Election in Ohio
1840 United States presidential election in Ohio![](//upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Flag_of_Ohio.svg/50px-Flag_of_Ohio.svg.png)
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← 1836 | October 30 - December 2, 1840 | 1844 → |
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| | | Nominee | William Henry Harrison | Martin Van Buren | | Party | Whig | Democratic | Home state | Ohio | New York | Running mate | John Tyler | none | Electoral vote | 21 | 0 | Popular vote | 148,157 | 124,782 | Percentage | 54.10% | 45.57% | |
County Results Congressional District Results Results Harrison 40-50% 50-60% 60-70% 70–80% | Van Buren 50-60% 60-70% 70-80% | |
President before election Martin Van Buren Democratic | Elected President William Henry Harrison Whig | |
Elections in Ohio |
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![](//upload.wikimedia.org/wikipedia/commons/thumb/e/ea/Seal_of_Ohio_%28B%26W%29.svg/150px-Seal_of_Ohio_%28B%26W%29.svg.png) |
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The 1840 United States presidential election in Ohio took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 21 representatives, or electors to the Electoral College, who voted for President and Vice President.
Ohio voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Ohio by a margin of 8.53%. Ohio was the home state of William Henry Harrison, Harrison improved his margin of victory from the last election over Van Buren by +4.22%
Results
See also
References
- ^ "1840 Presidential General Election Results - Ohio". U.S. Election Atlas. Retrieved December 23, 2013.
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