1836 United States presidential election in Georgia

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1836 United States presidential election in Georgia
Flag of the State of Georgia (non-official).svg
← 1832 November 3 – December 7, 1836 1840 →
  HLWhite.jpg Martin Van Buren circa 1837 crop.jpg
Nominee Hugh White Martin Van Buren
Party Whig Democratic
Home state Tennessee New York
Running mate John Tyler Richard Johnson
Electoral vote 11 0
Popular vote 24,481 22,778
Percentage 51.80% 48.20%

The 1836 United States presidential election in Georgia took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose 11 representatives, or electors to the Electoral College, who voted for president and vice president.

Georgia voted for Whig candidate Hugh White over the Democratic candidate, Martin Van Buren. White won Georgia by a margin of 3.6%.

This was the only election in which a Democrat won the presidency without carrying Georgia until 1964, nearly 130 years later. Alongside Lyndon Johnson in that year, only Bill Clinton in 1996 (though he carried the state four years prior) and Barack Obama in 2008 and 2012 have been elected without carrying it.[1]

Results[]

United States presidential election in Georgia, 1836[2]
Party Candidate Votes Percentage Electoral votes
Whig Hugh White 24,481 51.80% 11
Democratic Martin Van Buren 22,778 48.20% 0
Totals 47,259 100.0% 11

References[]

  1. ^ "Historical U.S. Presidential Elections 1789-2020". 270toWin.com. Retrieved 2021-12-25.
  2. ^ "1836 Presidential General Election Results - Georgia". U.S. Election Atlas. Retrieved 4 August 2012.


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