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The standard cell potential of the following cell Zn | Zn ²⁺( aq ) | Fe ²⁺( aq ) Fe is 0.32 V . Calculate the standard Gibbs energy change for the reaction : Zn ( s )+ Fe ²⁺( aq ) Zn ²⁺( aq )+ Fe ( s ) (Given : 1 ~F =96487 C )

Options

  1. A-61.75 ~kJ ~mol ⁻¹
  2. B+5.006 ~kJ ~mol ⁻¹
  3. C-5.006 ~kJ ~mol ⁻¹
  4. D+61.75 ~kJ ~mol ⁻¹

Correct answer

A. -61.75 ~kJ ~mol ⁻¹

Step-by-step solution

_r G^ =-n F E_ cell ^ For the given reaction, n=2 _ r G ^ =-2 96487 0.32=-61751.68 ~J ~mol ⁻¹=-61.751 ~kJ ~mol ⁻¹

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