1884 United States presidential election in Rhode Island

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United States presidential election in Rhode Island, 1884

← 1880 November 4, 1884 1888 →
  JamesGBlaine.png StephenGroverCleveland.png
Nominee James G. Blaine Grover Cleveland
Party Republican Democratic
Home state Maine New York
Running mate John A. Logan Thomas A. Hendricks
Electoral vote 4 0
Popular vote 19,030 12,391
Percentage 58.07% 37.81%

President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

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

Rhode Island voted for the Republican nominee, James G. Blaine, over the Democratic nominee, Grover Cleveland. Blaine won the state by a margin of 20.26%.

With 58.07% of the popular vote, Rhode Island would prove to be Blaine's fourth strongest victory in terms of percentage in the popular vote after Vermont, Minnesota and Kansas.[1]

Results[edit]

United States presidential election in Rhode Island, 1884[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican James Gillespie Blaine of Maine John Alexander Logan of Illinois 19,030 58.07% 4 100.00%
Democratic Grover Cleveland of New York Thomas Andrews Hendricks of Indiana 12,391 37.81% 0 0.00%
Prohibition John Pierce St. John of Kansas William Daniel of Maryland 928 2.83% 0 0.00%
Greenback Benjamin Franklin Butler of Massachusetts Absolom Madden West of Mississippi 423 1.29% 0 0.00%
Total 32,771 100.00% 4 100.00%

References[edit]

  1. ^ "1884 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ^ "1884 Presidential General Election Results - Rhode Island".