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Subject: Chemistry
, asked on 22/5/18
The molar conductivity of a 0.02 M KCL solution is 149.2 S cm
^{2}
mol
^{-1. }
Calculate the value of the cell constant if the electrical resistance of the solution is 480 ohm.
Answer
1
Aparna Sahu
Subject: Chemistry
, asked on 20/5/18
Please don't send link again becoz I can't able to get it from link it's very confusing.
Solve this:
Q. 17. The hydrogen electrode is dipped in a solution of pH 3 at 25
$\xb0$
C. The potential would be (the value of 2.303 RT/F is 0.059 V)
1) 0.177 V
2) 0.087 V
3) 0.059 V
4) - 0.177 V
Answer
1
Aparna Sahu
Subject: Chemistry
, asked on 20/5/18
Solve this:
Q.17. The hydrogen electrode is dipped in a solution of pH 3 at 25
$\xb0$
C. The potential would be (the value of 2.303 RT/F is 0.059 V)
1) 0.177 V
2) 0.087 V
3) 0.059 V
4) - 0.177 V
Answer
1
Queens Land
Subject: Chemistry
, asked on 20/5/18
Solve this:
Q. Exactly 0.2 mole electrons passed through two electrolytic cells in series containing
$CuS{O}_{4}andZnS{O}_{4}$
, respectively. How many grams of each metal will be deposited on the respective cathodes
in the two cells?
Answer
1
Queens Land
Subject: Chemistry
, asked on 20/5/18
Calculate the ΔG° and equilibrium constant of the reaction at 27°C
$\mathrm{Mg}+{\mathrm{Cu}}^{+2}\rightleftharpoons {\mathrm{Mg}}^{+2}+\mathrm{Cu}\phantom{\rule{0ex}{0ex}}{\mathrm{E}}_{{\mathrm{Mg}}^{2+}/\mathrm{Mg}}^{0}=2.37\mathrm{V},{\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{0}=+0.34\mathrm{V}$
Answer
1
Queens Land
Subject: Chemistry
, asked on 19/5/18
Solve this
$\mathit{Q}\mathbf{.}\mathbf{}\mathbf{}Whichofthefollowingmetaldonotevolve{H}_{2}gasfromdil.acid?\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(1\right)A\left({E}_{{A}^{+}/A}^{\xb0}=0.25V\right)\left(2\right)B\left({E}_{B/{B}^{+}}^{\xb0}=+0.22V\right)\phantom{\rule{0ex}{0ex}}\left(3\right)C\left({E}_{{C}^{+}/C}^{\xb0}=-0.30V\right)\left(4\right)D\left({E}_{D/{D}^{+}}^{\xb0}=+0.44V\right)$
Answer
1
Queens Land
Subject: Chemistry
, asked on 19/5/18
Solve this:
Q. If the standard electrode potential of
$C{u}^{2+}$
/Cu electrode is 0.34 V, what is the electrode potential of 0.1 M concentration of
$C{u}^{2+}$
?
(1) 3.99 V
(2) 0.3105 V
(3) 0.222 V
(4) 0.176 V
Answer
1
Queens Land
Subject: Chemistry
, asked on 19/5/18
Please answer
Calculate the EMF of the cell
$Fe\left(s\right)+2{H}^{+}\left(1M\right)\to F{e}^{+2}\left(0.001M\right)+{H}_{2}\left(g\right)\left(1atm\right)\left(given:{E}_{F{e}^{2+}/Fe}^{\xb0}=-0.44V\right)$
Answer
1
Aaradhya Kondawar
Subject: Chemistry
, asked on 18/5/18
can u pls explain this Para, "the conductivity of all electrolytes increases with temperature. The conductivity of a solution varies slightly with pressure due to change in viscosity of the medium . The viscosity of a dilute solution decreases with increasing pressure this is accompanied by an increase in the equivalent conductivity provided the pressure is not too high the effect is most marked in a weak electrolyte than a strong electrolyte because in case of weak electrolyte increase of pressure tends to favour ionization but this factor is not appreciable in case of strong electrolytes."
^m or ^eq is directly proportional to pressure.
Answer
1
Aaradhya Kondawar
Subject: Chemistry
, asked on 18/5/18
What is meant by " Salt bridge avoids liquid junction potential which acts as internal resistance for the cell."?
Answer
1
Vaibhav Vikas Pawar
Subject: Chemistry
, asked on 18/5/18
Answer the following:
Why is fluorine gas the strongest oxidising agent while fluoride ion is the weakest reducing agent?
Why is lithium ion the weakest oxidising agent while lithium metal is the most powerful reducing agent in an aqueous?
Answer
3
Aparna Sahu
Subject: Chemistry
, asked on 18/5/18
Solve qus no. 6
Q.6. To obtain 2.70 g of silver (at. mass 108) from an aqueous solution of
$AgN{O}_{3}$
it is required to pass
A. 4750 coulomb
B. 12062 coulomb
C. 2412.5 coulomb
D. 241.5 coulomb
Answer
3
Deepshika Subramani
Subject: Chemistry
, asked on 17/5/18
Solve this :
$67.\mathrm{The}\mathrm{reduction}\mathrm{potential}\mathrm{at}\mathrm{pH}=14\mathrm{for}\mathrm{the}{\mathrm{Cu}}^{2+}/\mathrm{Cu}\mathrm{couple}\mathrm{is}[\mathrm{Given},\mathrm{E}{\xb0}_{{\mathrm{Cu}}^{2+}/\mathrm{cu}}=0.34\mathrm{V};{\mathrm{K}}_{\mathrm{sp}}\mathrm{Cu}(\mathrm{OH}{)}_{2}=1\times {10}^{-19}\left]\phantom{\rule{0ex}{0ex}}\right(\mathrm{a})0.34\mathrm{V}\phantom{\rule{0ex}{0ex}}(\mathrm{b})-0.34\mathrm{V}\phantom{\rule{0ex}{0ex}}(\mathrm{c})0.22\mathrm{V}\phantom{\rule{0ex}{0ex}}(\mathrm{d})-0.22\mathrm{V}$
Answer
1
Tulsi Mehta
Subject: Chemistry
, asked on 11/5/18
Plz explain. This part
Answer
2
Mitali
Subject: Chemistry
, asked on 10/5/18
Please solve and kindly tell..When and why do we take into account the pressure of the reacting specied in the nernst equation....??! Like we have in this question . Also are there any.other modifications in the nernst equation...??
$\mathbf{8}\mathbf{.}Considerthefollowingcellreaction,\phantom{\rule{0ex}{0ex}}2Fe\left(s\right)+{O}_{2}\left(g\right)+4{H}^{+}\left(aq\right)\to 2F{e}^{2+}\left(aq\right)+2{H}_{2}O\left(l\right),\phantom{\rule{0ex}{0ex}}E\xb0=1.67V\phantom{\rule{0ex}{0ex}}At\left[F{e}^{2+}\right]={10}^{-3}M,P\left({O}_{2}\right)=0.1atmandpH=3,thecellpotentialat25\xb0Cis\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(a\right)1.47V\left(b\right)1.77V\phantom{\rule{0ex}{0ex}}\left(c\right)1.87V\left(d\right)1.57V$
Answer
1
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What are you looking for?

^{2}mol^{-1. }Calculate the value of the cell constant if the electrical resistance of the solution is 480 ohm.Solve this:

Q. 17. The hydrogen electrode is dipped in a solution of pH 3 at 25$\xb0$C. The potential would be (the value of 2.303 RT/F is 0.059 V)

1) 0.177 V

2) 0.087 V

3) 0.059 V

4) - 0.177 V

Q.17. The hydrogen electrode is dipped in a solution of pH 3 at 25$\xb0$C. The potential would be (the value of 2.303 RT/F is 0.059 V)

1) 0.177 V

2) 0.087 V

3) 0.059 V

4) - 0.177 V

Q. Exactly 0.2 mole electrons passed through two electrolytic cells in series containing $CuS{O}_{4}andZnS{O}_{4}$, respectively. How many grams of each metal will be deposited on the respective cathodes

in the two cells?

$\mathrm{Mg}+{\mathrm{Cu}}^{+2}\rightleftharpoons {\mathrm{Mg}}^{+2}+\mathrm{Cu}\phantom{\rule{0ex}{0ex}}{\mathrm{E}}_{{\mathrm{Mg}}^{2+}/\mathrm{Mg}}^{0}=2.37\mathrm{V},{\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{0}=+0.34\mathrm{V}$

$\mathit{Q}\mathbf{.}\mathbf{}\mathbf{}Whichofthefollowingmetaldonotevolve{H}_{2}gasfromdil.acid?\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(1\right)A\left({E}_{{A}^{+}/A}^{\xb0}=0.25V\right)\left(2\right)B\left({E}_{B/{B}^{+}}^{\xb0}=+0.22V\right)\phantom{\rule{0ex}{0ex}}\left(3\right)C\left({E}_{{C}^{+}/C}^{\xb0}=-0.30V\right)\left(4\right)D\left({E}_{D/{D}^{+}}^{\xb0}=+0.44V\right)$

Q. If the standard electrode potential of $C{u}^{2+}$/Cu electrode is 0.34 V, what is the electrode potential of 0.1 M concentration of $C{u}^{2+}$?

(1) 3.99 V

(2) 0.3105 V

(3) 0.222 V

(4) 0.176 V

Calculate the EMF of the cell

$Fe\left(s\right)+2{H}^{+}\left(1M\right)\to F{e}^{+2}\left(0.001M\right)+{H}_{2}\left(g\right)\left(1atm\right)\left(given:{E}_{F{e}^{2+}/Fe}^{\xb0}=-0.44V\right)$

^m or ^eq is directly proportional to pressure.

Q.6. To obtain 2.70 g of silver (at. mass 108) from an aqueous solution of $AgN{O}_{3}$ it is required to pass

A. 4750 coulomb

B. 12062 coulomb

C. 2412.5 coulomb

D. 241.5 coulomb

Solve this :$67.\mathrm{The}\mathrm{reduction}\mathrm{potential}\mathrm{at}\mathrm{pH}=14\mathrm{for}\mathrm{the}{\mathrm{Cu}}^{2+}/\mathrm{Cu}\mathrm{couple}\mathrm{is}[\mathrm{Given},\mathrm{E}{\xb0}_{{\mathrm{Cu}}^{2+}/\mathrm{cu}}=0.34\mathrm{V};{\mathrm{K}}_{\mathrm{sp}}\mathrm{Cu}(\mathrm{OH}{)}_{2}=1\times {10}^{-19}\left]\phantom{\rule{0ex}{0ex}}\right(\mathrm{a})0.34\mathrm{V}\phantom{\rule{0ex}{0ex}}(\mathrm{b})-0.34\mathrm{V}\phantom{\rule{0ex}{0ex}}(\mathrm{c})0.22\mathrm{V}\phantom{\rule{0ex}{0ex}}(\mathrm{d})-0.22\mathrm{V}$

$\mathbf{8}\mathbf{.}Considerthefollowingcellreaction,\phantom{\rule{0ex}{0ex}}2Fe\left(s\right)+{O}_{2}\left(g\right)+4{H}^{+}\left(aq\right)\to 2F{e}^{2+}\left(aq\right)+2{H}_{2}O\left(l\right),\phantom{\rule{0ex}{0ex}}E\xb0=1.67V\phantom{\rule{0ex}{0ex}}At\left[F{e}^{2+}\right]={10}^{-3}M,P\left({O}_{2}\right)=0.1atmandpH=3,thecellpotentialat25\xb0Cis\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(a\right)1.47V\left(b\right)1.77V\phantom{\rule{0ex}{0ex}}\left(c\right)1.87V\left(d\right)1.57V$