1836 United States presidential election in Ohio

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1836 United States presidential election in Ohio
Flag of Ohio.svg
  1832 November 3 – December 7, 1836 1840  
  William Henry Harrison by James Reid Lambdin, 1835 crop.jpg Martin Van Buren circa 1837 crop.jpg
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate Francis Granger Richard Johnson
Electoral vote210
Popular vote104,95896,238
Percentage51.87%47.56%

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

Contents

Ohio voted for Whig candidate William Henry Harrison over Democratic candidate Martin Van Buren. Harrison won Ohio by a narrow margin of 4.31%. Ohio was the home state of William Henry Harrison.

Results

1836 United States presidential election in Ohio [1]
PartyCandidateRunning matePopular voteElectoral vote
Count%Count%
Whig William Henry Harrison of Ohio Francis Granger of New York 104,95851.87%21100.00%
Democratic Martin Van Buren of New York Richard M. Johnson of Kentucky 96,23847.56%00.00%
N/AOthersOthers1,1370.56%00.00%
Total202,333100.00%21100.00%

See also

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References

  1. "1836 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved December 23, 2013.