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1884 United States presidential election in Illinois

←  1880 November 4, 1884 1888 →
 
Nominee James G. Blaine Grover Cleveland
Party Republican Democratic
Home state Maine New York
Running mate John A. Logan Thomas A. Hendricks
Electoral vote 22 0
Popular vote 337,469 312,351
Percentage 50.17% 46.43%

County Results[ citation needed]

President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

The 1884 United States presidential election in Illinois took place on November 4, 1884, as part of the 1884 United States presidential election. Voters chose 22 representatives, or electors to the Electoral College, who voted for president and vice president.

Illinois voted for the Republican nominee, James G. Blaine, over the Democratic nominee, Grover Cleveland. Blaine won the state by a narrow margin of 3.74%. [1]

This is the first of only three elections in the history of the party that a Democrat won the presidency without winning Illinois (the others being 1916 and 1976). [2]

Results

1884 United States presidential election in Illinois [3]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican James Gillespie Blaine of Maine John Alexander Logan of Illinois 337,469 50.17% 22 100.00%
Democratic Grover Cleveland of New York Thomas Andrews Hendricks of Indiana 312,351 46.43% 0 0.00%
Prohibition John Pierce St. John of Kansas William Daniel of Maryland 12,074 1.79% 0 0.00%
Greenback Benjamin Franklin Butler of Massachusetts Absolom Madden West of Mississippi 10,776 1.60% 0 0.00%
Total 672,670 100.00% 22 100.00%

See also

References

  1. ^ "True Republican 12 December 1888 — Illinois Digital Newspaper Collections". idnc.library.illinois.edu. Retrieved December 3, 2021.
  2. ^ "Historical U.S. Presidential Elections 1789-2020". 270toWin.com. Retrieved December 17, 2021.
  3. ^ "1884 Presidential General Election Results - Illinois". U.S. Election Atlas. Retrieved December 23, 2013.