United States presidential election in Pennsylvania, 1840

October 30 – December 2, 1840

President before election

Martin Van Buren
Democratic

Elected President

William Henry Harrison
Whig

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The 1840 United States presidential election in Pennsylvania took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 30 representatives, or electors to the Electoral College, who voted for President and Vice President.

Pennsylvania voted for the Whig candidate, William Henry Harrison, over the Democratic candidate, Martin Van Buren. Harrison won Pennsylvania by a margin of 0.12%.

Results

United States presidential election in Pennsylvania, 1840^[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Whig	William Henry Harrison	144,010	50.00%	30
	Democratic	Martin Van Buren	143,676	49.88%	0
	Liberty	James G. Birney	340	0.12%	0
Totals			288,026	100.0%	30

State Results of the 1840 U.S. presidential election

Candidates	William Henry Harrison Martin Van Buren James G. Birney

General articles	Election timeline

Local results	Alabama Arkansas Connecticut Delaware Georgia Illinois Indiana Kentucky Louisiana Maine Maryland Massachusetts Michigan Mississippi Missouri New Hampshire New Jersey New York North Carolina Ohio Pennsylvania Rhode Island South Carolina Tennessee Vermont Virginia

Other 1840 elections	House Senate Gubernatorial

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