﻿ Calculate the cell e.m.f. at 25° C for the following cell:Mg(s) | Mg2+ (0.01 M) || Sn2+ (0.1 M) | Sn (s)[Given E°Mg2+/Mg = – 2.34 V, E°Sn2+/Sn = – 0.136V, 1 F = 96,500 C mol–1]Calculate the maximum work that can be accomplished by the operation of this cell. from Chemistry Electrochemistry Class 12 Nagaland Board

## Book Store

Currently only available for.
CBSE Gujarat Board Haryana Board

## Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12
Calculate the cell e.m.f. at 25° C for the following cell:
Mg(s) | Mg2+ (0.01 M) || Sn2+ (0.1 M) | Sn (s)
[Given E°Mg2+/Mg = – 2.34 V, E°Sn2+/Sn = – 0.136V, 1 F = 96,500 C mol–1]
Calculate the maximum work that can be accomplished by the operation of this cell.

Mg(s) | Mg2+ (0.01 M) || Sn2+ (0.1 M) | Sn(s)
Mg(s) → Mg2+ (aq) + 2e
(oxidation at anode)
Sn2+(aq) + 2e → Sn(s)
(reduction at cathode)

684 Views

How would you determine the standard electrode potential of the system Mg2+/Mg?

Use standard hydrogen electrode as anode and Mg2+ | Mg as a cathode we can measure the standard electrodepotential of systemMg2+ | Mg. Standard hydrogen electrode, represented by Pt(s), H2(g) (1 atm) | H+ (aq) and dip the electrode of Magnesium wire in a 1M MgSO4 solution .The standard hydrogen electrode is always zero.
Use formula
Eocell = Eo right  – Eoleft
The standard hydrogen electrode is always zero.
So that the value of
Eoleft =0
Hence
Eocell = Eo Mg|Mg2+
Or
Eo Mg|Mg2+= Eocell
2203 Views

Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.

For hydrogen electrode, ,
given that      pH = 10
use formula [H+] = 10–pH
so that  [H+] = 10−10 M
Electrode reaction will
H+  + e –  →1/2 H2
Use the formula

1279 Views

Can you store copper sulphate solutions in a Zinc pot?

No. Because zinc is more reactive than copper and thus holes will be developed in zinc pot.

Cu2+(aq) + Zn(s) → Zn2+ (aq) + Cu(s)
1384 Views

Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0.002 M) $\to$ Ni2+ (0.160 M) + 2Ag(s)
Given that

$\mathrm{Ni}\left(\mathrm{s}\right)+2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Ni}}^{2+}\left(\mathrm{aq}\right)+2\mathrm{As}\left(\mathrm{s}\right)$

or

The equation is also written as

or

= 1.05 V – 0.0295 x log 80

= 1.05 V – 0.0295 x 1.9031

= 1.05 V – 0.056 = 0.99 V.

1585 Views

Consult the table of standard electrode potentials and suggest three substance that can oxidize ferrous ions under suitable conditions.