United States presidential election in Rhode Island, 1888

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United States presidential election in Rhode Island, 1888

← 1884 November 6, 1888 1892 →

  Pach Brothers - Benjamin Harrison.jpg StephenGroverCleveland.png
Nominee Benjamin Harrison Grover Cleveland
Party Republican Democratic
Home state Indiana New York
Running mate Levi P. Morton Allen G. Thurman
Electoral vote 4 0
Popular vote 21,969 17,530
Percentage 53.88% 42.99%

President before election

Grover Cleveland

Elected President

Benjamin Harrison

The 1888 United States presidential election in Rhode Island took place on November 6, 1888, as part of the 1888 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, Benjamin Harrison, over the Democratic nominee, incumbent President Grover Cleveland. Harrison won the state by a margin of 10.89%.


United States presidential election in Rhode Island, 1888[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Benjamin Harrison of Indiana Levi Parsons Morton of New York 21,969 53.88% 4 100.00%
Democratic Grover Cleveland of New York Allen Granberry Thurman of Ohio 17,530 42.99% 0 0.00%
Prohibition Clinton Bowen Fisk of New Jersey John Anderson Brooks of Missouri 1,251 3.07% 0 0.00%
Labor Alson Streeter of Illinois Charles E. Cunningham of Arkansas 18 0.04% 0 0.00%
N/A Others Others 7 0.02% 0 0.00%
Total 40,775 100.00% 4 100.00%


  1. ^ "1888 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.
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