1

### AIPMT 2006

A hypothetical electrochemical cell is shown below.

$A\left| {{A^ + }\left( {xM} \right)} \right|\left| {{B^ + }\left( {yM} \right)} \right|B$

The emf measured is + 0.20 V. The cell reaction is
A
A + B+ $\to$ A+ + B
B
A+ + B $\to$ A + B+
C
A+ + e$-$ $\to$ A;  B+ + e$-$ $\to$ B
D
the cell reaction cannot be predicted.

## Explanation

From the given expression:

At anode : A $\to$ A+ + e

At cathode : B+ + e $\to$ B

Overall reaction is : A + B+ $\to$ A+ + B
2

### AIPMT 2006

EoFe2+/Fe = $-$ 0.441 V and EoFe3+/Fe2+ = 0.771 V, the standard EMF of the reaction Fe + 2Fe3+ $\to$ 3Fe2+ will be
A
0.111 V
B
0.330 V
C
1.653 V
D
1.212 V

## Explanation

At anode : Fe $\to$ Fe2+ + 2e-; Eo = -0.441 V

At cathode : Fe3+ + e $\to$ Fe2+; Eo = 0.771 V

Fe + 2Fe3+ $\to$ 3Fe2+; Eo = ?

To get the above equation, (ii) × 2 – (i)

2Fe3+ + 2e $\to$ 2Fe2+; Eo = 0.771 V

Fe $\to$ Fe2+ + 2e-; Eo = - 0.441 V
------------------------------------------------
Fe + 2Fe3+ $\to$ 3Fe2+

Eo = 0.771 + 0.441 = 1.212 V
3

### AIPMT 2005

The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270 kg of aluminium metal from bauxite by the Hall process is
(Atomic mass : Al = 27)
A
270 kg
B
540 kg.
C
90 kg
D
180 kg

## Explanation

2Al2O3 + 3C $\to$ 4Al + 3CO2

From the above equation,

3 mol × 12 g mol–1 = 36 g of carbon is consumed

to give 4 mol × 27 g mol–1 = 108 g of Al

So, 108 g of Al is produced by 36 g of carbon

$\therefore$ 270000 g of Al is produced by

= ${{36} \over {108}} \times 270000$ of C

= 90000 g of C

= 90 kg of C
4

### AIPMT 2005

4.5 g of aluminium (at. mass 27 amu) is deposited at cathode from Al3+ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H ions in solution by the same quantity of electric charge will be
A
44.8 L
B
22.4 L
C
11.2 L
D
5.6 L

## Explanation

${{{M_{Al}}} \over {{M_H}}} = {{{E_{Al}}} \over {{E_H}}}$

$\Rightarrow$ ${{4.5} \over {{M_H}}} = {{{{27} \over 3}} \over 1}$

$\Rightarrow$MH = 0.5 g

We know, Volume of 2 g H2 at STP = 22.4 L

$\therefore$ Volume of 0.5 g H2 at STP

= ${{22.4 \times 0.5} \over 2}$ = 5.6 L