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1840 United States presidential election in Ohio
A presidential election was held in Ohio on October 30, 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.
A presidential election was held in Ohio on October 30, 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%
| 1840 United States presidential election in Ohio | |||||||
|---|---|---|---|---|---|---|---|
| Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 148,157 | 54.10% | 21 | 100.00% | |
| Democratic | Martin Van Buren of New York | Richard Mentor Johnson of Kentucky | 124,782 | 45.57% | 0 | 0.00% | |
| Liberty | James G. Birney of New York | Thomas Earle of Pennsylvania | 903 | 0.33% | 0 | 0.00% | |
| Total | 273,842 | 100.00% | 21 | 100.00% |
- United States presidential elections in Ohio
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