Question

For the given cell: \[ {Fe}^{2+}(aq) + {Ag}^+(aq) \to {Fe}^{3+}(aq) + {Ag}(s) \] The standard cell potential of the above reaction is given. The standard reduction potentials are given as: \[ {Ag}^+ + e^- \to {Ag} \quad E^\circ = x \, {V} \] \[ {Fe}^{2+} + 2e^- \to {Fe} \quad E^\circ = y \, {V} \] \[ {Fe}^{3+} + 3e^- \to {Fe} \quad E^\circ = z \, {V} \] The correct answer is:

• A. \( x + y - z \)
  (2) \( x + 2y - 3z \)
  (3) \( y - 2x \)
  (4) \( x + 2y \)
• B. \( x + 2y - 3z \)
• C. \( y - 2x \)
• D. \( x + 2y \)

Hint:

For an electrochemical cell, the standard cell potential is the difference between the reduction potentials at the cathode and anode. Remember to reverse the potential for oxidation reactions at the anode.

Answer 1

The Correct Option is B

The standard cell potential is given by: \[ E^\circ_{{cell}} = E^\circ_{{cathode}} - E^\circ_{{anode}} \] At the cathode, the reduction half-reaction is \( {Ag}^+ + e^- \to {Ag} \), so the cathode potential is \( E^\circ_{{cathode}} = x \, {V} \). At the anode, the oxidation half-reaction is \( {Fe} \to {Fe}^{2+} + 2e^- \), which is the reverse of \( {Fe}^{2+} + 2e^- \to {Fe} \). So, the anode potential is \( E^\circ_{{anode}} = -y \, {V} \). Thus, the standard cell potential is: \[ E^\circ_{{cell}} = x - (-y) = x + y \] Therefore, the correct answer is \( x + 2y - 3z \), corresponding to option \( \boxed{(2)} \).