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What is the potential of an electrode wh...

What is the potential of an electrode which originally contained `0.1 MNO_(3)^(-)` and `0.4MH^(+)` and which has been treated by `60%` of the cadmium necessary to reduce all the `NO_(3)^(-)` to `NO(g)` at 1 atm.
Given,
`NO_(3)^(-) +4H^(+)+3e^(-)rarr NO+2H_(2)O, E^(@)=0.95V`
and `log2=0.3010`

A

0.52 V

B

0.44

C

0.86 V

D

0.78 V

Text Solution

Verified by Experts

The correct Answer is:
C
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What is the potential of an electrode which originally contained 0.1 M NO_(3)^(-) and 0.4 M H^(+) and which has been treated by 8% of the cadmium necessary to reduce all the NO_(3)^(-) to NO(g) at 1 bar ? Given : NO_(3)^(-)+4H^(+)+3e^(-)rarr NO+2H_(2)O , E^(@)=0.96 , log 2 = 0.3

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Read the following passage for the evaluation of E^(@) when different number of electrons are involved. Consider addition of the following half reactions (1) Fe^(3+)(aq)+3e^(-)rarr Fe(s) E_(1)^(@)=0.45V (2) Fe(s)rarr Fe^(2+)(aq)+2e^(-) E_(2)^(@)=-0.04V (3) Fe^(3+)(aq)+e^(-)rarr Fe^(2+)(aq) E_(3)^(@)= ? Because half-reactions (1) and (2) contains a different number of electrons, the net reaction (3) is another half-reaction and E_(3)^(@) can't be obtained simply by adding E_(1)^(@) and E_(3)^(@) . The free - energy changes however, are additive because G is a state function : Delta G_(3)^(@)=Delta G_(1)^(@)+Delta G_(2)^(@) For the reactions (1) MnO_(4)^(-)+8H^(+)+5e^(-)rarr Mn^(2)+4H_(2)O E^(@)=1.51 V (2) MnO_(2)+4H^(+)+2e^(-)rarr Mn^(2+)2H_(2)O E^(@)=1.32 then for the reaction (3) MnO_(4)^(-)+4H^(+)+3e^(-)rarrMnO_(2)+2H_(2)O, E^(@) is

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