1896 United States presidential election in Nebraska
|Elections 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. Nebraska 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 William Jennings Bryan of Nebraska and his running mate Arthur Sewall of Maine. Four electors cast their Vice Presidential ballots for Thomas E. Watson.
|United States presidential election in Nebraska, 1896|
|Democratic||William Jennings Bryan||115,007||51.53%||8|
|National Democratic||John M. Palmer||2,885||1.29%||0|
|National Prohibition||Charles Bentley||797||0.36%||0|
|Socialist Labor||Charles Matchett||186||0.08%||0|
- Dave Leip’s U.S. Election Atlas; Presidential General Election Results – Nebraska