See main article: 1860 United States presidential election.
| Election Name: | 1860 United States presidential election in Iowa |
| Country: | Iowa |
| Flag Year: | 1846 |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1856 United States presidential election in Iowa |
| Previous Year: | 1856 |
| Next Election: | 1864 United States presidential election in Iowa |
| Next Year: | 1864 |
| Election Date: | November 6, 1860 |
| Image1: | Abraham Lincoln by Alexander Hesler, 1860-restored (cropped).png |
| Nominee1: | Abraham Lincoln |
| Party1: | Republican Party (United States) |
| Home State1: | Illinois |
| Running Mate1: | Hannibal Hamlin |
| Electoral Vote1: | 4 |
| Popular Vote1: | 70,302 |
| Percentage1: | 54.61% |
| Nominee2: | Stephen A. Douglas |
| Party2: | Democratic Party (United States) |
| Home State2: | Illinois |
| Running Mate2: | Herschel V. Johnson |
| Electoral Vote2: | 0 |
| Popular Vote2: | 55,639 |
| Percentage2: | 43.22% |
| Map Size: | 300px |
| President | |
| Before Election: | James Buchanan |
| Before Party: | Democratic Party (United States) |
| After Election: | Abraham Lincoln |
| After Party: | Republican Party (United States) |
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%.