United States presidential election in Nebraska, 1884

United States presidential election in Nebraska, 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 5 0
Popular vote 76,912 54,391
Percentage 57.31% 40.53%
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, Rhode Island would prove to be Blaine's fourth strongest victory in terms of percentage in the popular vote after Vermont, Minnesota, Kansas and Rhode Island[1].

Results

United States presidential election in Nebraska, 1884[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican James Gillespie Blaine of Maine John Alexander Logan of Illinois 76,912 57.31% 5 100.00%
Democratic Grover Cleveland of New York Thomas Andrews Hendricks of Indiana 54,391 40.53% 0 0.00%
Prohibition John Pierce St. John of Kansas William Daniel of Maryland 2,899 2.16% 0 0.00%
Total 134,202 100.00% 5 100.00%

References

  1. ↑ "1884 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
  2. ↑ "1884 Presidential General Election Results - Nebraska".
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