See main article: 1864 United States presidential election.
| Election Name: | 1864 United States presidential election in Iowa |
| Country: | Iowa |
| Flag Year: | 1864 |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1860 United States presidential election in Iowa |
| Previous Year: | 1860 |
| Next Election: | 1868 United States presidential election in Iowa |
| Next Year: | 1868 |
| Election Date: | November 8, 1864 |
| Turnout: | 19.70% of the total population 0.63 pp[1] |
| Image1: | Abraham Lincoln November 1863.jpg |
| Nominee1: | Abraham Lincoln |
| Party1: | National Union Party (United States) |
| Home State1: | Illinois |
| Running Mate1: | Andrew Johnson |
| Electoral Vote1: | 8 |
| Popular Vote1: | 88,500 |
| Percentage1: | 64.12% |
| Nominee2: | George B. McClellan |
| Party2: | Democratic Party (United States) |
| Home State2: | New Jersey |
| Running Mate2: | George H. Pendleton |
| Electoral Vote2: | 0 |
| Popular Vote2: | 49,525 |
| Percentage2: | 35.88% |
| Map Size: | 312px |
| President | |
| Before Election: | Abraham Lincoln |
| Before Party: | Republican Party (United States) |
| After Election: | Abraham Lincoln |
| After Party: | National Union Party (United States) |
The 1864 United States presidential election in Iowa took place on November 8, 1864, as part of the 1864 United States presidential election. Iowa voters chose eight representatives, or electors, to the Electoral College, who voted for president and vice president.[2]
Iowa was won by the National Union candidate Republican incumbent President Abraham Lincoln of Illinois and his running mate former Senator and Military Governor of Tennessee Andrew Johnson. They defeated the Democratic candidate 4th Commanding General of the United States Army George B. McClellan of New Jersey and his running mate Representative George H. Pendleton of Ohio. Lincoln won the state by a margin of 28.24%.[2]