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

← 1876November 2, 18801884 →
 
NomineeJames A. GarfieldWinfield S. Hancock
PartyRepublicanDemocratic
Home stateOhioPennsylvania
Running mateChester A. ArthurWilliam H. English
Electoral vote40
Popular vote18,19510,779
Percentage62.24%36.87%

County Results
Garfield
  50-60%
  60-70%


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

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1880 United States presidential election in Rhode Island[2]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
RepublicanJames Abram Garfield of OhioChester Alan Arthur of New York18,19562.24%4100.00%
DemocraticWinfield Scott Hancock of PennsylvaniaWilliam Hayden English of Indiana10,77936.87%00.00%
GreenbackJames Baird Weaver of IowaBarzillai Jefferson Chambers of Texas2360.81%00.00%
ProhibitionNeal Dow of MaineHenry Adams Thompson of Ohio200.07%00.00%
Anti-MasonicJohn Wolcott Phelps of VermontSamuel Clarke Pomeroy of Kansas40.01%00.00%
N/AOthersOthers10.01%00.00%
Total29,235100.00%4100.00%

See also

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References

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  1. ^ "1880 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  2. ^ "1880 Presidential General Election Results - Rhode Island". Dave Leip's Atlas of U.S. Presidential Elections.