1888 United States presidential election in Rhode Island
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.
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 County Results
Harrison 50-60%
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| Elections in Rhode Island |
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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%.
Results
| 1888 United States presidential election in Rhode Island[1] | ||||||||
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| 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% | ||||
References
- "1888 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.
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