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The standard reduction potential E^(@) o...

The standard reduction potential `E^(@)` of the following systems are:-
`{:(System,E^(@)("volts")),((i)MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O,1.51),((ii)Sn^(4+)+2e^(-)rarrSn^(2+)0.15,),((iii)Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)rarr2Cr^(3+)+7H_(2)O,1.33),((iv)Ce^(4+)+e^(-)rarrCe^(3+)1.61,):}`
The oxidising power of teh various species decreases in the order

A

`Ce^(4+) gt Cr_(2)O^(2-_(7)) gt Sn^(4+) gt MnO^(-_(4)`

B

`Ce^(4+) gt MnO^(-_(4) gt Cr_(2)O^(2-_(7)) gt Sn^(4+)`

C

`Cr_(2)O^(2-_(7) gt Sn^(4+) gt Ce^(4+) gt MnO^(-_(4))`

D

`MnO^(-_(4) gt Ce^(4+) gt Sn^(4+) gt Cr_(2)O^(2-_(7)`

Text Solution

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The correct Answer is:
B
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For reaction : Cr_(2)O_(7)^(-2)+14H^(+)rarr2Cr^(+3)+7H_(2)O , How many e^(-)s are required

MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4H_(2)O , E^(@) = 1.51V MnO_(2) + 4H^(+) + 2e^(-) righarrow Mn^(2+) + 2H_(2)O E^(@) = 1.23V E_(MnO_(4)^(-)|MnO_(2)

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reaction and their standard potentials are given below: MnO_(4)^(-)(aq) +8H^(+)(aq) +5e^(-) rarr Mn^(2+)(aq) +4H_(2)O(l) E^(@) = 1.51V Cr_(2)O_(7)^(2-)(aq) +14H^(+) (aq) +6e^(-) rarr 2Cr^(3+)(aq) +7H_(2)O(l), E^(@) = 1.38V Fe^(3+) (aq) +e^(-) rarr Fe^(2+) (aq), E^(@) = 0.77V CI_(2)(g) +2e^(-) rarr 2CI^(-)(aq), E^(@) = 1.40V Identify the only incorrect statement regarding quantitative estimation of aqueous Fe(NO_(3))_(2)

Use the given standard reduction potentials to determine the reduction potential for this half-reactions. MnO_(4)^(-)(aq) + 3e^(-) + 4H^(+) rightarrow MnO_(2)(S) + 2H_(2)O(l)

Standard electrode potentials of few half-cell reactions are given below : {:(MnO_4^(-)+8H^(+)+5e^(-) to Mn^(2+)+ 4H_2O,,E^@=1.51V),(Cr_2O_7^(2-)+14H^(+)+6e^(-)to 2Cr^(3+)+7H_2O,,E^@=1.33V),(Fe^(3+)+e^(-)to Fe^(2+),,E^@=0.77V),(Cl_2+2e^(-) to 2Cl^(-),,E^@=1.36):} Based on the above information match the column I with column II and mark the appropriate choice.

For Cr_(2)O_(7)^(2)+14 H^(+)+6e^(-)rarr2Cr^(3+)+7H_(2)O E^(@)=1.33 V.At 298 k, [Cr_(2)O_(7)^(2-)] =4.5 millimole [Cr^(3+)]=15 millimole ,E is 1.067 v The pH of the solution is nearly eual to

On the basis of Eo values identify which amongst the following is the strongest oxidising agent Cl_(2)(g)+2e^(-)rarr2Cl^(-)E^(@)=+1.36V , MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)OE^(@)=+1.51V Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)rarr2Cr^(3+)+7H_(2)O E^(@)=+1.33V (ii) The following figure 2, represents variation of (Λ_(m)) vs √c for an electrolyte. Here Λ_(m) is the molar conductivity and c is the concentration of the electrolyte. a) Define molar conductivity b) Identify the nature of electrolyte on the basis of the above plot. Justify your answer. c) Determine the value of Λmo for the electrolyte. d) Show how to calculate the value of A for the electrolyte using the above graph.

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