A presidential election was held in Rhode Island on November 2, 1840 as part of the 1840 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for President and Vice President.
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.
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Text from the Wikipedia article “1840 United States presidential election in Rhode Island”, available under CC BY-SA 4.0.