A presidential election was held in Ohio on November 4, 1836 as part of the 1836 United States presidential election.[1] Voters chose twenty-one representatives, or electors to the Electoral College, who voted for President and Vice President.

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[2]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Whig William Henry Harrison of Ohio Francis Granger of New York 104,958 51.87% 21 100.00%
Democratic Martin Van Buren of New York Richard Mentor Johnson of Kentucky 96,238 47.56% 0 0.00%
N/A Others Others 1,137 0.56% 0 0.00%
Total 202,333 100.00% 21 100.00%

See also

References

  1. ↑ "Presidential Elections". Weekly Messenger. November 12, 1836.
  2. ↑ "1836 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved December 23, 2013.


source: https://en.wikipedia.org/wiki/1836_United_States_presidential_election_in_Ohio