12 - Electrochemistry Questions Answers

How Is the value of n calculated for nerst equation??

 

Asked By: AAYUSH TANEJA
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Joshi sir comment

n is the no. of electrons in anode or cathode reaction.

If no. of electrons are different then make these no. common

for ex. if reactions are 

             Al ---------------> Al+++ + 3e-

and     Cu++ + 2e- ------------> Cu

then multiply (1) to 2 and (2) to 3

after that electrons will be 6 in both reactions so n = 6

consider two half cells based on the Rxn

Ag+(aq) +e- ----> Ag(s)

the left half cell contain Ag+ ions at unit concn , &the right half cell initially had the same concn of Ag+ ions , but just enough NaCl (aq) had been added to completely precipitate the Ag+(aq) as AgCl . if emf ofthe cell is 0.29 V , then log10Ksp would have been ??                                                                                                                                    ( ANS -9.804)

(AMU 2012)

Asked By: SARIKA SHARMA
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Joshi sir comment

Ecell = EC - E- (0.0591/1) logs

on taking the values 0.29 = -0.0591 logs

or  logs = -0.29/0.0591 = -4.907

now Ksp = s2

so logKsp = 2logs = -9.8

 

 

Emf of the cell Ni/Ni+2||Au+3/Au at 298 K will b?    [EoNi/Ni+2 = 0.25 V,EoAu/Au+3= -1.5V]

Asked By: AYANSH
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Joshi sir comment

 

Ecell = E0cell - (0.0591/n)  log {[Ni++]3 / [Au+++]2}

Reactions are 3Ni ------> 3Ni++   +   6e-

and      2Au+++ +   6e- ---------->   2Au  

now put E0cell = E0c - E0= 1.5 - (-0.25)      here you should remember that you have to take only reduction potentials not oxidation potentials

                                               = 1.75

n = 6  and concentrations if given otherwise take these two equal to zero.

Q IN THAT Fe QUES ANSWER GIVEN IS -VE 0.405 BUT ACC. TO THIS METHOD ANSWER DO NOT MATCH

Q I STILL DIDNOT  UNDERSTAND  CO & CO2 RELATION

Asked By: SARIKA
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Joshi sir comment

 

Fe                ----->  Fe++  + 2e-         

so E for the new reaction = [1F(0.77) + 3F(0.04)]/2F = 0.89/2 = 0.445

so it will be -0.445 for the reverse, you are asking for reduction and i submitted the answer for oxidation so the duality created.

 

according to the given reaction we get 2L monooxide by 1L dioxide,

initially litres of CO2  = x so it will form 2x litres of CO

and 1-x litres CO are present from the start so total litres of CO = 2x+1-x = 1+x

 

 IN QUES OF RXn Fe 2+  + 2e- ---> Fe  THE E* GIVEN IS -VE 0.405 &I WANT TO KNOW IN THE LAST STEP WHY3/2F(0.4) IS USED

IN RXn CO2 +C---> 2CO IN INITIAL WE TAKE XFOR CO2 BUT  FOR FINAL Y 2X +1 -X=1.2 IS USED.

Asked By: SARIKA
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Joshi sir comment

for any reaction obtained by other two reactions formula used will be nE = n'E' + n''E''

 

in second question 2x is used because x litre CO2 will give 2x litre CO 

 

sub atomic particle

Asked By: PANKAJKUMARSHARMA
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Joshi sir comment

Very soon, we will submit

 

Q  what vol. of O2 OBTAINED AT ANODE AT STP IF Aq SOLn OF CuSO4 IS ELECTROLYSED BY PASSG 100A CURRENT FOR 60 MIN? ANS 20.89L

Asked By: SARIKA
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Joshi sir comment

Total charge flown in the circuit = 100*60*60 = 360000 coulomb

so no of faradays of electricity = 360000/96500 = 3600/965 = 3.73 F

so amount of O2 = 3.73*5.6 = 20.89 litre

Here 5.6 litre is eq. volume corresponding to 8 gm. of O2

The change in standard gibbs free energy of given cell at NTP

Zn I Zn(2+) II OH- I Ag2O I Ag         Eo =1.104V

Asked By: NIKHIL VARSHNEY
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Joshi sir comment

Zn  ----->    Zn++   + 2e-

Ag2O  +  H2O + 2e-  ------>    2Ag  +  2OH-

so Gibbs free energy change = nFE = 2*96500*1.104

A CURRENT OF 0.25 A  IS PASSED THROUGH 400ML OF A2.0 M SOLN OF NaCl FOR 35 MINUTES.WHAT WILL BE THE PH OF THE SOLN AFTER THE CURRENT IS SWITCHED OFF?

Asked By: NIKHIL VARSHNEY
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Joshi sir comment

q = it = 0.25*35*60 = 525 coul. 

so no. of eq.( of H2)  formed = 525/96500 = 21/3860, same will be the no. of moles

total initial no. of moles of NaCl = 2*0.4 = 0.8 moles

remaining no. of moles of H+ = 0.8 - (21/3860)

so pH = -log[H+

N2 gas is formed by electrolysis of magnesium nitride .The volume of one equivalent of N2 gas evolved at NTP is

Asked By: NIKHIL VARSHNEY
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Joshi sir comment

Mg3N2    −−−−>     3 Mg + N2

so n factor of N = 3 , for N2 it will be 6, so one equivalent of N2 = 22.4/6 = 11.2/3         

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