Calculate the quantity of electricity (i.e. charge) delivered by a Daniel cell initially containing 1L each of ion and which is operated until its potential drops to 1V.
(Given : \(E_{Zn2+/Zn}^0 = -0.76V; E_{Cu2+/Cu}^0 = +0.34V\))
(1) 0.029 x 105C
(2) 2.239 x 105C
(3) 1.92 x 105C
(4) 3.123 x 105C
Some half cell reaction & their standard potential are given which combination would result in a cell with the largest potential.
(A) i and ii
(B) ii and iii
(C) iv and v
(D) ii and v
A hydrogen electrode is immersed in a solution with pH = 0 (HCl). By how much will the potential (reduction) change if an equivalent amount of NaOH is added to the solution.
(1) increase by 0.41 V
(2) increase by 59 mV
(3) decrease by 0.41 V
(4) decrease by 59 mV
The solubility product of silver iodide is and the standard potential (reduction) of Ag, electrode is + 0.800 volts at The standard potential of Ag, AgI/ electrode (reduction) from these data is-
(1) – 0.30 V
(2) + 0.15 V
(3) + 0.10 V
(4) – 0.15 V
How much time is required for the complete decomposition of 2 moles of water using a current of 2 ampere-
1. 26.805 h
2. 53.61 h
3. 107.22 h
4. None of these
The standard redox potentials Eo of the following systems are-
System | Eo(Volts) |
\(MnO_{4}^{-}+8 H^{+}+5 e \rightarrow M n^{2+}+4 H_{2}O\) | 1.51 |
Sn4+ + 2e- → S n2+ | 0.15 |
Cr2O72- + 14H+ + 6e- → 2Cr3+ + 7H2O | 1.33 |
C e4+ + e- → Ce3+ | 1.61 |
The oxidizing power of the various species decreases in the order-
1.
2.
3.
4.
Calculate the maximum work that can be obtained from the Daniell cell given below -
Given that
(1) – 212300 J
(2) – 202100 J
(3) – 513100 J
(4) – 232120 J
The metal that cannot be produced on reduction of its oxide by aluminium is :
(1) K
(2) Mn
(3) Cr
(4) Fe
We have taken a saturated solution of AgBr, Ksp of AgBr is 12 × 10-14. If 10-7 mole of AgNO3 is added to 1 litre of this solution then the conductivity of this solution in terms of 10-10 Sm-1 units will be:
1. 39
2. 55
3. 15
4. 41
\(Z n →Z n^{2+}+2 e ; E^{\circ}=+0.76 \mathrm{~V}\)
\(F e → F e^{2+}+2 e ; E^{\circ}=+0.41 \mathrm{~V}\)
1. – 0.35 V
2. + 0.35 V
3. + 1.17 V
4. – 1.17 V