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1896 United States presidential election in Nebraska information


1896 United States presidential election in Nebraska
1896 United States presidential election in Nebraska
← 1892 November 3, 1896 1900 →
 
Nominee William Jennings Bryan William McKinley
Party Democratic Republican
Alliance Populist
Home state Nebraska Ohio
Running mate Arthur Sewall
(Democratic)
Thomas E. Watson
(Populist)
Garret Hobart
Electoral vote 8 0
Popular vote 115,007 103,064
Percentage 51.53% 46.18%

County Results

President before election

Grover Cleveland
Democratic

Elected President

William McKinley
Republican

The 1896 United States presidential election in Nebraska took place on November 3, 1896. All contemporary 45 states were part of the 1896 United States presidential election. Voters chose eight electors to the Electoral College, which selected the president and vice president.

Nebraska was won by the Democratic nominees, former U.S. Representative and Nebraska native William Jennings Bryan and his Democratic running mate Arthur Sewall of Maine. Four electors cast their vice presidential ballots for Thomas E. Watson, who was Bryan's running mate on the Populist Party. They defeated the Republican nominees, former Governor of Ohio William McKinley and his running mate Garret Hobart of New Jersey. Bryan won his home state by a margin of 5.35%.

As a result of his win, Bryan became the first Democratic presidential candidate to win Nebraska. Bryan would later lose the state to McKinley in 1900 but would later win it again against William Howard Taft in 1908. This is just one of two occasions where the Republican candidate won without the state, the other being in 1908.

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