1836 United States presidential election in Arkansas

United States presidential election in Arkansas, 1836
	November 3 – December 7, 1836
	November 3 – December 7, 1836
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, 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

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References

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

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