United States presidential election in Rhode Island, 1840

United States presidential election in Rhode Island, 1840

October 30 - December 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

President before election Martin Van Buren Democratic Elected President William Henry Harrison Whig

Elections in Rhode Island
Federal government Presidential Elections 1792 1796 1800 1804 1808 1812 1816 1820 1824 1828 1832 1836 1840 1844 1848 1852 1856 1860 1864 1868 1872 1876 1880 1884 1888 1892 1896 1900 1904 1908 1912 1916 1920 1924 1928 1932 1936 1940 1944 1948 1952 1956 1960 1964 1968 1972 1980 1984 1988 1992 1996 2000 2004 2008 2012 2016 Presidential Primaries Democratic 2008 Republican 2008 2012 U.S. Senate elections 1970 1972 1976 1978 1982 1984 1988 1990 1994 1996 2000 2002 2006 2008 2012 2014 U.S. House elections 2006 2008 2010 2012 2014 2016
State government Gubernatorial Elections 1994 1998 2002 2006 2010 2014

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.

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%.

Results

References

State Results of the 1840 U.S. presidential election
Candidates	William Henry Harrison Martin Van Buren James G. Birney
General articles	Election timeline
Local results	Alabama Arkansas Connecticut Delaware Georgia Illinois Indiana Kentucky Louisiana Maine Maryland Massachusetts Michigan Mississippi Missouri New Hampshire New Jersey New York North Carolina Ohio Pennsylvania Rhode Island South Carolina Tennessee Vermont Virginia
Other 1840 elections	House Senate Gubernatorial