1836 United States presidential election in Ohio

Main article: 1836 United States presidential election

1836 United States presidential election in Ohio

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Nominee William Henry Harrison Martin Van Buren Party Whig Democratic Home state Ohio New York Running mate Francis Granger Richard Johnson Electoral vote 21 0 Popular vote 104,958 96,238 Percentage 51.87% 47.56%

County Results Harrison   50–60%   60–70%   70–80% Van Buren   50–60%   60–70%   70–80% No Votes   No data

President before election Andrew Jackson Democratic Elected President Martin Van Buren Democratic

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 M. 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

• United States presidential elections in Ohio

References

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