The following electrochemical cell has been set upPt(1) | Fe3+, Fe2+ (a = 1) || Ce4+, Ce3+ (a = 1) | Pt (2)E°(Fe2+) = 0.77 V, and E°(Ce4+,Ce3+) = 1.61 VIf an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will the current increase or decrease with time? from Chemistry Electrochemistry Class 12 Nagaland Board
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The following electrochemical cell has been set up
Pt(1) | Fe3+, Fe2+ (a = 1) || Ce4+, Ce3+ (a = 1) | Pt (2)
(Fe2+) = 0.77 V, and E°(Ce4+,Ce3+) = 1.61 V
If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will the current increase or decrease with time?


For the electrochemical cell
Pt(1) | Fe3+, Fe2+ (a = 1) || Ce4+, Ce3+ (a = 1) | Pt (2)
the cell regions are

  Right-half cell: reduction
                     Ce4++e-   Ce3+
 
Left-half cell: oxidation
                      Fe2+   Fe3++e-
______________________________
Add Ce4++Fe2+ Ca3++Fe3+

The net cell potential is
E° Cell = E° R – E° L = 1.61 V – 0.77 V = 0.84 V.
Since E°cell is positive, the cell reaction will be spontaneous.
The current in the external circuit will flow from Pt (1) (which serves as anode to Pt(2) which serves as cathode.

With the passage of time, Ecell will decrease and so is the current in the external circuit.

220 Views

Can you store copper sulphate solutions in a Zinc pot?

Answer:

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)  Ni2+ (0.160 M) + 2Ag(s)
Given that EoCell = 1.05 V

Answer:

Ni(s)+2Ag+(aq)Ni2+(aq)+2As(s)
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.

1585 Views

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

Answer:

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
1279 Views

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

Answer:


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 .

879 Views

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

Answer:

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