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Write the Nernst equation and emf of the following cells at 298 k.
Pt(s) | Br2 (l) | Br (0.010 M) || H+ (0.030 M) | + H2(g) (1 bar) | Pt(s).

For given cell Anode reaction:

2Br-(aq)  Br2(l) + 2e-

Cathode reaction:

2H+(aq) + 2e-  H2(g)

Overall cell reaction:
                     2Br-(aq) + 2H+(aq)  Br2(l) + H2(g)

Here,  n = 2,  

E0cell = E0cathode -E0anode = 0- 1.08 V = -1.08 V.

The Nernst equation for Ecell and 298 k can be written as:

Ecell  =E0cell - 0.059nlog Br2+ H2Br-H+2       = 1.08 - 0.0592 log 1[10-2]2 [ 3 × 10-2] 2        =-1.08 - 0.0295 log 110-4×9×10-4        = -1.08 - 0.0295 (log 108 - log 9)         = - 1.08 -  0.0295 (7.0458)          = -1.08- 0.2078 = -1.2878 V.

The negative value of E
cell indicates the cell has been arranged in a reverse way, i.e., hydrogen electrode will act as anode and bromine electrode act as cathode. The cell should be represented as Pt | H2 (1 bar), H+ (0.03 M) || Br (0.01 M) | Br2(l), Pt
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Depict the galvanic cell in which the reaction
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.


The electrochemical cell can be depicted as

The electrochemical cell can be depicted as
(ii) Electrons move from

(ii) Electrons move from anode (zinc electrode) to cathode (silver electrode) in the external circuit. Zinc ions go into solution at anode and Agions get deposited at cathode. Thus electrons in the external and metal ions in the internal circuit act as carrier of current in.
(iii) Overall reaction is obtained by anode and cathode reactions.
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)

299 Views

Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of AgNO3 with platinum electrodes.
(iii) A dilute solution of H2SO4 with platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.



At cathode: Ag+ are preferably discharged as compared to H+ as its
At cathode: Ag+ are preferably discharged as compared to H+ as its reduction potential is higher
Ag to the power of plus plus straight e to the power of minus space rightwards arrow space Ag left parenthesis straight s right parenthesis
At anode: Silver electrode (being reactive) dissolves to produce Ag+

At cathode: Ag+ are preferably discharged as compared to H+ as its

As the electrodes are inert, these do not take part in the reaction. At cathode, Ag+ ions are discharged in preference to H+ ions and OH are discharged in preference to NO3 ions at anode

At cathode: Ag+ are preferably discharged as compared to H+ as its
Here, only H+ are discharged at cathode as there is no other positive ion. Out of OH and SO42– ions, OH ions are discharged at anode preferably

At cathode: Ag+ are preferably discharged as compared to H+ as its

The platinum electrodes are inert and do not take part in the reaction. The following reactions occur at platinum (inert) electrodes.

At cathode: Ag+ are preferably discharged as compared to H+ as its

174 Views

How much charge is required for the following reductions:
 1 mol of Al3+ to Al?


Al3++3e-Al

Formula required charge n × F
n = difference of charge on ions   
F is constant and equal to 96487 Coulombs
Here n = 3
Hence required charge = 3 × 96487 Coulombs
                                  = 289461 Coulombs
                                  = 2.89  ×10 –5 Coulombs
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Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variaion with concentration.


Conductivity: The reciprocal of resistance of an electrolyte in aqueous solution is known as its conductivity. It is equal to 1 over straight R.
Molar Conductivity: Molar conductivity of a solution is defined as the conductance of all ions present in one mole of electrolyte in the solution. If M is the molar concentration in mol L–1, then.

Conductivity: The reciprocal of resistance of an electrolyte in aqueo

Fig: Molar conductivity versus C1/2 for acetic acid (week electrolyte) and potassium chloride (strong electrolyte in aqueous solutions)
The curve shown below gives the change in conductance against square root of concentrations. We observe that for strong electrolytes like KCl, the conductance does not change much with decrease in square root of straight C semicolon whereas in the case of weak electrolyte like acetic acid (CH3COOH) it increases much with decrease in square root of straight C.
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