United States presidential election in Arkansas, 1836

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

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.

Arkansas voted for the Democratic candidate, Martin Van Buren, over Whig candidate Hugh White. 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.