1884 United States presidential election in Rhode Island

1884 United States presidential election in Rhode Island

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%

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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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.

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]

• United States presidential elections in Rhode Island