1880 United States presidential election in Rhode Island
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| Elections in Rhode Island |
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The 1880 United States presidential election in Rhode Island took place on November 2, 1880, as part of the 1880 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 A. Garfield, over the Democratic nominee, Winfield Scott Hancock. Garfield won the state by a margin of 25.37%.
Results
| United States presidential election in Rhode Island, 1880[1] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Republican | James Abram Garfield of Ohio | Chester Alan Arthur of New York | 18,195 | 62.24% | 4 | 100.00% | ||
| Democratic | Winfield Scott Hancock of Pennsylvania | William Hayden English of Indiana | 10,779 | 36.87% | 0 | 0.00% | ||
| Greenback | James Baird Weaver of Iowa | Barzillai Jefferson Chambers of Texas | 236 | 0.81% | 0 | 0.00% | ||
| Prohibition | Neal Dow of Maine | Henry Adams Thompson of Ohio | 20 | 0.07% | 0 | 0.00% | ||
| Anti-Masonic | John Wolcott Phelps of Vermont | Samuel Clarke Pomeroy of Kansas | 4 | 0.01% | 0 | 0.00% | ||
| N/A | Others | Others | 1 | 0.01% | 0 | 0.00% | ||
| Total | 29,235 | 100.00% | 4 | 100.00% | ||||
References
- ^ {{cite web|title=1880 Presidential General Election Results - Rhode Island|url=http://uselectionatlas.org/RESULTS/state.php?year=1880&fips=44&f=1&off=0&elect=0&minper=0
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