1860 United States presidential election in Iowa
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| |  |  | | 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% | |
 County Results | Lincoln 40-50% 50-60% 60-70% 70-80% 80-90% 90-100% | Douglas 40-50% 50-60% 60-70% 70-80% 80-90% | Tie ~50% | |
President before election
James Buchanan
Democratic | Elected President
Abraham Lincoln
Republican | |
| Elections in Iowa |
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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 the Republican nominees Illinois Representative Abraham Lincoln and his running mate Senator Hannibal Hamlin of Maine. They defeated the Democratic nominees Senator Stephen A. Douglas of Illinois and his running mate 41st Governor of Georgia Herschel V. Johnson. Lincoln won the state by a margin of 11.39%.
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% |
See also
References
- ^ "1860 Presidential Election Results Iowa".
