1876 United States presidential election in Rhode Island

Main article: 1876 United States presidential election

The 1876 United States presidential election in Rhode Island took place on November 7, 1876, as part of the 1876 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

1876 United States presidential election in Rhode Island

← 1872 November 7, 1876 1880 →

Nominee Rutherford B. Hayes Samuel J. Tilden Party Republican Democratic Home state Ohio New York Running mate William A. Wheeler Thomas A. Hendricks Electoral vote 4 0 Popular vote 15,787 10,712 Percentage 59.29% 40.23%

County Results Hayes   50–60%   60–70%

President before election Ulysses S. Grant Republican Elected President Rutherford B. Hayes Republican

Rhode Island voted for the Republican nominee, Rutherford B. Hayes, over the Democratic nominee, Samuel J. Tilden. Hayes won the state by a margin of 19.06%.

With 59.29% of the popular vote, Rhode Island would be Hayes' fourth strongest victory in terms of percentage in the popular vote after Vermont, Nebraska and Kansas.^[1]

This was the only election between 1860 and 1888 where the Democratic candidate earned more than 40% statewide, as well as the only election in the same period where the Democratic candidate earned more than 40% in at least one Rhode Island county.

Results

1876 United States presidential election in Rhode Island[2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Republican	Rutherford B. Hayes of Ohio	William A. Wheeler of New York	15,787	59.29%	4	100.00%
	Democratic	Samuel J. Tilden of New York	Thomas A. Hendricks of Indiana	10,712	40.23%	0	0.00%
	Greenback	Peter Cooper of New York	Samuel Fenton Cary of Ohio	68	0.26%	0	0.00%
	Prohibition	Green Clay Smith of Kentucky	Gideon Tabor Stewart of Ohio	60	0.23%	0	0.00%
Total				26,627	100.00%	4	100.00%

See also

• United States presidential elections in Rhode Island