United States presidential election in Iowa, 1860
United States presidential election in Iowa, 1860
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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.
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[edit]
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% |
References[edit]
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Categories:
- United States presidential election, 1860, by state
- United States presidential elections in Iowa
- Iowa elections, 1860
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