Question

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 \[ (F=96500\ \mathrm{C\,mol^{-1}},\; E^\circ_{\mathrm{Ag^+/Ag}}=0.80\ \mathrm{V},\; E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}}=0.77\ \mathrm{V},\; R=8.3\ \mathrm{J\,mol^{-1}K^{-1}}) \]

• A. \(-0.508;\ 2.895\)
• B. \(2.895;\ 5.08\)
• C. \(-2.895;\ 0.508\)
• D. \(2.895;\ 0.508\)

Hint:

Remember, \[ \boxed{ \Delta G^\circ=-nFE^\circ_{\text{cell}} } \] and \[ \boxed{ \Delta G^\circ=-2.303RT\log K. } \]

Answer 1

The Correct Option is C

Step 1: Calculate the standard cell potential. Cathode: \[ \mathrm{Ag^++e^-\rightarrow Ag}, \qquad E^\circ=0.80\text{ V}. \] Anode: \[ \mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^-}, \] whose oxidation potential is \[ -0.77\text{ V}. \] Therefore, \[ E^\circ_{\text{cell}} = 0.80-0.77 = 0.03\text{ V}. \]

Step 2: Calculate \(\Delta_rG^\circ\). Since \[ n=1, \] \[ \Delta_rG^\circ = -nFE^\circ_{\text{cell}} = -(1)(96500)(0.03) = -2895\text{ J mol}^{-1} = -2.895\text{ kJ mol}^{-1}. \]

Step 3: Calculate \(\log K_c\). Using \[ \Delta_rG^\circ=-2.303RT\log K_c, \] \[ \log K_c = \frac{2895}{2.303\times8.3\times298} \approx0.508. \] Hence, \[ \boxed{\Delta_rG^\circ=-2.895\text{ kJ mol}^{-1},\qquad \log K_c=0.508.} \] Thus, \[ \boxed{(C)} \] is the correct answer.