1884 United States presidential election in Nebraska

1884 United States presidential election in Nebraska

← 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 5 0
Popular vote 76,912 54,391
Percentage 57.31% 40.53%

County Results

President before election

Chester A. Arthur
Republican

Elected President

Grover Cleveland
Democratic

The 1884 United States presidential election in Nebraska took place on November 4, 1884, as part of the 1884 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for president and vice president.

Nebraska voted for the Republican nominee, James G. Blaine, over the Democratic nominee, Grover Cleveland. Blaine won the state by a margin of 16.78%.

With 57.31% of the popular vote, Nebraska would prove to be Blaine's fifth strongest victory in terms of percentage in the popular vote after Vermont, Minnesota, Kansas and Rhode Island.[1]

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

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