The e.m.f. of the cell \[Zn|Z{{n}^{2+}}(0.01M)||F{{e}^{2+}}(0.001M)|Fe\] at 298 K is 0.2905 then the value of equilibrium for the cell reaction is [IIT-JEE Screening 2004]

\[\frac{0.26}{{{10}^{0.0295}}}\]

\[\frac{0.32}{{{e}^{0.0295}}}\]

\[\frac{0.32}{{{10}^{0.0295}}}\]

\[\frac{0.32}{{{10}^{0.0591}}}\]

The e.m.f. of the cell \[Zn|Z{{n}^{2+}}(0.01M)||F{{e}^{2+}}(0.001M)|Fe

1.	The e.m.f. of the cell \[Zn\|Z{{n}^{2+}}(0.01M)\|\|F{{e}^{2+}}(0.001M)\|Fe\] at 298 K is 0.2905 then the value of equilibrium for the cell reaction is [IIT-JEE Screening 2004]
A.	\[\frac{0.32}{{{e}^{0.0295}}}\]
B.	\[\frac{0.32}{{{10}^{0.0295}}}\]
C.	\[\frac{0.26}{{{10}^{0.0295}}}\]
D.	\[\frac{0.32}{{{10}^{0.0591}}}\]
Answer» C. \[\frac{0.26}{{{10}^{0.0295}}}\]

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The e.m.f. of the cell \[Zn|Z{{n}^{2+}}(0.01M)||F{{e}^{2+}}(0.001M)|Fe\] at 298 K is 0.2905 then the value of equilibrium for the cell reaction is [IIT-JEE Screening 2004]

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