1840_United_States_presidential_election_in_Rhode_Island

1840 United States presidential election in Rhode Island

1840 United States presidential election in Rhode Island

Election in Rhode Island

The 1840 United States presidential election in Rhode Island took place between October 30 and December 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.

Main article: 1840 United States presidential election
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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.[1]

Results

More information Party, Candidate ...

See also

References

  1. [1]
    "1840 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved March 5, 2018.
  2. [2]
    "1840 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved December 23, 2013.


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