1884 United States presidential election in Alabama

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United States presidential election in Alabama, 1884

← 1880 November 4, 1884 1888 →
Turnout12.03% of the total population Increase 0.14 pp[1]
  StephenGroverCleveland.png James G. Blaine - Brady-Handy.jpg
Nominee Grover Cleveland James G. Blaine
Party Democratic Republican
Home state New York Ohio
Running mate Thomas A. Hendricks John A. Logan
Electoral vote 10 0
Popular vote 92,736 59,444
Percentage 60.37% 38.69%

Alabama Presidential Election Results by County, 1884.svg
County Results

President before election

James 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 1884 United States presidential election. Alabama voters chose ten representatives, or electors, to the Electoral College, who voted for president and vice president.[2]

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-Ohio), running with Senator John A. Logan, with 38.69% of the vote.[2]

Results[edit]

United States presidential election in North Carolina, 1868[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Democratic Grover Cleveland of New York Thomas A. Hendricks of Indiana 92,736 60.37% 10 100.00%
Republican James G. Blaine of Ohio John A. Logan of Illinois 59,444 38.69% 0 0.00%
Greenback Benjamin Butler of Massachusetts Absolom West of Mississippi 762 0.50% 0 0.00%
Prohibition John St. John of Kansas William Daniel of Maryland 610 0.40% 0 0.00%
Write-in Write-in of United States of America Write-in of United States of America 72 0.05% 0 0.00%
Total 153,624 100.00% 10 100.00%

References[edit]

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