1840 United States presidential election in Rhode Island

1840 United States presidential election in Rhode Island
November 2, 1840
 
Nominee William Henry Harrison Martin Van Buren
Party Whig Democratic
Home state Ohio New York
Running mate John Tyler none
Electoral vote 4 0
Popular vote 5,278 3,301
Percentage 61.22% 38.29%
County Results
Harrison
  50–60%
  60–70%
  70–80%

President before election

Martin Van Buren
Democratic

 Elected President

William Henry Harrison
Whig

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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