1

### JEE Main 2017 (Online) 8th April Morning Slot

Consider the following standard electrode potentials (Eo in volts) in aqueous solution :

Element M3+ /M M+ /M
A1 -1.66 + 0.55
T1 +1.26 - 0.34

Based on these data, which of the following statements is correct ?
A
T1+ is more stable than A13+
B
A1+ is more stable than A13+
C
T1 + is more stable than A1+
D
T13+ is more stable than A13+
2

### JEE Main 2017 (Online) 9th April Morning Slot

To find the standard potential of M3+/M electrode,the following cell is constituted : Pt/M/M3+(0.001 mol L−1 )/Ag+(0.01 mol L−1 )/Ag

The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3+ + 3e$\to$ M at 298 K will be :

(Given $E_{A{g^ + }\,/\,Ag}^ -$ at 298 K = 0.80 Volt)
A
0.38 Volt
B
0.32 Volt
C
1.28 Volt
D
0.66 Volt
3

### JEE Main 2018 (Offline)

How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane?
(Atomic weight of B = 10.8 u)
A
1.6 hours
B
6.4 hours
C
0.8 hours
D
3.2 hours

## Explanation

Required reaction :

B2H6 + 3O2 $\to$ B2 O3 + 3 H2 O

Here molar mass of B2H6 =10.8 $\times$ 2 + 6 = 27.6 gm

Given weight of B2H6 = 27.66 g

$\therefore\,\,\,\,$No of moles of B2H6 = ${{27.6} \over {27.66}} \simeq 1$ mole.

For combustion of 1 mole B2H6 3 moles O2 required.

This 3 mole of O2 is obtained by electrolysis of H2O.

2H2O($l$) $\to$ O2 (g) + 4 H+ (aq) + 4 e$-$

moles $\times$ nf = ${{It} \over {96500}}$

Here moles of O2 = 3.

Nf of O2 = 4 (in H2 change of O = $-$2

and in O2 change of 0 = O.

So change in charge = 2 .

for two atoms of O2 change in charge = 2 $\times$ 2 = 4)

$\therefore\,\,\,\,$ 3 $\times$ 4 = ${{100 \times t} \over {96500}}$

$\Rightarrow \,\,\,\,$ t = 12 $\times$ 965 sec.

$\Rightarrow \,\,\,\,$ t = ${{12 \times 965} \over {60 \times 60}}$ hr

= 3.2 hr
4

### JEE Main 2018 (Online) 15th April Morning Slot

When an electric currents passed through acidified water, 112 mL of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is :
A
1.0
B
0.5
C
0.1
D
2.0

## Explanation

Reaction at cathode :

2H+ + 2e$-$ $\to$ H2

We know,

$\omega$ = zIt = ${{EIt} \over {96500}}$
<
no. of moles of H2 = ${{112} \over {22400}}$

$\therefore\,\,\,$ mass (w) of H2 = ${{112} \over {22400}}$ $\times$ 2

$\therefore\,\,\,\,$ ${{112} \over {22400}}$ $\times$ 2 = ${{1 \times I \times 965} \over {96500}}$

$\Rightarrow$$\,\,\,\,$ $I$ = 1 A