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.

Main article: 1860 United States presidential election
1860 United States presidential election in Iowa
November 6, 1860
 
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
Democratic

 Elected President

Abraham Lincoln
Republican

Elections in Iowa

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

Results
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%
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References
  1. "1860 Presidential Election Results Iowa".
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