1884 United States presidential election in Rhode Island
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The 1884 United States presidential election in Rhode Island occurred on November 4, 1884. This election was part of the 1884 United States presidential election. Voters in Rhode Island selected four representatives, or electors, to the Electoral College, which elects the president and vice president.
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November 4, 1884
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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 became Blaine's fourth strongest victory in terms of percentage, following Vermont, Minnesota and Kansas.[1]
Results
| 1884 United States presidential election in Rhode Island[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% | ||||
