\(\ce{MnO4^- (aq) + Y + 5Fe^2+ (aq) -> Mn^2+ (aq) + 5Fe^2+ (aq) + 4H2O (l)}\)

In the equation above, Y is:

Explanation (acidic medium redox balancing):

• Note: On the products side the iron should be \(\ce{Fe^3+}\), not \(\ce{Fe^2+}\). The correct overall reaction in acid is: \[\ce{MnO4^- + Y + 5Fe^2+ -> Mn^2+ + 5Fe^3+ + 4H2O}\]
• Half-reactions
  • \(\ce{MnO4^- -> Mn^2+}\): balance O with water and H with \(\ce{H+}\), then charge with electrons:
    \[\ce{MnO4^- + 8H+ + 5e- -> Mn^2+ + 4H2O}\]
  • \(\ce{Fe^2+ -> Fe^3+ + e-}\)
• Combine: multiply the iron half-reaction by 5 to cancel 5 electrons and add: \[\ce{MnO4^- + 8H+ + 5Fe^2+ -> Mn^2+ + 5Fe^3+ + 4H2O}\]

Therefore, \(Y = \mathbf{\ce{8H+_{(aq)}}}\).