1896 United States presidential election in Texas explained

See main article: 1896 United States presidential election.

Election Name:1896 United States presidential election in Texas
Country:Texas
Flag Year:1839
Type:presidential
Ongoing:no
Previous Election:1892 United States presidential election in Texas
Previous Year:1892
Next Election:1900 United States presidential election in Texas
Next Year:1900
Election Date:November 3, 1896
Image1:William Jennings Bryan 2 (cropped).jpg
Nominee1:William Jennings Bryan
Party1:Democratic Party (United States)
Alliance1:Populist
Home State1:Nebraska
Running Mate1:Arthur Sewall
Electoral Vote1:15
Popular Vote1:370,434
Percentage1:68.00%
Nominee2:William McKinley
Party2:Republican Party (United States)
Home State2:Ohio
Running Mate2:Garret Hobart
Electoral Vote2:0
Popular Vote2:167,520
Percentage2:30.75%
Map Size:350px
President
Before Election:Grover Cleveland
Before Party:Democratic Party (United States)
After Election:William McKinley
After Party:Republican Party (United States)

The 1896 United States presidential election in Texas took place on November 3, 1896. All contemporary 45 states were part of the 1896 United States presidential election. State voters chose 15 electors to the Electoral College, which selected the president and vice president.

Texas 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.

Bryan would later win Texas again in both 1900 and 1908.

Results

1896 United States presidential election in Texas[1]
PartyCandidateVotesPercentageElectoral votes
DemocraticWilliam Jennings Bryan370,43468.00%15
RepublicanWilliam McKinley167,52030.75%0
National DemocraticJohn M. Palmer5,0460.93%0
ProhibitionJoshua Levering1,7860.33%0
Totals544,786100.00%15
Voter turnout

See also

Notes and References

  1. Dave Leip's U.S. Election Atlas; Presidential General Election Results – Texas