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Calculate the reduction potential for th...

Calculate the reduction potential for the following half cells at `25^@C`
`Pt|Fe^(2+) (0.1M)-Fe^(3+) (0.01M), E_(Fe^(3+),Fe^(2+))^@=+0.77V`

Text Solution

Verified by Experts

`Fe^(3+)+e leftrightarrow Fe^(2+)` (reduction)
`E_(Fe^(3+),Fe^(2+))=E_(Fe^(3+),Fe^(2+))-0.0591/n log""([Fe^(2+)])/([Fe^(3+)])`
`=0.77- 0.0591/1 log"" 0.1/0.01=0.7109V`
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BODY BOOKS PUBLICATION-ELECTROMOTIVE FORCE -EXERCISE
  1. Calculate the reduction potential for the following half cells at 25^@...

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  2. Is 1M H^+ solution under hydrogen gas at 1 atm capable of oxidising Ag...

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  3. The potential of hydrogen electrode is -118mV. The concentration of H^...

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  4. E^@ for the half cell Zn^(2+)|Zn is -0.76V emf of the cell Zn|Zn^(2+...

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  5. The standard reduction potentials at 298K for the following half react...

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  6. The standard reduction potentials E^@ for the half reactions are as ...

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  7. From the following E^@ values of half cells, (i) A+e to A^(-) , E^@=...

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  8. From the following E^@ values of half cells (i) A to A^(+)+e, E^@=+1...

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  9. From the following E^@ values of half cells (i) A^(3-)to A^(2-) +e,...

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  10. IF the following half cells have the E^@ values as Fe^(3+)+e to Fe^(...

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  11. E^@ (red.) values of the half cells Mg^(2+)//Mg and Cl2//Cl^- are resp...

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  12. For the cell reaction Zn(s)+Mg^(2+) (1M)=Zn^(2+) (1M) +Mg, the emf has...

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  13. E^@ for F2+2e=2F^(-) is 2.8V, E^@ for 1/2 F2+e=F^(-) is

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  14. Delta G^@ of the cell reaction AgCl(s)+1/2H2(g)=Ag(s)+H^+ +Cl^(-) is ...

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  15. The value of equilibrium constant for the feasible cell reaction is

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  16. E^@ for the reaction Fe+Zn^(2+)=Zn+Fe^(2+) is -0.35 V. The given cell...

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  17. A galvanic cell is composed of two hydrogen electrodes, one of which i...

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  18. 1/2 H2(g)+AgCl(s)=H^+ (aq)+Cl^(-) (aq)+Ag(s) occurs in the galvanic c...

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  19. For the cell Tl|Tl^(+) (0.001M) ||Cu^(2+) (0.1M) |Cu. E(cell) at 25^@C...

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  20. (i) E^@ (Cu^(2+),Cu)=0.34V (ii) E^@(Cu^+,Cu)=+0.52V (iii) E^@[O2(...

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  21. Given below the half cell reactions Mn^(2+)+2e to Mn, E^@=-1.18V 2...

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