See main article: 1880 United States presidential election.
| Election Name: | 1880 United States presidential election in Florida |
| Country: | Florida |
| Flag Image: | Flag of Florida (1868–1900).svg |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1876 United States presidential election in Florida |
| Previous Year: | 1876 |
| Next Election: | 1884 United States presidential election in Florida |
| Next Year: | 1884 |
| Election Date: | November 2, 1880 |
| Turnout: | 19.15% of the total population 5.77 pp[1] |
| Image1: | WinfieldScottHancock2 (cropped 3x4).jpg |
| Nominee1: | Winfield S. Hancock |
| Party1: | Democratic Party (United States) |
| Home State1: | Pennsylvania |
| Running Mate1: | William H. English |
| Electoral Vote1: | 4 |
| Popular Vote1: | 27,964 |
| Percentage1: | 54.17% |
| Nominee2: | James A. Garfield |
| Party2: | Republican Party (United States) |
| Home State2: | Ohio |
| Running Mate2: | Chester A. Arthur |
| Electoral Vote2: | 0 |
| Popular Vote2: | 23,654 |
| Percentage2: | 45.83% |
| President | |
| Before Election: | Rutherford B. Hayes |
| Before Party: | Republican Party (United States) |
| After Election: | James Garfield |
| After Party: | Republican Party (United States) |
| Map Size: | 400px |
The 1880 United States presidential election in Florida took place on November 2, 1880, as part of the 1880 United States presidential election. Florida voters chose four representatives, or electors, to the Electoral College, who voted for president and vice president.[2]
Florida was won by General Winfield Scott Hancock (D–Pennsylvania), running with former Representative William Hayden English, with 54.17% of the popular vote, against Representative James Garfield (R-Ohio), running with the 10th chairman of the New York State Republican Executive Committee Chester A. Arthur, with 41.05% of the vote.[2]