Login
Get Advice on Live Video Call
Earn $ Cash $ with
consultations on Bissoy App

Redox reactins play a pivoted role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which

Monjur Alahi

Redox reactins play a pivoted role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below is a set of half-cell reactions (acidic medium) along with their `E^(@)` values with respect to normal hydrogen electrode. Using this data, obtained the correct explanations to questions `15-16`
`I_(2)+2e^(-)to2I^(-),E^(@)=1.36V`
`Mn^(3+)+e^(-)toMn^(2+),E^(@)=1.50V`
`Fe^(3+)+e^(-)toFe^(2+),E^(@)=0.77V`
`O_(2)+4H^(+)+4e^(-)to2H_(2)O,E^(@)=1.23V`
Q. Electrolysis of dilute aqueous `NaCl` solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mole of `H_(2)` gas at the cathode is: (1 faraday `=96500C`)
A. `9.65xx10^(4)sec`
B. `19.3xx10^(4)sec`
C. `28.95xx10^(4)sec`
D. `38.6xx10^(4)sec`


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - B
Faraday law equivalents of `H_(2)` produced `=(Ixxt(sec))/(96500)`
`0.01xx2=(10xx10^(3)xxt)/(96500)=96500xx2=t`
`19.3xx10^(3)sec=t`

0 like 0 dislike
1 views
The standard reduction potentials of some half cell reactions are given below: `PbO_2+4H^++2e^(-)hArrPb^(2+)+2H_(2)O , E^@=1.455 V` `MnO_4^(-)+8H^++5e

Correct Answer - A,B,C On basis of given SRP values

2 Answers 1 views
The standard potential for the reaction, `Ag^(+)(aq.)+Fe^(2+)(aq.)toFe^(3+)(aq.)+Ag(s)` is 0.028V. What is the standard free energy change for this re

`-2.702kJ" "mol^(-1)`

2 Answers 1 views
What is the single electrode potential of a half-cell for zinc electrode dipping in 0.01 M `ZnSO_(4)` solution at `25^(@)C` ? The standard electrode p

We know that, `E_(o x)=E_(o x)^(@)-0.0591/n log_(10) ["ion"]` Putting the value of `E_(o x)^(@)=0.763 V, n=2` and `[Zn^(2+)]=0.01 M`, `E_(o x)=0.763-0.0591/2 log_(10) [0.01]` `=0.763-0.02955xx(-2)` `=(0.763+0.0591)` volt `=0.8221` volt

2 Answers 1 views
The standard oxidation potential potential of zinc is 0.76 volt and of silver is - 0.80 volt. Calculate the emf of the cell : `Zn|underset(0.25 M)(Zn(

The cell reaction is `Zn+2Ag^(+) rarr 2Ag+Zn^(2+)` `E_(o x)^(@)` of `Zn =0.76` volt `E_("red")^(@)` of `Ag=0.80` volt `E_(cell)^(@)=E_(o x)^(@)` of `Zn+E_("red")^(@)` of `Ag=0.76+0.80=1.56` volt We know that, `E_(cell)=E_(cell)^(@)-0.0591/n"log" (["Products"])/(["Reactants"])` `=E_(cell)^(@)-0.0591/2"log"...

2 Answers 5 views
Prove that for two half reactions having potentials `E_(1)` and `E_(2)` which are combined to yield a third half reaction, having a potential `E_(3)`,

`DeltaG_(3)=DeltaG_(1)+DeltaG_(2)` `-n_(3)FE_(3)=-n_(1)FE_(1)-n_(2)FE_(2)` or `n_(3)E_(3)=n_(1)E_(1)+n_(2)E_(2)` or `E_(3)=(n_(1)E_(1)+n_(2)E_(2))/n_(3)`

2 Answers 1 views
The standard oxidation potential for the half-cell `NO_(2)^(-)(g)+H_(2)OtoNO_(3)^(-)(aq)+2H^(+)(aq)+2e` is `-0.78V`. Calculate the reduction in 9 mola

Correct Answer - `0.836"volt", 1.1937"volt"` `E_("oxidation")=-0.78-(0.0591)/(2)log9^(2)=-0.78-(0.0591)/(2)xx2xxlog9=-0.835"volt"` `E_("reduction")=-E_("oxidation")=0.836"volt"` In neutral medium `E_("Oxidation")=-0.78-(0.0591)/(2)log(10^(-7))^(2)=-1.1937"volt"`

2 Answers 1 views
Using the standard electrode potential values given below, decide which of the statements, `I,II,III` and `IV` are correct. Choose the right answer fr

Correct Answer - C Lower standard reduction potential related metal ions can displace higher standard reduction potential related metals ions.

2 Answers 2 views
Calculate the standard cell potential of the galvanic cell in which the following reaction takes place: `2Cr(s)+3Cd^(2+)(aq)to2Cr^(3+)(aq)+3Cd(s)` Als

`E_(cell)^(ɵ)=E_(Cd^(2+)//Cd)^(@)-E_(Cr^(3-)//Cr)^(@)` `=-0.40-(-0.74)` `=-0.40+0.74=+0.34V` the half cell reaction are `2Cr(s)to2Cr^(3+)+6e^(-)` `underline(3Cd^(2+)+6e^(-)to3Cd(s))` `underline(2Cr(s)+3Cd^(2+)to2Cr^(3+)+3Cd(s))` `therefore` No. of electrons `n=6` `triangle_(r)G^(ɵ)=nFE^(ɵ)` `=-6xx96500xx0.34` `=-196.86kJmol^(-1)`

2 Answers 1 views
SOP : Create a mark list. The mark list should display: Fields as Name, Math, Physics, Chemistry, Biology, Total, Percentage.

Step 1: Open LibreOffice Calc and add fields like Name, Math, Physics, Chemistry, Biology, Total, Percentage. Step 2: Enter 10 records in it. Step 3: To Calculate Total Click inside the cell where...

2 Answers 51 views
A standard cell of emf 1.08 V is balanced by the potential difference across 91 cm of a meter long wire supplied by a cell of emf 2 V through a series

Correct Answer - `0.03Omegacm^(-1)`

2 Answers 1 views

Log in to ask questions, provide answers, or leave comments.


Login

Install the Bissoy app to consult with a doctor.


Download