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1876 United States presidential election in Nebraska

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1876 United States presidential election in Nebraska

← 1872 November 7, 1876 1880 →
 
Nominee Rutherford B. Hayes Samuel J. Tilden
Party Republican Democratic
Home state Ohio New York
Running mate William A. Wheeler Thomas A. Hendricks
Electoral vote 3 0
Popular vote 31,915 17,413
Percentage 64.70% 35.30%

County Results

President before election

Ulysses S. Grant
Republican

Elected President

Rutherford B. Hayes
Republican

The 1876 United States presidential election in Nebraska took place on November 7, 1876, as part of the 1876 United States presidential election. Voters chose three representatives, or electors to the Electoral College, who voted for president and vice president.

Nebraska voted for the Republican nominee, Ohio Governor Rutherford B. Hayes, over the Democratic nominee, New York Governor Samuel J. Tilden by a margin of 29.4%.

With 64.70% of the popular vote, Nebraska would be Hayes' second strongest victory in terms of percentage in the popular vote after Vermont.[1]

Results

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1876 United States presidential election in Nebraska[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Rutherford B. Hayes of Ohio William A. Wheeler of New York 31,915 64.70% 3 100.00%
Democratic Samuel J. Tilden of New York Thomas A. Hendricks of Indiana 17,413 35.30% 0 0.00%
Total 49,328 100.00% 3 100.00%

See also

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References

[edit]
  1. ^ "1876 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  2. ^ "1876 Presidential General Election Results - Nebraska".