Redox reactions play a pivotal 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.
For the cell: Pt `|H_(2)(0.4"atm")H^(+)(pH=1)||H^(+)(pH=2)|H_(2)(0.1"atm")|` Pt, the measured potential at `25^(@)C` is:
[Given: 2.303RT/F=0.06V, log 2=0.3]

For the cell Pt||H_2(0.4atm)|H^+(pH=1)||(H^+(pH=2)|H_2(0.1atm)|Pt The measured potential at 25^(@)C is

Redox reactions play a pivotal 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. E^(@)(SRP) of different half cells are given below: E_(Cu^(2+)|Cu)^(@)=0.34V, E_(Zn^(2+)|Zn)^(@)=-0.76V E_(Ag^(+)|Ag=0.8V, E_(Mg^(2+)|Mg))=-2.37V In which cell, DeltaG^(@) per "mole" of electron is most negative?

Redox reactions play a pivotal role I chemistry and bilogy . The values of standard redox potential (E^(@)) of two half - cell reaction decided which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper get deposited. Given below are a set of half 0 cell reactions ( acidic medium ) along with their E^(@) values ( with respect to normal hydrogen electrode ) Using this data : I_2 + 2e^(-) to 2I^(-) E^(@) = 0.54 , Cl_2 + 2e^(-) to 2Cl^(-) " "E=1.36V Mn^(3+) + e^(-) to Mn^(2+) E^(@) = 1.50 , Fe^(3+) + e^(-) to Fe^(2+) E = 0.77V O_2 +4H^(+) + 4e^(-) to 2H_2O" " E^(@) = 1.23 , While Fe^(3+) is stable , Mn^(3+) is not stable in acid solution because :

Redox reactions play a pivotal role I chemistry and bilogy . The values of standard redox potential (E^(@)) of two half - cell reaction decided which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper get deposited. Given below are a set of half 0 cell reactions ( acidic medium ) along with their E^(@) values ( with respect to normal hydrogen electrode ) Using this data : I_2 + 2e^(-) to 2I^(-) E^(@) = 0.54 , Cl_2 + 2e^(-) to 2Cl^(-) " "E=1.36V Mn^(3+) + e^(-) to Mn^(2+) E^(@) = 1.50 , Fe^(3+) + e^(-) to Fe^(2+) E = 0.77V O_2 +4H^(+) + 4e^(-) to 2H_2O" " E^(@) = 1.23 , Among the following , identify the correct statement :

Redox reactions play a vital role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cells reactions decide which way the reaction is expected to proceed. A simple example is a Daniell cell in which zince goes into solution and copper gets deposited. Given below are set of half-cell reactions (acidic medium ) along with their E^(@) in V with respect to normal hydrogen electrode values. {:(l_(2)+2e^(-)rarr2l^(-),E^(@)=0.54),(Cl_(2)+2e^(-)rarr2Cl^(-)" ",E^(@)=1.36),(Mn^(3+)+e^(-)rarrMn^(2+),E^(2)=1.50),(Fe^(3+)+e^(-)rarrFe^(2+)" ",E^(@)=0.77),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23):} while Fe^(3+) is stable, Mn^(3+) is not stable in acid solution because :