- Organic Chemistry
- Aldehydes and Ketones
- Alkyl Halides, Alcohols and ethers
- Amines and other nitrogen compounds
- Aromatic Chemistry
- Carbohydrates, Amino acids, protein, Vitamin and Fat
- Carboxylic acids and its derivatives
- Chemistry in daily life
- General Mechanism in organic compounds
- Hydrocarbons
- Nomenclature and isomerism
12 - Electrochemistry Questions Answers
How Is the value of n calculated for nerst equation??
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)
Ecell = EC - EA - (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]
Ecell = E0cell - (0.0591/n) log {[Ni++]3 / [Au+++]2}
Reactions are 3Ni ------> 3Ni++ + 6e-
and 2Au+++ + 6e- ----------> 2Au
now put E0cell = E0c - E0a = 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
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.
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
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
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
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?
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
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