1880 United States presidential election in Florida
Election in Florida
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| Turnout | 19.15% of the total population  5.77 pp[1] | |||||||||||||||||||||||||
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 County Results
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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
| United States presidential election in Florida, 1880[2] | ||||||||
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| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Democratic | Winfield Scott Hancock of Pennsylvania | William Hayden English of Indiana | 27,964 | 54.17% | 4 | 100.00% | ||
| Republican | James Garfield of Ohio | Chester A. Arthur of New York | 23,654 | 45.83% | 0 | 0.00% | ||
| Total | 51,618 | 100.00% | 4 | 100.00% | ||||
See also
- 1880 United States House of Representatives elections in Florida
- 1880 Florida gubernatorial election
References
- ^ "1884 Presidential Election Results Florida Total Population Turnout".
- ^ a b c "1880 Presidential Election Results Florida".
State and district results of the 1880 United States presidential election | ||
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