1880 United States presidential election in Florida![](//upload.wikimedia.org/wikipedia/commons/thumb/4/49/Flag_of_Florida_%281868%E2%80%931900%29.svg/50px-Flag_of_Florida_%281868%E2%80%931900%29.svg.png)
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Turnout | 19.15% of the total population 5.77 pp[1] |
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| | | Nominee | Winfield S. Hancock | James A. Garfield | | Party | Democratic | Republican | Home state | Pennsylvania | Ohio | Running mate | William H. English | Chester A. Arthur | Electoral vote | 4 | 0 | Popular vote | 27,964 | 23,654 | Percentage | 54.17% | 45.83% | |
![](//upload.wikimedia.org/wikipedia/commons/thumb/0/06/Florida_Presidential_Election_Results_1880.svg/400px-Florida_Presidential_Election_Results_1880.svg.png) County Results Hancock 50-60% 60-70% 70-80% 80-90% 90-100% | Garfield 50-60% 60-70% 70-80% | |
President before election Rutherford B. Hayes Republican | Elected President James Garfield Republican | |
Elections in Florida |
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![](//upload.wikimedia.org/wikipedia/commons/thumb/2/2b/Seal_of_Florida.svg/150px-Seal_of_Florida.svg.png) |
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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]
Results
See also
References
- ^ "1884 Presidential Election Results Florida Total Population Turnout".
- ^ a b c "1880 Presidential Election Results Florida".
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