1836 United States presidential election in Arkansas

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

United States presidential election in Arkansas, 1836

November 3 – December 7, 1836
 
Nominee Martin Van Buren Hugh White
Party Democratic Whig
Home state New York Tennessee
Running mate Richard Johnson John Tyler
Electoral vote 3 0
Popular vote 2,380 1,334
Percentage 64.08% 35.92%

President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic

Arkansas, having been admitted to the Union as the 25th state on June 15, 1836, voted for the Democratic candidate, Martin Van Buren, over Whig candidate Hugh White during its first presidential election. Van Buren won Arkansas by a margin of 28.16%.

Results

United States presidential election in Arkansas, 1836[1]
Party Candidate Votes Percentage Electoral votes
Democratic Martin Van Buren 2,380 64.08% 3
Whig Hugh White 1,334 35.92% 0
Totals 3,714 100.0% 3

References

  1. "1836 Presidential General Election Results - Arkansas". U.S. Election Atlas. Retrieved 4 August 2012.


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