1884 United States presidential election in Florida

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United States presidential election in Florida, 1884

← 1880 November 4, 1884 1888 →
Turnout22.26% of the total population Increase 3.11 pp[1]
  StephenGroverCleveland.png James G. Blaine - Brady-Handy.jpg
Nominee Grover Cleveland James G. Blaine
Party Democratic Republican
Home state New York Maine
Running mate Thomas A. Hendricks John A. Logan
Electoral vote 4 0
Popular vote 31,769 28,031
Percentage 52.96% 46.73%

President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

The 1884 United States presidential election in Florida took place on November 4, 1884, as part of the 1884 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 Grover Cleveland, the 28th governor of New York, (DNew York), running with the former governor of Indiana Thomas A. Hendricks, with 52.96% of the popular vote, against Secretary of State James G. Blaine (R-Maine), running with Senator John A. Logan, with 46.73% of the vote.[2]

Results[edit]

United States presidential election in Florida, 1884[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Grover Cleveland of New York Thomas A. Hendricks of Indiana 31,769 52.96% 4 100.00%
Republican James G. Blaine of Maine John A. Logan of Illinois 28,031 46.73% 0 0.00%
Write-in Write-in of Write-in of 118 0.20% 0 0.00%
Prohibition John St. John of Kansas William Daniel of Maryland 72 0.12% 0 0.00%
Total 59,990 100.00% 4 100.00%

References[edit]

  1. ^ "1884 Presidential Election Results Florida Total Population Turnout".
  2. ^ a b c "1884 Presidential Election Results Florida".