1884 United States presidential election in Rhode Island

Main article: 1884 United States presidential election

The 1884 United States presidential election in Rhode Island occurred on November 4, 1884. This election was part of the 1884 United States presidential election. Voters in Rhode Island selected four representatives, or electors, to the Electoral College, which elects the president and vice president.

1884 United States presidential election in Rhode Island

← 1880 November 4, 1884 1888 →

Nominee James G. Blaine Grover Cleveland Party Republican Democratic Home state Maine New York Running mate John A. Logan Thomas A. Hendricks Electoral vote 4 0 Popular vote 19,030 12,391 Percentage 58.07% 37.81%

County Results Blaine   50–60%   60–70%

President before election Chester A. Arthur Republican Elected President Grover Cleveland Democratic

Rhode Island voted for the Republican nominee, James G. Blaine, over the Democratic nominee, Grover Cleveland. Blaine won the state by a margin of 20.26%.

With 58.07% of the popular vote, Rhode Island became Blaine's fourth strongest victory in terms of percentage, following Vermont, Minnesota and Kansas.^[1]

Results

1884 United States presidential election in Rhode Island[2]
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Republican	James Gillespie Blaine of Maine	John Alexander Logan of Illinois	19,030	58.07%	4	100.00%
	Democratic	Grover Cleveland of New York	Thomas Andrews Hendricks of Indiana	12,391	37.81%	0	0.00%
	Prohibition	John Pierce St. John of Kansas	William Daniel of Maryland	928	2.83%	0	0.00%
	Greenback	Benjamin Franklin Butler of Massachusetts	Absolom Madden West of Mississippi	423	1.29%	0	0.00%
Total				32,771	100.00%	4	100.00%

See also

• United States presidential elections in Rhode Island