1848 United States presidential election in Rhode Island explained

See main article: 1848 United States presidential election.

Election Name:1848 United States presidential election in Rhode Island
Country:Rhode Island
Type:presidential
Ongoing:no
Previous Election:1844 United States presidential election in Rhode Island
Previous Year:1844
Next Election:1852 United States presidential election in Rhode Island
Next Year:1852
Election Date:November 7, 1848
Image1:Zachary Taylor cropped.jpg
Nominee1:Zachary Taylor
Party1:Whig Party (United States)
Home State1:Louisiana
Running Mate1:Millard Fillmore
Electoral Vote1:4
Popular Vote1:6,779
Percentage1:60.77%
Nominee2:Lewis Cass
Party2:Democratic Party (United States)
Home State2:Michigan
Running Mate2:William O. Butler
Electoral Vote2:0
Popular Vote2:3,646
Percentage2:32.68%
Image3:Portrait of Martin Van Buren (cropped).jpg
Nominee3:Martin Van Buren
Party3:Free Soil Party
Home State3:New York
Running Mate3:Charles Francis Adams, Sr.
Electoral Vote3:0
Popular Vote3:730
Percentage3:6.54%
Map Size:250px
President
Before Election:James K. Polk
Before Party:Democratic Party (United States)
Before Color:FF3333
After Election:Zachary Taylor
After Party:Whig Party (United States)
After Color:FF3333

The 1848 United States presidential election in Rhode Island took place on November 7, 1844, as part of the 1848 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, Zachary Taylor, over Democratic candidate Lewis Cass and Free Soil candidate former president Martin Van Buren. Taylor won the state by a margin of 28.09%.

With 60.77% of the popular vote, Rhode Island would prove to be Taylor's strongest state in the election in terms of percentage in the popular vote.[1]

See also

Notes and References

  1. Web site: 1848 Presidential Election Statistics. Dave Leip’s Atlas of U.S. Presidential Elections. 2018-03-05.