1884 United States presidential election in Alabama

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1884 United States presidential election in Alabama
Flag of Alabama (1861, obverse).svg
  1880 November 4, 1884 1888  
Turnout12.03% of the total population Increase2.svg 0.14 pp [1]
  StephenGroverCleveland.png Unsuccessful 1884.jpg
Nominee Grover Cleveland James G. Blaine
Party Democratic Republican
Home state New York Maine
Running mate Thomas A. Hendricks John A. Logan
Electoral vote100
Popular vote92,73659,444
Percentage60.37%38.69%

Alabama Presidential Election Results 1884.svg
County results

President before election

James A. Garfield
Republican

Elected President

Grover Cleveland
Democratic

The 1884 United States presidential election in Alabama took place on November 4, 1884, as part of the nationwide presidential election. Alabama voters chose ten representatives, or electors, to the Electoral College, who voted for president and vice president. [2]

Contents

Alabama was won by Grover Cleveland, the 28th governor of New York, (DNew York), running with the former governor of Indiana Thomas A. Hendricks, with 60.37% of the popular vote, against Secretary of State James G. Blaine (R-Maine), running with Senator John A. Logan, with 38.69% of the vote. [2]

Results

1884 United States presidential election in Alabama [2]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Democratic Grover Cleveland of New York Thomas A. Hendricks of Indiana 92,73660.37%10100.00%
Republican James G. Blaine of Maine John A. Logan of Illinois 59,44438.69%00.00%
Greenback Benjamin Butler of Massachusetts Absolom M. West of Mississippi 7620.50%00.00%
Prohibition John St. John of Kansas William Daniel of Maryland 6100.40%00.00%
Write-in Write-in of United StatesWrite-in of United States720.05%00.00%
Total153,624100.00%10100.00%

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References

  1. "1884 Presidential Election Results Alabama Total Population Turnout".
  2. 1 2 3 "1884 Presidential Election Results Alabama".