1
IIT-JEE 2011 Paper 2 Offline
+3
-0.75
Consider the following cell reaction:
2Fe(s) + O2(g) + 4H+(aq) $$\to$$ 2Fe2+ (aq) + 2H2O (l); Eo = 1.67 V
At [Fe2+] = 10-3 M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25oC is
A
1.47 V
B
1.77 V
C
1.87 V
D
1.57 V
2
IIT-JEE 2010 Paper 1 Offline
+4
-1
The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is :
M(s) | M+ (aq ; 0.05 molar) || M+ (aq ; 1 molar) | M(s)
For the above electrolytic cell the magnitude of the cell potential | Ecell | = 70 mV.

For the above cell :
A
Ecell < 0 ; $$\Delta G > 0$$
B
Ecell > 0 ; $$\Delta G < 0$$
C
Ecell < 0 ; $$\Delta G^o > 0$$
D
Ecell > 0 ; $$\Delta G^o > 0$$
3
IIT-JEE 2010 Paper 1 Offline
+4
-1
The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is :
M(s) | M+ (aq ; 0.05 molar) || M+ (aq ; 1 molar) | M(s)
For the above electrolytic cell the magnitude of the cell potential | Ecell | = 70 mV.

If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would be :
A
35 mV
B
70 mV
C
140 mV
D
700 mV
4
IIT-JEE 2008 Paper 2 Offline
+3
-1

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H$$_2$$ gas at the cathode is (1 Faraday = 96500 C mol$$^{-1}$$].

A
9.65 $$\times$$ 10$$^4$$ sec
B
19.3 $$\times$$ 10$$^4$$ sec
C
28.95 $$\times$$ 10$$^4$$ sec
D
38.6 $$\times$$ 10$$^4$$ sec
EXAM MAP
Medical
NEET