United States presidential election in Rhode Island, 1884

Main article: United States presidential election, 1884

United States presidential election in Rhode Island, 1884

November 4, 1884

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%

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

The 1884 United States presidential election in Rhode Island took place on November 4, 1884, as part of the 1884 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 G. Blaine, over the Democratic nominee, Grover Cleveland. Blaine won the state by a margin of 20.26%.

Results

United States presidential election in Rhode Island, 1884
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%