What is the Nernst equation for the Potential of an electrode? Can Nernst equation be applied to the cell relation ? Apply this equation to a general reaction.aA+bB→cC+dD from Chemistry Electrochemistry Class 12 Nagaland Board
Zigya Logo

Chapter Chosen


Book Chosen

Chemistry I

Subject Chosen


Book Store

Download books and chapters from 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
What is the Nernst equation for the Potential of an electrode? Can Nernst equation be applied to the cell relation ? Apply this equation to a general reaction.

The concentration of all species involved in the species involved in the electrode reaction is unity.This need not be always true.
Nernst shows that for the electrode reaction:
Mn+(aq) + ne- M(s)
the electrode potential at any concentration measured with respect to standard hydrogen electrode can be represented by:
EMn+M = EMn+M- - RTnFIn [M][Mn+]
but concentration of solid M is taken as unity as we have
EMn+M = EMn+M- - RTnFIn [1][Mn+]

R is gas constant (8.314 JK–1 mol–1),
F is Faraday constant (96487 C mol–1), T is temperature in kelvin and [Mn+] is the concentration of the species, Mn

Let us take a electrode reaction
The Nernst equation of this electrode
E = E°-2.303 RTnFlogaproductareactant
Instead of activity, we can take molar concentration.
E = E°-0.05916nlogZnZn2+
For pure solid and liquid molar concentration is taken as unity.
E = E°-0.059162log1Zn2+
Yes,  Nernst equation can be applied to the cell reaction.

E = E°-0.05916nlogCcDdAa Bb



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-EFe3+Fe2+0 = 0.77V
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 .


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

EH+/H2  = EoH+/H2 - RTnFlnH2[H+]2              = 0-8.314×2982×96500ln IH+2              = 0.05915 log [H+]              =-0.05915 pH             =-0.05915 × 10 =- 0.59 V

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
Eocell = Eo Mg|Mg2+
Eo Mg|Mg2+= Eocell

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)

Calculate the emf of the cell in which the following reaction takes place:
Ni(s) + 2Ag+ (0.002 M)  Ni2+ (0.160 M) + 2Ag(s)
Given that EoCell = 1.05 V


Ecell = Eocell + RT2FlnAg+(aq)2Ni2+ (aq)
or      Ecell = Eocell+0.0592 logAg+(aq)2Ni2+ (aq)
The equation is also written as

or    Ecell = Eocell-0.0592logNi2+(aq)Ag+ (aq)2         = 1.05 V - 0.0295 log 0.1600.002  

= 1.05 V – 0.0295 x log 80

= 1.05 V – 0.0295 x 1.9031

= 1.05 V – 0.056 = 0.99 V.