1840 United States presidential election in Rhode Island
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A presidential election was held in Rhode Island on November 2, 1840 as part of the 1840 United States presidential election.[1] Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.
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November 2, 1840
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 County Results
Harrison 50â60% 60â70% 70â80%
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Rhode Island voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Rhode Island by a margin of 22.93%.
With 61.22% of the popular vote, Rhode Island would be Harrison's third strongest state in the 1840 election after Kentucky and Vermont.[2]
Results
| 1840 United States presidential election in Rhode Island[3] | ||||||||
|---|---|---|---|---|---|---|---|---|
| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 5,278 | 61.22% | 4 | 100.00% | ||
| Democratic | Martin Van Buren of New York | Richard Mentor Johnson of Kentucky | 3,301 | 38.29% | 0 | 0.00% | ||
| Liberty | James G. Birney of New York | Thomas Earle of Pennsylvania | 42 | 0.49% | 0 | 0.00% | ||
| Total | 8,621 | 100.00% | 4 | 100.00% | ||||
