1880 United States presidential election in Rhode Island

Main article: 1880 United States presidential election

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%

County Results Garfield   50-60%   60-70%

President before election Rutherford B. Hayes Republican Elected President James A. Garfield Republican

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%.

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.

This was the last time until 1964 that a Democrat won the town of North Kingstown.

Results

1880 United States presidential election in Rhode Island
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

• United States presidential elections in Rhode Island