1884 United States presidential election in Rhode Island

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

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]

  1. ^ "1884 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.

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