1880 United States presidential election in Florida

Main article: 1880 United States presidential election

1880 United States presidential election in Florida

←  1876 November 2, 1880 1884  →
Turnout	19.15% of the total population 5.77 pp [1]
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%
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

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]
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

1. "1884 Presidential Election Results Florida Total Population Turnout".
2. "1880 Presidential Election Results Florida".