{\lambda ^ \circ }N{a^ + } + \frac{1}{2}{\lambda ^ \circ }{K^ + }$$
$$\eqalign{ & = 74 + \frac{{50}}{2} + \frac{{73}}{2} \cr & = 135.5\,oh{m^{ - 1}}\,c{m^2}e{q^{ - 1}} \cr} $$

Inert electrode does not participate in the chemical reaction and acts only as source or sink for electrons and provides surface either for oxidation or for reduction reaction.

The oxidation state shows a change only in (D)

According to Faraday’s first law of electrolysis
$$\eqalign{ & W = \frac{{E \times i \times t}}{{96500}} \cr & {\text{Where }}E{\text{ = equivalent weight}} \cr & = \frac{{mol.{\text{ }}mass{\text{ of metal}}(M)}}{{{\text{oxidation state of metal }}(x)}} \cr & {\text{Substituting the value in the formula}} \cr & W = \frac{M}{x} \times \frac{{i \times t}}{{96500}} \cr & {\text{or}}\,\,x = \frac{M}{W} \times \frac{{i \times t}}{{96500}} \cr & = \frac{{10 \times 2 \times 60 \times 60}}{{96500 \times 0.250}} \cr & = 3 \cr & \left[ {{\text{Given : no}}{\text{. of }}moles = \frac{M}{W} = 0.250} \right] \cr & {\text{Hence oxidation state of metal is}}\left( { + 3} \right) \cr} $$

Reducing character decreases down the series. Hence the correct order is
$$Al < F{e^{2 + }} < B{r^ - }$$

Due to reduction of $$NO_3^ - $$  in preference to $${H^ + }$$ ion. $${H^ + }$$ ion is not reduced to give $${H_2}$$ gas.