1836 United States presidential election in Pennsylvania

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1836 United States presidential election in Pennsylvania
Flag of Pennsylvania.svg
  1832 November 3 – December 7, 1836 1840  
  Martin Van Buren circa 1837 crop.jpg William Henry Harrison by James Reid Lambdin, 1835 crop.jpg
Nominee Martin Van Buren William Henry Harrison
Party Democratic Whig
Home state New York Ohio
Running mate Richard Johnson Francis Granger
Electoral vote300
Popular vote91,45787,235
Percentage51.18%48.82%

Pennsylvania Presidential Election Results 1836.svg
County Results

President before election

Andrew Jackson
Democratic

Elected President

Martin Van Buren
Democratic

The 1836 United States presidential election in Pennsylvania took place between November 3 and December 7, 1836, as part of the 1836 United States presidential election. Voters chose 30 representatives, or electors to the Electoral College, who voted for President and Vice President.

Contents

Pennsylvania voted for the Democratic candidate, Martin Van Buren, over the Whig candidate, William Henry Harrison. Van Buren won Pennsylvania by a narrow margin of 2.36%. The result would ultimately prove decisive in Van Buren's victory; had Harrison won the state, then Van Buren would not have achieved a majority in the Electoral College, meaning that the election would have been decided in the House of Representatives.

Results

1836 United States presidential election in Pennsylvania [1]
PartyCandidateVotesPercentageElectoral votes
Democratic Martin Van Buren 91,45751.18%30
Whig William Henry Harrison 87,23548.82%0
Totals178,692100.0%30

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References

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