Question:medium

The following reaction takes place in a galvanic cell at \(298\) K \[ \mathrm{Fe^{2+}(aq)+Ag^+(aq)\rightarrow Fe^{3+}(aq)+Ag(s)} \] The \(\Delta_rG^\circ\) (in kJ mol\(^{-1}\)) and \(\log K_c\) values are respectively \[ (F=96500\ \mathrm{C\,mol^{-1}},\; E^\circ_{\mathrm{Ag^+/Ag}}=0.80\ \mathrm{V},\; E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}}=0.77\ \mathrm{V},\; R=8.3\ \mathrm{J\,mol^{-1}K^{-1}}) \]

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Remember, \[ \boxed{ \Delta G^\circ=-nFE^\circ_{\text{cell}} } \] and \[ \boxed{ \Delta G^\circ=-2.303RT\log K. } \]
Updated On: Jul 18, 2026
  • \(-0.508;\ 2.895\)
  • \(2.895;\ 5.08\)
  • \(-2.895;\ 0.508\)
  • \(2.895;\ 0.508\)
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The Correct Option is C

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