1832 United States presidential election in Ohio

The 1832 United States presidential election in Ohio took place between November 2 and December 5, 1832, as part of the 1832 United States presidential election. Voters chose 21 representatives, or electors to the Electoral College, who voted for President and Vice President.

1832 United States presidential election in Ohio

November 2 – December 5, 1832

Nominee Andrew Jackson Henry Clay Party Democratic National Republican Home state Tennessee Kentucky Running mate Martin Van Buren John Sergeant Electoral vote 21 0 Popular vote 81,246 76,539 Percentage 51.33% 48.35%

President before election Andrew Jackson Democratic Elected President Andrew Jackson Democratic

Ohio voted for the Democratic Party candidate, Andrew Jackson, over the National Republican candidate, Henry Clay, and the Anti-Masonic Party candidate, William Wirt. Jackson won Ohio by a narrow margin of 2.98%.

Results

1832 United States presidential election in Ohio[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic	Andrew Jackson	81,246	51.33%	21
	National Republican	Henry Clay	76,539	48.35%	0
	Anti-Masonic	William Wirt	509	0.32%	0
Totals			158,294	100.0%	21

See also

• United States presidential elections in Ohio

References

1. "1832 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved 12 April 2013.