1856 United States presidential election in Texas explained

See main article: 1856 United States presidential election.

Election Name:1856 United States presidential election in Texas
Country:Texas
Election Date:November 4, 1856
Type:presidential
Ongoing:no
Previous Election:1852 United States presidential election in Texas
Previous Year:1852
Next Election:1860 United States presidential election in Texas
Next Year:1860
Image1:James Buchanan (cropped).jpg
Nominee1:James Buchanan
Party1:Democratic Party (United States)
Home State1:Pennsylvania
Electoral Vote1:4
Popular Vote1:31,169
Percentage1:66.59%
Nominee2:Millard Fillmore
Party2:Know Nothing
Home State2:New York
Running Mate2:Andrew J. Donelson
Electoral Vote2:0
Popular Vote2:15,639
Percentage2:33.41%
Map Size:350px
President
Before Election:Franklin Pierce
Before Party:Democratic Party (United States)
After Election:James Buchanan
After Party:Democratic Party (United States)

The 1856 United States presidential election in Texas was held on Tuesday November 4, as part of the 1856 United States presidential election. State voters chose four electors to represent the state in the Electoral College, which chose the president and vice president.

Texas voted for the Democratic nominee James Buchanan, who received 67% of the vote. Texas was Buchanan's second-strongest state.

Republican Party nominee John C. Frémont was not on the ballot. Texas would never be won by a Republican candidate until Herbert Hoover narrowly won the state in 1928.[1]

Results

1856 United States presidential election in Texas[2]
PartyCandidateVotesPercentageElectoral votes
DemocraticJames Buchanan31,169 66.589%4
Know-NothingMillard Fillmore15,63933.411%0
Total46,808100.0%4

See also

Notes and References

  1. Web site: Republican Party . 2023-12-22 . Britannica Kids . en-US.
  2. Web site: David Leip's Atlas of U.S. Presidential Elections.