Home
Class 12
CHEMISTRY
E^(c-) of some elements are given as : ...

`E^(c-)` of some elements are given as `:`
`{:(I_(2)+2e^(-)rarr 2I^(c-),,,,E^(c-)=0.54V),(MnO_(4)^(c-)+8H^(o+)+5e^(-)rarr Mn^(2+)+4H_(2)O,,,,E^(c-)=1.52V),(Fe^(3+)+e^(-)rarr Fe^(2+),,,,E^(c-)=0.77V),(Sn^(4+)+2e^(-)rarr Sn^(2+),,,,E^(c-)=0.1V):}`
`a.` Select the strongest reductant and weakest oxidant among these elements.

Text Solution

Verified by Experts

The correct Answer is:
a

`Zn` has highest tendency to get oxidized amongst `Cr, H_(2),` and `Fe`. So it is the strongest reducing agent. Since the standard reduction potential of `(E^(c-)._(Zn^(2+)|Zn))(-0.76V)` is lowest or standard oxidation of `E^(c-)._(Zn|Zn^(2+))(+0.76V)` is highest.
Promotional Banner

Similar Questions

Explore conceptually related problems

The following half reaction occur in a galvanic cell Fe^(2+) rarr Fe^(3+)+e^(-) E^(@)= 0.77V MnO_(4)^(-)+8H^(+)+5e^(-) rarr Mn^(2+)+4H_(2)O , E^(@)=1.49V The e.m.f. of the cell is :

The electrode potential of some half cell reactions are : Cl_(2) + 2e^(-) rarr 2 Cl^(-)(aq), E^(@) = + 1.36 V O_(2)+2H^(+)(aq) + 2e^(-) rarr H_(2)O_(2) (aq) , E^(@) = + 0.65 V F_(2)+2e^(-) rarr 2F^(-)(aq), E^(@) = + 2.87 V H_(2)O+ 2e^(-) rarr H_(2)(g) + 2OH^(-) (aq), E^(@) = - 0.83 V The strongest oxidising agent is :

Given below are the half cell reactions : Mb^(2+)+2e^(-) rarr Mn, E^(@)= -1.18 V 2(Mn^(3+)+e^(-)rarr Mn^(2+)), E^(@)= +1.51V The E^(@) for 3Mn^(2+) rarr Mn + 2 Mn^(3+) will be

Standard electrode potential data are useful for understanding the stability of an oxidant in a redox titration . Some half reactions and their standard potentials are given below : MnO_(4)^(-)(aq)+8H^(+)(g)+5e^(-)toMn^(2+)(aq)+4H_(2)O(l)E^(@)=1*51V Cr_(2)O_(7)^(2-)(aq)+14H^(+)aq)+6e^(-)to2Cr^(3+)(aq)+7H_(2)+(l)E^(@)=1*38V Fe^(3+)(aq)e^(-)toFe^(2+)(aq) " " E^(@)=0*77V Cl_(2)(g)+2e^(-)to2Cl^(-)(aq) " " E^(@)=1*40V Identify the only incorrect statement regarding the quantitative estimation of aqueous Fe(NO_(3))_(2) .

lim_(x rarr 0) (e^(x) + e^(-x) -2)/x^(2)=

The rusting of iron takes place as follows : 2H^(+) + 2e^(-) + (1)/(2)O_(2)(g) rarr H_(2)O (l) E^(@) = + 1.23 V Fe^(2+) + 2e^(-) rarr Fe(s) E^(@) = - 0.44 V Calculate Delta G^(@) for the net process