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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

Elections in Rhode Island
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Ballot measures 2020 Question 1
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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%.

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

• United States presidential elections in Rhode Island