1896 United States presidential election in Nebraska
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.
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| Elections in Nebraska |
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Nebraska was won by the Democratic nominees, former U.S. Representative and Nebraska native William Jennings Bryan and his running mate Arthur Sewall of Maine. Four electors cast their Vice Presidential ballots for Thomas E. Watson.
As a result of his win, Bryan became the first Democratic presidential candidate to win Nebraska. Bryan would later lose the state to William McKinley in 1900 but would later win it again against William Howard Taft in 1908.
Results
| 1896 United States presidential election in Nebraska[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic | William Jennings Bryan | 115,007 | 51.53% | 8 | |
| Republican | William McKinley | 103,064 | 46.18% | 0 | |
| National Democratic | John M. Palmer | 2,885 | 1.29% | 0 | |
| Prohibition | Joshua Levering | 1,243 | 0.56% | 0 | |
| National Prohibition | Charles Bentley | 797 | 0.36% | 0 | |
| Socialist Labor | Charles Matchett | 186 | 0.08% | 0 | |
| Totals | 223,182 | 100.00% | 8 | ||
| Voter turnout | — | ||||
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References
- Dave Leip's U.S. Election Atlas; Presidential General Election Results – Nebraska
Notes
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