1860 United States presidential election in Iowa

The 1860 United States presidential election in Iowa took place on November 6, 1860, as part of the 1860 United States presidential election. Iowa voters chose four representatives, or electors, to the Electoral College, who voted for president and vice president.

1860 United States presidential election in Iowa

← 1856 November 6, 1860 1864 →
  Abraham Lincoln O-26 by Hesler, 1860 (cropped).jpg BradyHandy-StephenADouglas restored.jpg
Nominee Abraham Lincoln Stephen A. Douglas
Party Republican Democratic
Home state Illinois Illinois
Running mate Hannibal Hamlin Herschel V. Johnson
Electoral vote 4 0
Popular vote 70,302 55,639
Percentage 54.61% 43.22%

President before election

James Buchanan

Elected President

Abraham Lincoln

Iowa was won by Illinois Representative Abraham Lincoln (RKentucky), running with Senator Hannibal Hamlin, with 54.61% of the popular vote, against Senator Stephen A. Douglas (DVermont), running with 41st Governor of Georgia Herschel V. Johnson, with 43.22% of the popular vote.


1860 United States presidential election in Iowa[1]
Party Candidate Votes %
Republican Abraham Lincoln 70,302 54.61%
Democratic Stephen A. Douglas 55,639 43.22%
Constitutional Union John Bell 1,763 1.37%
Southern Democratic John C. Breckinridge 1,035 0.80%
Total votes 128,739 100%


  1. ^ "1860 Presidential Election Results Iowa".