1880 United States presidential election in Rhode Island

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.

1880 United States presidential election in Rhode Island

November 2, 1880
 
Nominee James A. Garfield Winfield S. Hancock
Party Republican Democratic
Home state Ohio Pennsylvania
Running mate Chester A. Arthur William H. English
Electoral vote 4 0
Popular vote 18,195 10,779
Percentage 62.24% 36.87%

President before election

Rutherford B. Hayes
Republican

Elected President

James A. Garfield
Republican

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%.

With 62.24% of the popular vote, Rhode Island would be Garfield's fourth strongest victory in terms of percentage in the popular vote after Vermont, Nebraska and Minnesota.[1]

Results

1880 United States presidential election in Rhode Island[2]
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%

See also

References


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