﻿ Calculate the potential (emf.) of the cellCd | Cd2+ (0.10 M) || H+(0.20 M) | Pt, H2 (0.5 atm)(Given : E° Cd2+ / Cd = – 0.403 V, R = 8.314 K–1 K–1 mol–1, F = 96500 C mol–1)  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 potential (emf.) of the cell
Cd | Cd2+ (0.10 M) || H+(0.20 M) | Pt, H2 (0.5 atm)
(Given : E° Cd2+ / Cd = – 0.403 V, R = 8.314 K–1 K–1 mol–1, F = 96500 C mol–1

Cell reaction will be

Equilibrium Constant from Nernst Equation: If the Daniell cell is short circuited, then we note that the reacton.
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
takes place and as time passes, the concentration of Zn2+ keeps on increasing while the concentration of Cu2+ keeps decreasing. At the same time, voltage of the cell as read on the voltmeter keeps on decreasing. After some time, we shall note that there is no change in the concentration of Cu2+ and Zn2+ ions and at the same time, voltmeter gives zero reading. This indicates that equilibrium has been attained. In this situation the Nernst equation may be written as

But at equilibrium [Zn2+] / [Cu2+] = K
and the above equation can be written as
(E° = 1.1 V)
K = 2 x 1037 at 298
In general E°cell = 2.303 RT / nF x log K

595 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 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

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

oxidation of ferrous ion means :

Fe2+--> Fe3+ +e-
Any substance which standard electrode potential is more than that of Fe+3 /F+2 can oxidise ferrous ions.
(refer to the table given in book)

The EMF of the substance whose reduction potentials greater than 0.77V will oxidised ferrous ion.
for example Br2, Cl2,and F2 .

879 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

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